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]

-2/5*e/a/d/(e*sin(d*x+c))^(5/2)+2/5*e*cos(d*x+c)/a/d/(e*sin(d*x+c))^(5/2)-4/5*cos(d*x+c)/a/d/e/(e*sin(d*x+c))^
(1/2)+4/5*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^
(1/2))*(e*sin(d*x+c))^(1/2)/a/d/e^2/sin(d*x+c)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, 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 30

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

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 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 1.10, size = 124, normalized size = 0.92 \[ \frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (\cos (c+d x)+i \sin (c+d x)) \left (2 \sqrt {1-e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{2 i (c+d x)}\right ) (\cos (c+d x)+1)+3 i \sin (c+d x)-9 \cos (c+d x)-6\right )}{15 a d e \sqrt {e \sin (c+d x)}} \]

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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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \sin \left (d x + c\right )}}{a e^{2} \cos \left (d x + c\right )^{2} - a e^{2} + {\left (a e^{2} \cos \left (d x + c\right )^{2} - a e^{2}\right )} \sec \left (d x + c\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 3.70, size = 187, normalized size = 1.39 \[ \frac {-\frac {2 e}{5 a \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {\frac {4 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {7}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {4 \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {6 \left (\sin ^{3}\left (d x +c \right )\right )}{5}+\frac {2 \sin \left (d x +c \right )}{5}}{e a \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

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*sin(d*x+c)+2)^(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*sin(d*x+c)+2)^(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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (c+d\,x\right )}{a\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {1}{\left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )} + \left (e \sin {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \]

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]

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

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