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

Optimal. Leaf size=85 \[ \frac{4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac{6 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{d e^2 \sqrt{\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.132392, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2670, 2680, 2640, 2639} \[ \frac{4 a^4 (e \cos (c+d x))^{3/2}}{d e^3 \left (a^2-a^2 \sin (c+d x)\right )}-\frac{6 a^2 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{d e^2 \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.061673, size = 64, normalized size = 0.75 \[ \frac{4\ 2^{3/4} a^2 \sqrt [4]{\sin (c+d x)+1} \, _2F_1\left (-\frac{3}{4},-\frac{1}{4};\frac{3}{4};\frac{1}{2} (1-\sin (c+d x))\right )}{d e \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.483, size = 120, normalized size = 1.4 \begin{align*} -2\,{\frac{ \left ( 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 ) -4\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) -2\,\sin \left ( 1/2\,dx+c/2 \right ) \right ){a}^{2}}{e\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}\sin \left ( 1/2\,dx+c/2 \right ) d}} \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))^(3/2),x)

[Out]

-2/e/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(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))-4*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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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