3.216 \(\int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=172 \[ \frac{26 a^3 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}-\frac{26 a^3 (e \cos (c+d x))^{5/2}}{35 d e}+\frac{26 a^3 e \sin (c+d x) \sqrt{e \cos (c+d x)}}{21 d}-\frac{26 \left (a^3 \sin (c+d x)+a^3\right ) (e \cos (c+d x))^{5/2}}{63 d e}-\frac{2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e} \]

[Out]

(-26*a^3*(e*Cos[c + d*x])^(5/2))/(35*d*e) + (26*a^3*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sq
rt[e*Cos[c + d*x]]) + (26*a^3*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(21*d) - (2*a*(e*Cos[c + d*x])^(5/2)*(a + a
*Sin[c + d*x])^2)/(9*d*e) - (26*(e*Cos[c + d*x])^(5/2)*(a^3 + a^3*Sin[c + d*x]))/(63*d*e)

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Rubi [A]  time = 0.188479, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2678, 2669, 2635, 2642, 2641} \[ \frac{26 a^3 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}-\frac{26 a^3 (e \cos (c+d x))^{5/2}}{35 d e}+\frac{26 a^3 e \sin (c+d x) \sqrt{e \cos (c+d x)}}{21 d}-\frac{26 \left (a^3 \sin (c+d x)+a^3\right ) (e \cos (c+d x))^{5/2}}{63 d e}-\frac{2 a (a \sin (c+d x)+a)^2 (e \cos (c+d x))^{5/2}}{9 d e} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^3,x]

[Out]

(-26*a^3*(e*Cos[c + d*x])^(5/2))/(35*d*e) + (26*a^3*e^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*Sq
rt[e*Cos[c + d*x]]) + (26*a^3*e*Sqrt[e*Cos[c + d*x]]*Sin[c + d*x])/(21*d) - (2*a*(e*Cos[c + d*x])^(5/2)*(a + a
*Sin[c + d*x])^2)/(9*d*e) - (26*(e*Cos[c + d*x])^(5/2)*(a^3 + a^3*Sin[c + d*x]))/(63*d*e)

Rule 2678

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

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])

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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rubi steps

\begin{align*} \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^3 \, dx &=-\frac{2 a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}{9 d e}+\frac{1}{9} (13 a) \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{2 a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}{9 d e}-\frac{26 (e \cos (c+d x))^{5/2} \left (a^3+a^3 \sin (c+d x)\right )}{63 d e}+\frac{1}{7} \left (13 a^2\right ) \int (e \cos (c+d x))^{3/2} (a+a \sin (c+d x)) \, dx\\ &=-\frac{26 a^3 (e \cos (c+d x))^{5/2}}{35 d e}-\frac{2 a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}{9 d e}-\frac{26 (e \cos (c+d x))^{5/2} \left (a^3+a^3 \sin (c+d x)\right )}{63 d e}+\frac{1}{7} \left (13 a^3\right ) \int (e \cos (c+d x))^{3/2} \, dx\\ &=-\frac{26 a^3 (e \cos (c+d x))^{5/2}}{35 d e}+\frac{26 a^3 e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}{9 d e}-\frac{26 (e \cos (c+d x))^{5/2} \left (a^3+a^3 \sin (c+d x)\right )}{63 d e}+\frac{1}{21} \left (13 a^3 e^2\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{26 a^3 (e \cos (c+d x))^{5/2}}{35 d e}+\frac{26 a^3 e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}{9 d e}-\frac{26 (e \cos (c+d x))^{5/2} \left (a^3+a^3 \sin (c+d x)\right )}{63 d e}+\frac{\left (13 a^3 e^2 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 \sqrt{e \cos (c+d x)}}\\ &=-\frac{26 a^3 (e \cos (c+d x))^{5/2}}{35 d e}+\frac{26 a^3 e^2 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d \sqrt{e \cos (c+d x)}}+\frac{26 a^3 e \sqrt{e \cos (c+d x)} \sin (c+d x)}{21 d}-\frac{2 a (e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^2}{9 d e}-\frac{26 (e \cos (c+d x))^{5/2} \left (a^3+a^3 \sin (c+d x)\right )}{63 d e}\\ \end{align*}

Mathematica [C]  time = 0.0734213, size = 66, normalized size = 0.38 \[ -\frac{32 \sqrt [4]{2} a^3 (e \cos (c+d x))^{5/2} \, _2F_1\left (-\frac{13}{4},\frac{5}{4};\frac{9}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{5 d e (\sin (c+d x)+1)^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^3,x]

[Out]

(-32*2^(1/4)*a^3*(e*Cos[c + d*x])^(5/2)*Hypergeometric2F1[-13/4, 5/4, 9/4, (1 - Sin[c + d*x])/2])/(5*d*e*(1 +
Sin[c + d*x])^(5/4))

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Maple [A]  time = 0.449, size = 251, normalized size = 1.5 \begin{align*} -{\frac{2\,{a}^{3}{e}^{2}}{315\,d} \left ( 1120\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{11}-2160\,\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-2800\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{9}+3240\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +784\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-840\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +1624\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+195\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -120\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -1162\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+217\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x)

[Out]

-2/315/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^3*e^2*(1120*sin(1/2*d*x+1/2*c)^11-2160*cos(1/2
*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-2800*sin(1/2*d*x+1/2*c)^9+3240*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+784*si
n(1/2*d*x+1/2*c)^7-840*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+1624*sin(1/2*d*x+1/2*c)^5+195*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-120*sin(1/2*d*x+1/2*c)^2*co
s(1/2*d*x+1/2*c)-1162*sin(1/2*d*x+1/2*c)^3+217*sin(1/2*d*x+1/2*c))/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(3/2)*(a*sin(d*x + c) + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (3 \, a^{3} e \cos \left (d x + c\right )^{3} - 4 \, a^{3} e \cos \left (d x + c\right ) +{\left (a^{3} e \cos \left (d x + c\right )^{3} - 4 \, a^{3} e \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \sqrt{e \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(-(3*a^3*e*cos(d*x + c)^3 - 4*a^3*e*cos(d*x + c) + (a^3*e*cos(d*x + c)^3 - 4*a^3*e*cos(d*x + c))*sin(d
*x + c))*sqrt(e*cos(d*x + c)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(3/2)*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(3/2)*(a*sin(d*x + c) + a)^3, x)