Integrand size = 45, antiderivative size = 490 \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (3 A b^2+20 a b B+3 a^2 (3 A+5 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^2 d}-\frac {2 \sqrt {a+b} \left (3 b^2 (A-5 B)-2 a b (6 A-10 B+15 C)+a^2 (9 A-5 B+15 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a d}-\frac {2 b \sqrt {a+b} C \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 (3 A b+5 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)} \] Output:
2/15*(a-b)*(a+b)^(1/2)*(3*A*b^2+20*B*a*b+3*a^2*(3*A+5*C))*cot(d*x+c)*Ellip ticE((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1 /2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d-2 /15*(a+b)^(1/2)*(3*b^2*(A-5*B)-2*a*b*(6*A-10*B+15*C)+a^2*(9*A-5*B+15*C))*c ot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(- (a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b) )^(1/2)/a/d-2*b*(a+b)^(1/2)*C*cot(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2) /(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+ c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/d+2/15*(3*A*b+5*B*a)*(a+b* cos(d*x+c))^(1/2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/5*A*(a+b*cos(d*x+c))^(3/ 2)*sin(d*x+c)/d/cos(d*x+c)^(5/2)
Result contains complex when optimal does not.
Time = 7.33 (sec) , antiderivative size = 1353, normalized size of antiderivative = 2.76 \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
Integrate[((a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x] ^2))/Cos[c + d*x]^(7/2),x]
Output:
-1/15*((-4*a*(-3*a^2*A*b + 3*A*b^3 - 5*a^3*B + 5*a*b^2*B - 15*a^2*b*C)*Sqr t[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[ (c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[ c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a ]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + d*x ]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(9*a^3*A + 3*a*A*b^2 + 20*a^2*b*B + 15* a^3*C - 15*a*b^2*C)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(( (a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Cs c[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d* x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/ ((a + b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot [(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2 )/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*Ellip ticPi[-(a/b), ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqr t[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(9*a^2*A*b + 3*A*b^3 + 20*a*b^2*B + 15*a^2*b*C)*((I *Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d* x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x])*Sec[c + d*x])/(a + b)] ) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b...
Time = 2.40 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.311, Rules used = {3042, 3526, 27, 3042, 3526, 27, 3042, 3532, 3042, 3288, 3477, 3042, 3295, 3473}
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.
| \(\displaystyle \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {2}{5} \int \frac {\sqrt {a+b \cos (c+d x)} \left (5 b C \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+3 A b+5 a B\right )}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {\sqrt {a+b \cos (c+d x)} \left (5 b C \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+3 A b+5 a B\right )}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (5 b C \sin \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+5 b B+5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b+5 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {3 (3 A+5 C) a^2+20 b B a+3 A b^2+15 b^2 C \cos ^2(c+d x)+\left (5 B a^2+12 A b a+30 b C a+15 b^2 B\right ) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 (3 A+5 C) a^2+20 b B a+3 A b^2+15 b^2 C \cos ^2(c+d x)+\left (5 B a^2+12 A b a+30 b C a+15 b^2 B\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {3 (3 A+5 C) a^2+20 b B a+3 A b^2+15 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (5 B a^2+12 A b a+30 b C a+15 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3532 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (3 A+5 C) a^2+20 b B a+3 A b^2+\left (5 B a^2+12 A b a+30 b C a+15 b^2 B\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+15 b^2 C \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx\right )+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (3 A+5 C) a^2+20 b B a+3 A b^2+\left (5 B a^2+12 A b a+30 b C a+15 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+15 b^2 C \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3288 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\int \frac {3 (3 A+5 C) a^2+20 b B a+3 A b^2+\left (5 B a^2+12 A b a+30 b C a+15 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 b C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3477 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 (3 A+5 C)+20 a b B+3 A b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\left (a^2 (9 A-5 (B-3 C))-2 a b (6 A-10 B+15 C)+3 b^2 (A-5 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {30 b C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-\left (a^2 (9 A-5 (B-3 C))-2 a b (6 A-10 B+15 C)+3 b^2 (A-5 B)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (3 a^2 (3 A+5 C)+20 a b B+3 A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 b C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3295 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 (3 A+5 C)+20 a b B+3 A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \cot (c+d x) \left (a^2 (9 A-5 (B-3 C))-2 a b (6 A-10 B+15 C)+3 b^2 (A-5 B)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}-\frac {30 b C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3473 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (-\frac {2 \sqrt {a+b} \cot (c+d x) \left (a^2 (9 A-5 (B-3 C))-2 a b (6 A-10 B+15 C)+3 b^2 (A-5 B)\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (3 a^2 (3 A+5 C)+20 a b B+3 A b^2\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {30 b C \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (5 a B+3 A b) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 d \cos ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[((a + b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/C os[c + d*x]^(7/2),x]
Output:
(2*A*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + ( ((2*(a - b)*Sqrt[a + b]*(3*A*b^2 + 20*a*b*B + 3*a^2*(3*A + 5*C))*Cot[c + d *x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d* x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) - (2*Sqrt[a + b]*(3*b^2*(A - 5*B) + a^ 2*(9*A - 5*(B - 3*C)) - 2*a*b*(6*A - 10*B + 15*C))*Cot[c + d*x]*EllipticF[ ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x ]))/(a - b)])/(a*d) - (30*b*Sqrt[a + b]*C*Cot[c + d*x]*EllipticPi[(a + b)/ b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d)/3 + (2*(3*A*b + 5*a*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c *((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] && PosQ[(c + d)/b]
Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f _.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] ], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]
Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) ^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* (c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] )/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && PosQ[(c + d)/b]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S imp[(A - B)/(a - b) Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* x]]), x], x] - Simp[(A*b - a*B)/(a - b) Int[(1 + Sin[e + f*x])/((a + b*Si n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*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] && NeQ[A, B]
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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((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 - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{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] && GtQ[m, 0] && LtQ[n, -1]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2 Int[Sqrt[a + b*Sin[e + f*x]] /Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2 Int[(A*b^2 - a^2*C + b*(b* B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x ]]), x], x] /; FreeQ[{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]
Leaf count of result is larger than twice the leaf count of optimal. \(2321\) vs. \(2(441)=882\).
Time = 75.20 (sec) , antiderivative size = 2322, normalized size of antiderivative = 4.74
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(2322\) |
| default | \(\text {Expression too large to display}\) | \(2345\) |
Input:
int((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2 ),x,method=_RETURNVERBOSE)
Output:
-2/5*A/d/a*((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1 +cos(d*x+c)))^(1/2)*a^3*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/ 2))*(-3*cos(d*x+c)^4-6*cos(d*x+c)^3-3*cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x +c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b*Elliptic E(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-3*cos(d*x+c)^4-6*cos(d*x+c )^3-3*cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d* x+c))/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/ (a+b))^(1/2))*(-cos(d*x+c)^4-2*cos(d*x+c)^3-cos(d*x+c)^2)+(cos(d*x+c)/(1+c os(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^3*Elli pticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-cos(d*x+c)^4-2*cos(d*x +c)^3-cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d* x+c))/(1+cos(d*x+c)))^(1/2)*a^3*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a +b))^(1/2))*(3*cos(d*x+c)^4+6*cos(d*x+c)^3+3*cos(d*x+c)^2)+(cos(d*x+c)/(1+ cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b*E llipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(4*cos(d*x+c)^4+8*cos (d*x+c)^3+4*cos(d*x+c)^2)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b* cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticF(cot(d*x+c)-csc(d*x+c),(- (a-b)/(a+b))^(1/2))*(cos(d*x+c)^4+2*cos(d*x+c)^3+cos(d*x+c)^2)+(-3*cos(d*x +c)^2-cos(d*x+c)-1)*sin(d*x+c)*a^3+sin(d*x+c)*cos(d*x+c)*(-3*cos(d*x+c)^2- 3*cos(d*x+c)-3)*b*a^2+sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)-3)*a*b^2-b...
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c )^(7/2),x, algorithm="fricas")
Output:
integral((C*b*cos(d*x + c)^3 + (C*a + B*b)*cos(d*x + c)^2 + A*a + (B*a + A *b)*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)/cos(d*x + c)^(7/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x +c)**(7/2),x)
Output:
Timed out
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c )^(7/2),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/ 2)/cos(d*x + c)^(7/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c )^(7/2),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(3/ 2)/cos(d*x + c)^(7/2), x)
Timed out. \[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{7/2}} \,d x \] Input:
int(((a + b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^(7/2),x)
Output:
int(((a + b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^(7/2), x)
\[ \int \frac {(a+b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )}d x \right ) b c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \right ) a^{2}+2 \left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{3}}d x \right ) a b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) b^{2} \] Input:
int((a+b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2 ),x)
Output:
int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x),x)*b*c + in t((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**4,x)*a**2 + 2*int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**3,x)*a*b + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**2,x)*a* c + int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/cos(c + d*x)**2,x)*b **2