3.231 \(\int \frac{(a+a \sin (c+d x))^4}{(e \cos (c+d x))^{11/2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac{2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^4-a^4 \sin (c+d x)\right )}-\frac{2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^2-a^2 \sin (c+d x)\right )^2}+\frac{4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 d e^6 \sqrt{\cos (c+d x)}} \]

[Out]

(2*a^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*e^6*Sqrt[Cos[c + d*x]]) + (4*a^7*(e*Cos[c + d*x])
^(3/2))/(9*d*e^7*(a - a*Sin[c + d*x])^3) - (2*a^8*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a^2 - a^2*Sin[c + d*x])^2
) - (2*a^8*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a^4 - a^4*Sin[c + d*x]))

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Rubi [A]  time = 0.249945, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2670, 2680, 2681, 2683, 2640, 2639} \[ -\frac{2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^4-a^4 \sin (c+d x)\right )}-\frac{2 a^8 (e \cos (c+d x))^{3/2}}{15 d e^7 \left (a^2-a^2 \sin (c+d x)\right )^2}+\frac{4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}+\frac{2 a^4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 d e^6 \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(11/2),x]

[Out]

(2*a^4*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*e^6*Sqrt[Cos[c + d*x]]) + (4*a^7*(e*Cos[c + d*x])
^(3/2))/(9*d*e^7*(a - a*Sin[c + d*x])^3) - (2*a^8*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(a^2 - a^2*Sin[c + d*x])^2
) - (2*a^8*(e*Cos[c + d*x])^(3/2))/(15*d*e^7*(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 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))^4}{(e \cos (c+d x))^{11/2}} \, dx &=\frac{a^8 \int \frac{(e \cos (c+d x))^{5/2}}{(a-a \sin (c+d x))^4} \, dx}{e^8}\\ &=\frac{4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac{a^6 \int \frac{\sqrt{e \cos (c+d x)}}{(a-a \sin (c+d x))^2} \, dx}{3 e^6}\\ &=\frac{4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac{2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac{a^5 \int \frac{\sqrt{e \cos (c+d x)}}{a-a \sin (c+d x)} \, dx}{15 e^6}\\ &=\frac{4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac{2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}+\frac{a^4 \int \sqrt{e \cos (c+d x)} \, dx}{15 e^6}\\ &=\frac{4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac{2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}+\frac{\left (a^4 \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^4 \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{4 a^7 (e \cos (c+d x))^{3/2}}{9 d e^7 (a-a \sin (c+d x))^3}-\frac{2 a^6 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))^2}-\frac{2 a^5 (e \cos (c+d x))^{3/2}}{15 d e^7 (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.146502, size = 66, normalized size = 0.39 \[ \frac{4\ 2^{3/4} a^4 (\sin (c+d x)+1)^{9/4} \, _2F_1\left (-\frac{9}{4},-\frac{3}{4};-\frac{5}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{9 d e (e \cos (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^4/(e*Cos[c + d*x])^(11/2),x]

[Out]

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

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Maple [B]  time = 2.043, size = 514, normalized size = 3. \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))^4/(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)^4-272*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+176*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-144*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))+42*sin(1/2*d*x+1/2*c)^2*cos
(1/2*d*x+1/2*c)+144*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{11}{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))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^4/(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 (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^{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))^4/(e*cos(d*x+c))^(11/2),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^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))**4/(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 )}^{4}}{\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))^4/(e*cos(d*x+c))^(11/2),x, algorithm="giac")

[Out]

integrate((a*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(11/2), x)