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

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

[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)

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Rubi [A]  time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2669, 2635, 2640, 2639} \[ \frac {6 a e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e} \]

Antiderivative was successfully verified.

[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 2635

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] && Integer
Q[2*n]

Rule 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2669

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

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Mathematica [A]  time = 0.52, size = 80, normalized size = 0.80 \[ \frac {2 (e \sin (c+d x))^{5/2} \left (\sin ^{\frac {3}{2}}(c+d x) \left (5 b \sin ^2(c+d x)-7 a \cos (c+d x)\right )-21 a E\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )\right )}{35 d \sin ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[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))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b e^{2} \cos \left (d x + c\right )^{3} + a e^{2} \cos \left (d x + c\right )^{2} - b e^{2} \cos \left (d x + c\right ) - a e^{2}\right )} \sqrt {e \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*e^2*cos(d*x + c)^3 + a*e^2*cos(d*x + c)^2 - b*e^2*cos(d*x + c) - a*e^2)*sqrt(e*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.24, size = 171, normalized size = 1.71 \[ \frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 e}-\frac {e^{3} a \left (6 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (d x +c \right )\right )+2 \left (\sin ^{2}\left (d x +c \right )\right )\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2/7/e*b*(e*sin(d*x+c))^(7/2)-1/5*e^3*a*(6*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellip
ticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellipt
icF((-sin(d*x+c)+1)^(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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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