3.221 \(\int \frac{(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=127 \[ -\frac{6 a^6 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^3-a^3 \sin (c+d x)\right )}+\frac{4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}+\frac{6 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}} \]

[Out]

(6*a^3*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[c + d*x]]) + (4*a^5*(e*Cos[c + d*x])^
(3/2))/(5*d*e^5*(a - a*Sin[c + d*x])^2) - (6*a^6*(e*Cos[c + d*x])^(3/2))/(5*d*e^5*(a^3 - a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.202382, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2670, 2680, 2683, 2640, 2639} \[ -\frac{6 a^6 (e \cos (c+d x))^{3/2}}{5 d e^5 \left (a^3-a^3 \sin (c+d x)\right )}+\frac{4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}+\frac{6 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

(6*a^3*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[c + d*x]]) + (4*a^5*(e*Cos[c + d*x])^
(3/2))/(5*d*e^5*(a - a*Sin[c + d*x])^2) - (6*a^6*(e*Cos[c + d*x])^(3/2))/(5*d*e^5*(a^3 - a^3*Sin[c + d*x]))

Rule 2670

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

Rule 2680

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

Rule 2683

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

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 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]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx &=\frac{a^6 \int \frac{(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^3} \, dx}{e^6}\\ &=\frac{4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac{\left (3 a^4\right ) \int \frac{\sqrt{e \cos (c+d x)}}{a-a \sin (c+d x)} \, dx}{5 e^4}\\ &=\frac{4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac{6 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))}+\frac{\left (3 a^3\right ) \int \sqrt{e \cos (c+d x)} \, dx}{5 e^4}\\ &=\frac{4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac{6 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))}+\frac{\left (3 a^3 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 e^4 \sqrt{\cos (c+d x)}}\\ &=\frac{6 a^3 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{4 a^5 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))^2}-\frac{6 a^4 (e \cos (c+d x))^{3/2}}{5 d e^5 (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.0896046, size = 66, normalized size = 0.52 \[ \frac{4\ 2^{3/4} a^3 (\sin (c+d x)+1)^{5/4} \, _2F_1\left (-\frac{5}{4},-\frac{3}{4};-\frac{1}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{5 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

(4*2^(3/4)*a^3*Hypergeometric2F1[-5/4, -3/4, -1/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(5/4))/(5*d*e*(e*C
os[c + d*x])^(5/2))

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Maple [B]  time = 1.153, size = 332, normalized size = 2.6 \begin{align*}{\frac{2\,{a}^{3}}{5\,d{e}^{3}} \left ( 12\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -12\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+24\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+3\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +20\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) \left ( 4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-1} \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))^(7/2),x)

[Out]

2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e
^3*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin
(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1
/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x
+1/2*c)-20*sin(1/2*d*x+1/2*c)^5+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(
1/2*d*x+1/2*c),2^(1/2))+2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+20*sin(1/2*d*x+1/2*c)^3-sin(1/2*d*x+1/2*c))*
a^3/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}}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}\,{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), 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^{4} \cos \left (d x + c\right )^{4}}, 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))^(7/2),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^4
*cos(d*x + c)^4), 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))**(7/2),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}}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}\,{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))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), x)