3.230 \(\int \frac{(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=127 \[ -\frac{20 a^8 \sqrt{e \cos (c+d x)}}{21 d e^5 \left (a^4-a^4 \sin (c+d x)\right )}+\frac{4 a^7 (e \cos (c+d x))^{5/2}}{7 d e^7 (a-a \sin (c+d x))^3}+\frac{10 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt{e \cos (c+d x)}} \]

[Out]

(10*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*e^4*Sqrt[e*Cos[c + d*x]]) + (4*a^7*(e*Cos[c + d*x]
)^(5/2))/(7*d*e^7*(a - a*Sin[c + d*x])^3) - (20*a^8*Sqrt[e*Cos[c + d*x]])/(21*d*e^5*(a^4 - a^4*Sin[c + d*x]))

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Rubi [A]  time = 0.198481, antiderivative size = 127, 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 = {2670, 2680, 2642, 2641} \[ -\frac{20 a^8 \sqrt{e \cos (c+d x)}}{21 d e^5 \left (a^4-a^4 \sin (c+d x)\right )}+\frac{4 a^7 (e \cos (c+d x))^{5/2}}{7 d e^7 (a-a \sin (c+d x))^3}+\frac{10 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(9/2),x]

[Out]

(10*a^4*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(21*d*e^4*Sqrt[e*Cos[c + d*x]]) + (4*a^7*(e*Cos[c + d*x]
)^(5/2))/(7*d*e^7*(a - a*Sin[c + d*x])^3) - (20*a^8*Sqrt[e*Cos[c + d*x]])/(21*d*e^5*(a^4 - a^4*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 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))^4}{(e \cos (c+d x))^{9/2}} \, dx &=\frac{a^8 \int \frac{(e \cos (c+d x))^{7/2}}{(a-a \sin (c+d x))^4} \, dx}{e^8}\\ &=\frac{4 a^7 (e \cos (c+d x))^{5/2}}{7 d e^7 (a-a \sin (c+d x))^3}-\frac{\left (5 a^6\right ) \int \frac{(e \cos (c+d x))^{3/2}}{(a-a \sin (c+d x))^2} \, dx}{7 e^6}\\ &=\frac{4 a^7 (e \cos (c+d x))^{5/2}}{7 d e^7 (a-a \sin (c+d x))^3}-\frac{20 a^6 \sqrt{e \cos (c+d x)}}{21 d e^5 \left (a^2-a^2 \sin (c+d x)\right )}+\frac{\left (5 a^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{21 e^4}\\ &=\frac{4 a^7 (e \cos (c+d x))^{5/2}}{7 d e^7 (a-a \sin (c+d x))^3}-\frac{20 a^6 \sqrt{e \cos (c+d x)}}{21 d e^5 \left (a^2-a^2 \sin (c+d x)\right )}+\frac{\left (5 a^4 \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 e^4 \sqrt{e \cos (c+d x)}}\\ &=\frac{10 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt{e \cos (c+d x)}}+\frac{4 a^7 (e \cos (c+d x))^{5/2}}{7 d e^7 (a-a \sin (c+d x))^3}-\frac{20 a^6 \sqrt{e \cos (c+d x)}}{21 d e^5 \left (a^2-a^2 \sin (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 0.117539, size = 66, normalized size = 0.52 \[ \frac{8 \sqrt [4]{2} a^4 (\sin (c+d x)+1)^{7/4} \, _2F_1\left (-\frac{7}{4},-\frac{5}{4};-\frac{3}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{7 d e (e \cos (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(9/2),x]

[Out]

(8*2^(1/4)*a^4*Hypergeometric2F1[-7/4, -5/4, -3/4, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(7/4))/(7*d*e*(e*C
os[c + d*x])^(7/2))

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Maple [B]  time = 1.585, size = 401, normalized size = 3.2 \begin{align*} -{\frac{2\,{a}^{4}}{21\,{e}^{4}d} \left ( 40\,\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 ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-60\,\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 ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-128\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +30\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+128\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) -112\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}-5\,\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 ) +16\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) +112\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-4\,\sin \left ( 1/2\,dx+c/2 \right ) \right ) \left ( 8\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-12\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+6\, \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))^4/(e*cos(d*x+c))^(9/2),x)

[Out]

-2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*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^4*(40*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)^6-60*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(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)^4-128*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+30*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2
*c)^2+128*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-112*sin(1/2*d*x+1/2*c)^5-5*(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))+16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+
112*sin(1/2*d*x+1/2*c)^3-4*sin(1/2*d*x+1/2*c))*a^4/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 )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(9/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a^{4} \cos \left (d x + c\right )^{4} - 8 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} - 4 \,{\left (a^{4} \cos \left (d x + c\right )^{2} - 2 \, a^{4}\right )} \sin \left (d x + c\right )\right )} \sqrt{e \cos \left (d x + c\right )}}{e^{5} \cos \left (d x + c\right )^{5}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(9/2),x, algorithm="fricas")

[Out]

integral((a^4*cos(d*x + c)^4 - 8*a^4*cos(d*x + c)^2 + 8*a^4 - 4*(a^4*cos(d*x + c)^2 - 2*a^4)*sin(d*x + c))*sqr
t(e*cos(d*x + c))/(e^5*cos(d*x + c)^5), 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))**4/(e*cos(d*x+c))**(9/2),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 )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^4/(e*cos(d*x+c))^(9/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(9/2), x)