\(\int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx\) [34]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\frac {6 a e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e} \]

[Out]

-2/5*a*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/d+2/7*b*(e*sin(d*x+c))^(7/2)/d/e-6/5*a*e^2*(sin(1/2*c+1/4*Pi+1/2*d*x)
^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/2)/d/sin(d*
x+c)^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2748, 2715, 2721, 2719} \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\frac {6 a e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e} \]

[In]

Int[(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2),x]

[Out]

(6*a*e^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) - (2*a*e*Cos[c + d*x]
*(e*Sin[c + d*x])^(3/2))/(5*d) + (2*b*(e*Sin[c + d*x])^(7/2))/(7*d*e)

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[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])

Rubi steps \begin{align*} \text {integral}& = \frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}+a \int (e \sin (c+d x))^{5/2} \, dx \\ & = -\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}+\frac {1}{5} \left (3 a e^2\right ) \int \sqrt {e \sin (c+d x)} \, dx \\ & = -\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}+\frac {\left (3 a e^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{5 \sqrt {\sin (c+d x)}} \\ & = \frac {6 a e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\frac {2 (e \sin (c+d x))^{5/2} \left (-21 a E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+\sin ^{\frac {3}{2}}(c+d x) \left (-7 a \cos (c+d x)+5 b \sin ^2(c+d x)\right )\right )}{35 d \sin ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[(a + b*Cos[c + d*x])*(e*Sin[c + d*x])^(5/2),x]

[Out]

(2*(e*Sin[c + d*x])^(5/2)*(-21*a*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + Sin[c + d*x]^(3/2)*(-7*a*Cos[c + d*x] +
5*b*Sin[c + d*x]^2)))/(35*d*Sin[c + d*x]^(5/2))

Maple [A] (verified)

Time = 3.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.71

method	result	size
default	\(\frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 e}-\frac {e^{3} a \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (d x +c \right )\right )+2 \left (\sin ^{2}\left (d x +c \right )\right )\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(171\)
parts	\(-\frac {a \,e^{3} \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) E\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) F\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (d x +c \right )\right )+2 \left (\sin ^{2}\left (d x +c \right )\right )\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d e}\)	\(173\)

[In]

int((a+cos(d*x+c)*b)*(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(2/7/e*b*(e*sin(d*x+c))^(7/2)-1/5*e^3*a*(6*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipt
icE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-3*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF
((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^4+2*sin(d*x+c)^2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.27 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\frac {21 i \, \sqrt {2} a \sqrt {-i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} a \sqrt {i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (5 \, b e^{2} \cos \left (d x + c\right )^{2} + 7 \, a e^{2} \cos \left (d x + c\right ) - 5 \, b e^{2}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )}{35 \, d} \]

[In]

integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/35*(21*I*sqrt(2)*a*sqrt(-I*e)*e^2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x +
c))) - 21*I*sqrt(2)*a*sqrt(I*e)*e^2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x
+ c))) - 2*(5*b*e^2*cos(d*x + c)^2 + 7*a*e^2*cos(d*x + c) - 5*b*e^2)*sqrt(e*sin(d*x + c))*sin(d*x + c))/d

Sympy [F]

\[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\int \left (e \sin {\left (c + d x \right )}\right )^{\frac {5}{2}} \left (a + b \cos {\left (c + d x \right )}\right )\, dx \]

[In]

integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))**(5/2),x)

[Out]

Integral((e*sin(c + d*x))**(5/2)*(a + b*cos(c + d*x)), x)

Maxima [F]

\[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)*(e*sin(d*x + c))^(5/2), x)

Giac [F]

\[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))*(e*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)*(e*sin(d*x + c))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \]

[In]

int((e*sin(c + d*x))^(5/2)*(a + b*cos(c + d*x)),x)

[Out]

int((e*sin(c + d*x))^(5/2)*(a + b*cos(c + d*x)), x)