3.223 \(\int \frac{(a+a \sin (c+d x))^3}{(e \cos (c+d x))^{11/2}} \, dx\)

Optimal. Leaf size=165 \[ \frac{2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^3-a^3 \sin (c+d x)\right )}+\frac{2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac{2 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 d e^6 \sqrt{\cos (c+d x)}} \]

[Out]

(-2*a^3*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*e^6*Sqrt[Cos[c + d*x]]) + (2*a^6*(e*Cos[c + d*x]
)^(3/2))/(9*d*e^7*(a - a*Sin[c + d*x])^3) + (2*a^5*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a - a*Sin[c + d*x])^2) +
(2*a^6*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a^3 - a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.243165, antiderivative size = 165, 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 = {2670, 2681, 2683, 2640, 2639} \[ \frac{2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^3-a^3 \sin (c+d x)\right )}+\frac{2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac{2 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 d e^6 \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(11/2),x]

[Out]

(-2*a^3*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*e^6*Sqrt[Cos[c + d*x]]) + (2*a^6*(e*Cos[c + d*x]
)^(3/2))/(9*d*e^7*(a - a*Sin[c + d*x])^3) + (2*a^5*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a - a*Sin[c + d*x])^2) +
(2*a^6*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(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 2681

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)/(a*f*g*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
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] && LtQ[m, -1] && NeQ[2*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))^{11/2}} \, dx &=\frac{a^6 \int \frac{\sqrt{e \cos (c+d x)}}{(a-a \sin (c+d x))^3} \, dx}{e^6}\\ &=\frac{2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{a^5 \int \frac{\sqrt{e \cos (c+d x)}}{(a-a \sin (c+d x))^2} \, dx}{3 e^6}\\ &=\frac{2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac{a^4 \int \frac{\sqrt{e \cos (c+d x)}}{a-a \sin (c+d x)} \, dx}{15 e^6}\\ &=\frac{2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac{2 a^4 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}-\frac{a^3 \int \sqrt{e \cos (c+d x)} \, dx}{15 e^6}\\ &=\frac{2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac{2 a^4 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}-\frac{\left (a^3 \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 e^6 \sqrt{\cos (c+d x)}}\\ &=-\frac{2 a^3 \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e^6 \sqrt{\cos (c+d x)}}+\frac{2 a^6 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}+\frac{2 a^4 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.139316, size = 66, normalized size = 0.4 \[ \frac{2\ 2^{3/4} a^3 (\sin (c+d x)+1)^{9/4} \, _2F_1\left (-\frac{9}{4},\frac{1}{4};-\frac{5}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{9 d e (e \cos (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3/(e*Cos[c + d*x])^(11/2),x]

[Out]

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

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Maple [B]  time = 1.977, size = 514, normalized size = 3.1 \begin{align*} \text{result too large to display} \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))^(11/2),x)

[Out]

-2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2*c)^4-8*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^5*(48*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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)^8-96*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*
c)-96*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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)^6+192*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+72*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)^4-152*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*
x+1/2*c)-24*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+56*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+36*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))-48*sin(1/2*d*x+1/2*c)^2*cos(
1/2*d*x+1/2*c)-36*sin(1/2*d*x+1/2*c)^3-11*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{11}{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))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(11/2), 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^{6} \cos \left (d x + c\right )^{6}}, 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))^(11/2),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^6
*cos(d*x + c)^6), 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))**(11/2),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{11}{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))^(11/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(11/2), x)