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

Optimal. Leaf size=145 \[ \frac{2 a^2 \sin (c+d x)}{3 d e^5 \sqrt{e \cos (c+d x)}}+\frac{2 a^2 \sin (c+d x)}{9 d e^3 (e \cos (c+d x))^{5/2}}-\frac{2 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{3 d e^6 \sqrt{\cos (c+d x)}}+\frac{4 \left (a^2 \sin (c+d x)+a^2\right )}{9 d e (e \cos (c+d x))^{9/2}} \]

[Out]

(-2*a^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(3*d*e^6*Sqrt[Cos[c + d*x]]) + (2*a^2*Sin[c + d*x])/(9
*d*e^3*(e*Cos[c + d*x])^(5/2)) + (2*a^2*Sin[c + d*x])/(3*d*e^5*Sqrt[e*Cos[c + d*x]]) + (4*(a^2 + a^2*Sin[c + d
*x]))/(9*d*e*(e*Cos[c + d*x])^(9/2))

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Rubi [A]  time = 0.112072, antiderivative size = 145, 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 = {2676, 2636, 2640, 2639} \[ \frac{2 a^2 \sin (c+d x)}{3 d e^5 \sqrt{e \cos (c+d x)}}+\frac{2 a^2 \sin (c+d x)}{9 d e^3 (e \cos (c+d x))^{5/2}}-\frac{2 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{3 d e^6 \sqrt{\cos (c+d x)}}+\frac{4 \left (a^2 \sin (c+d x)+a^2\right )}{9 d e (e \cos (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Sqrt[e*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(3*d*e^6*Sqrt[Cos[c + d*x]]) + (2*a^2*Sin[c + d*x])/(9
*d*e^3*(e*Cos[c + d*x])^(5/2)) + (2*a^2*Sin[c + d*x])/(3*d*e^5*Sqrt[e*Cos[c + d*x]]) + (4*(a^2 + a^2*Sin[c + d
*x]))/(9*d*e*(e*Cos[c + d*x])^(9/2))

Rule 2676

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

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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

Mathematica [C]  time = 0.160718, size = 66, normalized size = 0.46 \[ \frac{2^{3/4} a^2 (\sin (c+d x)+1)^{9/4} \, _2F_1\left (-\frac{9}{4},\frac{5}{4};-\frac{5}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{9 d e (e \cos (c+d x))^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.622, size = 488, normalized size = 3.4 \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))^2/(e*cos(d*x+c))^(11/2),x)

[Out]

-2/9/(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)+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))-12*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*sin(1/2*
d*x+1/2*c))*a^2/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 )}^{2}}{\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))^2/(e*cos(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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