Integrand size = 23, antiderivative size = 129 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\frac {10 a e^4 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{21 d \sqrt {e \sin (c+d x)}}-\frac {10 a e^3 \cos (c+d x) \sqrt {e \sin (c+d x)}}{21 d}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e} \] Output:
10/21*a*e^4*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^(1/2))*sin(d*x+c)^(1/2) /d/(e*sin(d*x+c))^(1/2)-10/21*a*e^3*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/d-2/7* a*e*cos(d*x+c)*(e*sin(d*x+c))^(5/2)/d+2/9*b*(e*sin(d*x+c))^(9/2)/d/e
Time = 1.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.84 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\frac {e^3 \left (-120 a \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+(21 b-138 a \cos (c+d x)-28 b \cos (2 (c+d x))+18 a \cos (3 (c+d x))+7 b \cos (4 (c+d x))) \sqrt {\sin (c+d x)}\right ) \sqrt {e \sin (c+d x)}}{252 d \sqrt {\sin (c+d x)}} \] Input:
Integrate[(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(7/2),x]
Output:
(e^3*(-120*a*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + (21*b - 138*a*Cos[c + d *x] - 28*b*Cos[2*(c + d*x)] + 18*a*Cos[3*(c + d*x)] + 7*b*Cos[4*(c + d*x)] )*Sqrt[Sin[c + d*x]])*Sqrt[e*Sin[c + d*x]])/(252*d*Sqrt[Sin[c + d*x]])
Time = 0.56 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3148, 3042, 3115, 3042, 3115, 3042, 3121, 3042, 3120}
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.
| \(\displaystyle \int (e \sin (c+d x))^{7/2} (a+b \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{7/2} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle a \int (e \sin (c+d x))^{7/2}dx+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int (e \sin (c+d x))^{7/2}dx+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{7} e^2 \int (e \sin (c+d x))^{3/2}dx-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}\right )+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {5}{7} e^2 \int (e \sin (c+d x))^{3/2}dx-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}\right )+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}\right )+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {5}{7} e^2 \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}\right )+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle a \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}\right )+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {5}{7} e^2 \left (\frac {e^2 \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}\right )+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle a \left (\frac {5}{7} e^2 \left (\frac {2 e^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{5/2}}{7 d}\right )+\frac {2 b (e \sin (c+d x))^{9/2}}{9 d e}\) |
Input:
Int[(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(7/2),x]
Output:
(2*b*(e*Sin[c + d*x])^(9/2))/(9*d*e) + a*((-2*e*Cos[c + d*x]*(e*Sin[c + d* x])^(5/2))/(7*d) + (5*e^2*((2*e^2*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Si n[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) - (2*e*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*d)))/7)
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
Time = 3.64 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.98
| method | result | size |
| default | \(\frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9 e}-\frac {e^{4} a \left (-6 \sin \left (d x +c \right )^{5}+5 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-4 \sin \left (d x +c \right )^{3}+10 \sin \left (d x +c \right )\right )}{21 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(127\) |
| parts | \(-\frac {a \,e^{4} \left (-6 \sin \left (d x +c \right )^{5}+5 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-4 \sin \left (d x +c \right )^{3}+10 \sin \left (d x +c \right )\right )}{21 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {9}{2}}}{9 d e}\) | \(129\) |
Input:
int((a+cos(d*x+c)*b)*(e*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
(2/9/e*b*(e*sin(d*x+c))^(9/2)-1/21*e^4*a*(-6*sin(d*x+c)^5+5*(1-sin(d*x+c)) ^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1 /2),1/2*2^(1/2))-4*sin(d*x+c)^3+10*sin(d*x+c))/cos(d*x+c)/(e*sin(d*x+c))^( 1/2))/d
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (15 \, a \sqrt {-\frac {1}{2} i \, e} e^{3} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 15 \, a \sqrt {\frac {1}{2} i \, e} e^{3} {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (7 \, b e^{3} \cos \left (d x + c\right )^{4} + 9 \, a e^{3} \cos \left (d x + c\right )^{3} - 14 \, b e^{3} \cos \left (d x + c\right )^{2} - 24 \, a e^{3} \cos \left (d x + c\right ) + 7 \, b e^{3}\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{63 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))^(7/2),x, algorithm="fricas")
Output:
2/63*(15*a*sqrt(-1/2*I*e)*e^3*weierstrassPInverse(4, 0, cos(d*x + c) + I*s in(d*x + c)) + 15*a*sqrt(1/2*I*e)*e^3*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + (7*b*e^3*cos(d*x + c)^4 + 9*a*e^3*cos(d*x + c)^3 - 14*b*e^3*cos(d*x + c)^2 - 24*a*e^3*cos(d*x + c) + 7*b*e^3)*sqrt(e*sin(d*x + c)))/d
Timed out. \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))**(7/2),x)
Output:
Timed out
\[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))^(7/2),x, algorithm="maxima")
Output:
integrate((b*cos(d*x + c) + a)*(e*sin(d*x + c))^(7/2), x)
\[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))^(7/2),x, algorithm="giac")
Output:
integrate((b*cos(d*x + c) + a)*(e*sin(d*x + c))^(7/2), x)
Timed out. \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^{7/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \] Input:
int((e*sin(c + d*x))^(7/2)*(a + b*cos(c + d*x)),x)
Output:
int((e*sin(c + d*x))^(7/2)*(a + b*cos(c + d*x)), x)
\[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{7/2} \, dx=\frac {\sqrt {e}\, e^{3} \left (2 \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{4} b +9 \left (\int \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{3}d x \right ) a d \right )}{9 d} \] Input:
int((a+b*cos(d*x+c))*(e*sin(d*x+c))^(7/2),x)
Output:
(sqrt(e)*e**3*(2*sqrt(sin(c + d*x))*sin(c + d*x)**4*b + 9*int(sqrt(sin(c + d*x))*sin(c + d*x)**3,x)*a*d))/(9*d)