\(\int (a+b \sec (c+d x))^{3/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [824]

Optimal result

Integrand size = 34, antiderivative size = 312 \[ \int (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (20 a b B+3 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^2 d}+\frac {2 (a-b) \sqrt {a+b} (15 a B-5 b B-3 a C+9 b C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

-2/15*(a-b)*(a+b)^(1/2)*(20*B*a*b+3*C*a^2+9*C*b^2)*cot(d*x+c)*EllipticE((a 
+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a 
+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/d+2/15*(a-b)*(a+b)^(1/2)*(1 
5*B*a-5*B*b-3*C*a+9*C*b)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b) 
^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+ 
c))/(a-b))^(1/2)/b/d+2/15*(5*B*b+3*C*a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/ 
d+2/5*C*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 14.70 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.46 \[ \int (a+b \sec (c+d x))^{3/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (2 (a+b) \left (20 a b B+3 a^2 C+9 b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) (3 a (5 B+C)+b (5 B+9 C)) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (20 a b B+3 a^2 C+9 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 b d (b+a \cos (c+d x))^2 \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x)}+\frac {\cos (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac {2 \left (20 a b B+3 a^2 C+9 b^2 C\right ) \sin (c+d x)}{15 b}+\frac {2}{15} \sec (c+d x) (5 b B \sin (c+d x)+6 a C \sin (c+d x))+\frac {2}{5} b C \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))} \] Input:

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

Output:

(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(2*(a 
 + b)*(20*a*b*B + 3*a^2*C + 9*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] 
*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[ 
Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(a + b)*(3*a*(5*B + C) + b*(5*B 
+ 9*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/(( 
a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a 
 + b)] + (20*a*b*B + 3*a^2*C + 9*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])* 
Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(15*b*d*(b + a*Cos[c + d*x])^2*Sqrt[ 
Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(3/2)) + (Cos[c + d*x]*(a + b*Sec[c + d*x 
])^(3/2)*((2*(20*a*b*B + 3*a^2*C + 9*b^2*C)*Sin[c + d*x])/(15*b) + (2*Sec[ 
c + d*x]*(5*b*B*Sin[c + d*x] + 6*a*C*Sin[c + d*x]))/15 + (2*b*C*Sec[c + d* 
x]*Tan[c + d*x])/5))/(d*(b + a*Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 4544, 27, 3042, 4544, 27, 3042, 4546, 27, 3042, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {\left (3 C a^2+20 b B a+9 b^2 C\right ) \sec ^2(c+d x)+\left (15 B a^2+12 b C a+5 b^2 B\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\left (3 C a^2+20 b B a+9 b^2 C\right ) \sec ^2(c+d x)+\left (15 B a^2+12 b C a+5 b^2 B\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {\left (3 C a^2+20 b B a+9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (15 B a^2+12 b C a+5 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {\left (15 B a^2-3 C a^2-20 b B a+12 b C a+5 b^2 B-9 b^2 C\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\left (3 a^2 C+20 a b B+9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 C+20 a b B+9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+(a-b) (15 a B-3 a C-5 b B+9 b C) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 C+20 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) (15 a B-3 a C-5 b B+9 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 C+20 a b B+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} (15 a B-3 a C-5 b B+9 b C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} (15 a B-3 a C-5 b B+9 b C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a^2 C+20 a b B+9 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

Input:

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

Output:

(2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a 
 + b]*(20*a*b*B + 3*a^2*C + 9*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a 
+ b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x] 
))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)* 
Sqrt[a + b]*(15*a*B - 5*b*B - 3*a*C + 9*b*C)*Cot[c + d*x]*EllipticF[ArcSin 
[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[ 
c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d))/3 + (2 
*(5*b*B + 3*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5
 

Defintions of rubi rules used

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 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4544

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot 
[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[( 
a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m 
)*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
 

rule 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   Int[Csc[e + f*x]*(( 
1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A 
, B, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1447\) vs. \(2(282)=564\).

Time = 57.45 (sec) , antiderivative size = 1448, normalized size of antiderivative = 4.64

method	result	size
default	\(\text {Expression too large to display}\)	\(1448\)
parts	\(\text {Expression too large to display}\)	\(1456\)

Input:

int((a+b*sec(d*x+c))^(3/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNV 
ERBOSE)
 

Output:

-2/15/d/b*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x+c) 
+b)*(20*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d* 
x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticE(-csc(d*x+ 
c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+20*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/ 
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^( 
1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+3*(-cos(d 
*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a* 
cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a 
-b)/(a+b))^(1/2))+3*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b*Ellipti 
cE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+9*(-cos(d*x+c)^2-2*cos(d*x+ 
c)-1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d 
*x+c)+1))^(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2) 
)+9*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/ 
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^3*EllipticE(-csc(d*x+c)+cot 
(d*x+c),((a-b)/(a+b))^(1/2))+15*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(1/(a+b)*( 
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^ 
2*b*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+20*(cos(d*x+c)^2 
+2*cos(d*x+c)+1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b...
 

Fricas [F]

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

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

Output:

integral((C*b*sec(d*x + c)^3 + B*a*sec(d*x + c) + (C*a + B*b)*sec(d*x + c) 
^2)*sqrt(b*sec(d*x + c) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**3,x)*b*c + int(sqrt(sec(c + d*x 
)*b + a)*sec(c + d*x)**2,x)*a*c + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x 
)**2,x)*b**2 + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x),x)*a*b