\(\int (a+b \cos (c+d x))^{3/2} (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [825]

Optimal result

Integrand size = 42, antiderivative size = 295 \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {(5 b B+4 a C) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{4 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (7 a b B+4 a^2 C+8 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {\left (4 a^2 B+3 b^2 B+12 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{4 d \sqrt {a+b \cos (c+d x)}}+\frac {(5 b B+4 a C) \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{4 d}+\frac {a B \sqrt {a+b \cos (c+d x)} \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

-1/4*(5*B*b+4*C*a)*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^( 
1/2)*(b/(a+b))^(1/2))/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+1/4*(7*B*a*b+4*C*a^ 
2+8*C*b^2)*((a+b*cos(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^ 
(1/2)*(b/(a+b))^(1/2))/d/(a+b*cos(d*x+c))^(1/2)+1/4*(4*B*a^2+3*B*b^2+12*C* 
a*b)*((a+b*cos(d*x+c))/(a+b))^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2 
)*(b/(a+b))^(1/2))/d/(a+b*cos(d*x+c))^(1/2)+1/4*(5*B*b+4*C*a)*(a+b*cos(d*x 
+c))^(1/2)*tan(d*x+c)/d+1/2*a*B*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)*tan(d*x+ 
c)/d
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.62 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.43 \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {8 b (a B+4 b C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2 B+b^2 B+20 a b C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {2 i (5 b B+4 a C) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} \csc (c+d x) \left (-2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (-2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )+b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right )}{a b \sqrt {-\frac {1}{a+b}}}+4 \sqrt {a+b \cos (c+d x)} (2 a B+(5 b B+4 a C) \cos (c+d x)) \sec (c+d x) \tan (c+d x)}{16 d} \] Input:

Integrate[(a + b*Cos[c + d*x])^(3/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S 
ec[c + d*x]^4,x]
 

Output:

((8*b*(a*B + 4*b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x) 
/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] + (2*(8*a^2*B + b^2*B + 20*a* 
b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/( 
a + b)])/Sqrt[a + b*Cos[c + d*x]] - ((2*I)*(5*b*B + 4*a*C)*Sqrt[-((b*(-1 + 
 Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*Csc[c + d* 
x]*(-2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c 
+ d*x]]], (a + b)/(a - b)] + b*(-2*a*EllipticF[I*ArcSinh[Sqrt[-(a + b)^(-1 
)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*EllipticPi[(a + b)/a, I 
*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)])) 
)/(a*b*Sqrt[-(a + b)^(-1)]) + 4*Sqrt[a + b*Cos[c + d*x]]*(2*a*B + (5*b*B + 
 4*a*C)*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(16*d)
 

Rubi [A] (verified)

Time = 2.80 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.05, number of steps used = 23, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 3508, 3042, 3468, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 3284}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 3508

\(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^{3/2} (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\)

\(\Big \downarrow \) 3468

\(\displaystyle \frac {1}{2} \int \frac {\left (b (a B+4 b C) \cos ^2(c+d x)+2 \left (B a^2+4 b C a+2 b^2 B\right ) \cos (c+d x)+a (5 b B+4 a C)\right ) \sec ^2(c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int \frac {\left (b (a B+4 b C) \cos ^2(c+d x)+2 \left (B a^2+4 b C a+2 b^2 B\right ) \cos (c+d x)+a (5 b B+4 a C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {b (a B+4 b C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 \left (B a^2+4 b C a+2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (5 b B+4 a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\left (-a b (5 b B+4 a C) \cos ^2(c+d x)+2 a b (a B+4 b C) \cos (c+d x)+a \left (4 B a^2+12 b C a+3 b^2 B\right )\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\left (-a b (5 b B+4 a C) \cos ^2(c+d x)+2 a b (a B+4 b C) \cos (c+d x)+a \left (4 B a^2+12 b C a+3 b^2 B\right )\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {-a b (5 b B+4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 a b (a B+4 b C) \sin \left (c+d x+\frac {\pi }{2}\right )+a \left (4 B a^2+12 b C a+3 b^2 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {1}{4} \left (\frac {-\frac {\int -\frac {\left (a b \left (4 B a^2+12 b C a+3 b^2 B\right )+a b \left (4 C a^2+7 b B a+8 b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-a (4 a C+5 b B) \int \sqrt {a+b \cos (c+d x)}dx}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {\left (a b \left (4 B a^2+12 b C a+3 b^2 B\right )+a b \left (4 C a^2+7 b B a+8 b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-a (4 a C+5 b B) \int \sqrt {a+b \cos (c+d x)}dx}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {a b \left (4 B a^2+12 b C a+3 b^2 B\right )+a b \left (4 C a^2+7 b B a+8 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-a (4 a C+5 b B) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {a b \left (4 B a^2+12 b C a+3 b^2 B\right )+a b \left (4 C a^2+7 b B a+8 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {a b \left (4 B a^2+12 b C a+3 b^2 B\right )+a b \left (4 C a^2+7 b B a+8 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\int \frac {a b \left (4 B a^2+12 b C a+3 b^2 B\right )+a b \left (4 C a^2+7 b B a+8 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx+a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {1}{4} \left (\frac {\frac {a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\frac {a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\frac {a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}+\frac {2 a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\frac {2 a b \left (4 a^2 C+7 a b B+8 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}+\frac {2 a b \left (4 a^2 B+12 a b C+3 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}-\frac {2 a (4 a C+5 b B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}+\frac {(4 a C+5 b B) \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) \sqrt {a+b \cos (c+d x)}}{2 d}\)

Input:

Int[(a + b*Cos[c + d*x])^(3/2)*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + 
 d*x]^4,x]
 

Output:

(a*B*Sqrt[a + b*Cos[c + d*x]]*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (((-2*a*( 
5*b*B + 4*a*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + 
b)])/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((2*a*b*(7*a*b*B + 4*a^2*C + 
 8*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/ 
(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]) + (2*a*b*(4*a^2*B + 3*b^2*B + 12*a* 
b*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/( 
a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/(2*a) + ((5*b*B + 4*a*C)*Sqrt[a 
+ b*Cos[c + d*x]]*Tan[c + d*x])/d)/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3286

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]
 

rule 3468

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c 
 + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 
1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n 
+ 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a 
*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) 
 - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - 
a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] 
/; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 
, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3508

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 0]
 

rule 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1402\) vs. \(2(283)=566\).

Time = 4.29 (sec) , antiderivative size = 1403, normalized size of antiderivative = 4.76

method	result	size
default	\(\text {Expression too large to display}\)	\(1403\)
parts	\(\text {Expression too large to display}\)	\(1722\)

Input:

int((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,me 
thod=_RETURNVERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*b^2*C*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(- 
2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1 
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+2*B*a^2*(-1/2*cos(1/2*d*x+1/2*c)/a*(-2*b 
*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+ 
1/2*c)^2)^2+3/4*b/a^2*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)* 
sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1+2*cos(1/2*d*x+1/2*c)^2)-1/8*b/a*(sin(1/2*d 
*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin( 
1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1 
/2*c),(-2*b/(a-b))^(1/2))+3/8/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d 
*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d 
*x+1/2*c)^2)^(1/2)*b*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-3/8* 
b^2/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b) 
)^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip 
ticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4 
+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/( 
a-b))^(1/2))-3/8/a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2 
*b+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2 
)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2,(-2*b/(a-b))^(1/2))*b^2)+2*a*(2...
 

Fricas [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 
4,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:

integrate((a+b*cos(d*x+c))**(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c 
)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 
4,x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(3/2)*s 
ec(d*x + c)^4, x)
 

Giac [F]

\[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 
4,x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c) + a)^(3/2)*s 
ec(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/cos(c 
 + d*x)^4,x)
 

Output:

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(3/2))/cos(c 
 + d*x)^4, x)
 

Reduce [F]

\[ \int (a+b \cos (c+d x))^{3/2} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) a b +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{4}d x \right ) b c +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) a c +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{4}d x \right ) b^{2} \] Input:

int((a+b*cos(d*x+c))^(3/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

Output:

int(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)*sec(c + d*x)**4,x)*a*b + int(sqr 
t(cos(c + d*x)*b + a)*cos(c + d*x)**3*sec(c + d*x)**4,x)*b*c + int(sqrt(co 
s(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**4,x)*a*c + int(sqrt(cos(c 
+ d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**4,x)*b**2