3.125 \(\int \frac{1}{(a+a \sec (c+d x)) (e \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ -\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 a d e^2 \sqrt{\sin (c+d x)}}-\frac{2 e}{5 a d (e \sin (c+d x))^{5/2}}+\frac{2 e \cos (c+d x)}{5 a d (e \sin (c+d x))^{5/2}}-\frac{4 \cos (c+d x)}{5 a d e \sqrt{e \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.248337, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3872, 2839, 2564, 30, 2567, 2636, 2640, 2639} \[ -\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{5 a d e^2 \sqrt{\sin (c+d x)}}-\frac{2 e}{5 a d (e \sin (c+d x))^{5/2}}+\frac{2 e \cos (c+d x)}{5 a d (e \sin (c+d x))^{5/2}}-\frac{4 \cos (c+d x)}{5 a d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[c + d*x])*(e*Sin[c + d*x])^(3/2)),x]

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2567

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a*Cos[e +
f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/(b*f*(n + 1)), x] + Dist[(a^2*(m - 1))/(b^2*(n + 1)), Int[(a*Cos[e +
f*x])^(m - 2)*(b*Sin[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && (Intege
rsQ[2*m, 2*n] || EqQ[m + n, 0])

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

Mathematica [C]  time = 1.06398, size = 124, normalized size = 0.92 \[ \frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (\cos (c+d x)+i \sin (c+d x)) \left (2 \sqrt{1-e^{2 i (c+d x)}} (\cos (c+d x)+1) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )+3 i \sin (c+d x)-9 \cos (c+d x)-6\right )}{15 a d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[c + d*x])*(e*Sin[c + d*x])^(3/2)),x]

[Out]

(Sec[(c + d*x)/2]^2*(Cos[c + d*x] + I*Sin[c + d*x])*(-6 - 9*Cos[c + d*x] + 2*Sqrt[1 - E^((2*I)*(c + d*x))]*(1
+ Cos[c + d*x])*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))] + (3*I)*Sin[c + d*x]))/(15*a*d*e*Sqrt[e*
Sin[c + d*x]])

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Maple [A]  time = 1.458, size = 187, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( -{\frac{2\,e}{5\,a} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}+{\frac{2}{5\,ae \left ( \sin \left ( dx+c \right ) \right ) ^{3}\cos \left ( dx+c \right ) } \left ( 2\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{7/2}{\it EllipticE} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) -\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},{\frac{\sqrt{2}}{2}} \right ) +2\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}-3\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{e\sin \left ( dx+c \right ) }}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(3/2),x)

[Out]

(-2/5/a*e/(e*sin(d*x+c))^(5/2)+2/5/e*(2*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(7/2)*Elliptic
E((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(7/2)*EllipticF((
-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+2*sin(d*x+c)^5-3*sin(d*x+c)^3+sin(d*x+c))/a/sin(d*x+c)^3/cos(d*x+c)/(e*sin(d
*x+c))^(1/2))/d

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Maxima [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(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(e*sin(d*x + c))/(a*e^2*cos(d*x + c)^2 - a*e^2 + (a*e^2*cos(d*x + c)^2 - a*e^2)*sec(d*x + c)), 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(1/(a+a*sec(d*x+c))/(e*sin(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{1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \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(1/(a+a*sec(d*x+c))/(e*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)*(e*sin(d*x + c))^(3/2)), x)