3.218 \(\int \frac{(a+a \sin (c+d x))^3}{\sqrt{e \cos (c+d x)}} \, dx\)

Optimal. Leaf size=136 \[ -\frac{6 a^3 \sqrt{e \cos (c+d x)}}{d e}-\frac{6 \left (a^3 \sin (c+d x)+a^3\right ) \sqrt{e \cos (c+d x)}}{5 d e}+\frac{6 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{e \cos (c+d x)}}-\frac{2 a (a \sin (c+d x)+a)^2 \sqrt{e \cos (c+d x)}}{5 d e} \]

[Out]

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

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Rubi [A]  time = 0.147357, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2678, 2669, 2642, 2641} \[ -\frac{6 a^3 \sqrt{e \cos (c+d x)}}{d e}-\frac{6 \left (a^3 \sin (c+d x)+a^3\right ) \sqrt{e \cos (c+d x)}}{5 d e}+\frac{6 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{e \cos (c+d x)}}-\frac{2 a (a \sin (c+d x)+a)^2 \sqrt{e \cos (c+d x)}}{5 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^3*Sqrt[e*Cos[c + d*x]])/(d*e) + (6*a^3*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(d*Sqrt[e*Cos[c + d
*x]]) - (2*a*Sqrt[e*Cos[c + d*x]]*(a + a*Sin[c + d*x])^2)/(5*d*e) - (6*Sqrt[e*Cos[c + d*x]]*(a^3 + a^3*Sin[c +
d*x]))/(5*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 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 \frac{(a+a \sin (c+d x))^3}{\sqrt{e \cos (c+d x)}} \, dx &=-\frac{2 a \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}{5 d e}+\frac{1}{5} (9 a) \int \frac{(a+a \sin (c+d x))^2}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{2 a \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}{5 d e}-\frac{6 \sqrt{e \cos (c+d x)} \left (a^3+a^3 \sin (c+d x)\right )}{5 d e}+\left (3 a^2\right ) \int \frac{a+a \sin (c+d x)}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{6 a^3 \sqrt{e \cos (c+d x)}}{d e}-\frac{2 a \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}{5 d e}-\frac{6 \sqrt{e \cos (c+d x)} \left (a^3+a^3 \sin (c+d x)\right )}{5 d e}+\left (3 a^3\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx\\ &=-\frac{6 a^3 \sqrt{e \cos (c+d x)}}{d e}-\frac{2 a \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}{5 d e}-\frac{6 \sqrt{e \cos (c+d x)} \left (a^3+a^3 \sin (c+d x)\right )}{5 d e}+\frac{\left (3 a^3 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{\sqrt{e \cos (c+d x)}}\\ &=-\frac{6 a^3 \sqrt{e \cos (c+d x)}}{d e}+\frac{6 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d \sqrt{e \cos (c+d x)}}-\frac{2 a \sqrt{e \cos (c+d x)} (a+a \sin (c+d x))^2}{5 d e}-\frac{6 \sqrt{e \cos (c+d x)} \left (a^3+a^3 \sin (c+d x)\right )}{5 d e}\\ \end{align*}

Mathematica [C]  time = 0.0305894, size = 64, normalized size = 0.47 \[ -\frac{16 \sqrt [4]{2} a^3 \sqrt{e \cos (c+d x)} \, _2F_1\left (-\frac{9}{4},\frac{1}{4};\frac{5}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{d e \sqrt [4]{\sin (c+d x)+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*2^(1/4)*a^3*Sqrt[e*Cos[c + d*x]]*Hypergeometric2F1[-9/4, 1/4, 5/4, (1 - Sin[c + d*x])/2])/(d*e*(1 + Sin[c
+ d*x])^(1/4))

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Maple [A]  time = 0.447, size = 178, normalized size = 1.3 \begin{align*} -{\frac{2\,{a}^{3}}{5\,d} \left ( 8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{7}-20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -12\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+15\,\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 ) +10\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -34\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+19\,\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((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x)

[Out]

-2/5/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)*a^3*(8*sin(1/2*d*x+1/2*c)^7-20*sin(1/2*d*x+1/2*c)^
4*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^5+15*(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))+10*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-34*sin(1/2*d*x+1/2*c)^3+19*si
n(1/2*d*x+1/2*c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{e \cos \left (d x + c\right )}}{e \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(-(3*a^3*cos(d*x + c)^2 - 4*a^3 + (a^3*cos(d*x + c)^2 - 4*a^3)*sin(d*x + c))*sqrt(e*cos(d*x + c))/(e*c
os(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((a+a*sin(d*x+c))**3/(e*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3/(e*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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