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

Optimal. Leaf size=224 \[ \frac{4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{15 a^2 d e^2 \sqrt{\sin (c+d x)}}-\frac{4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac{16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac{4 \cos (c+d x)}{15 a^2 d e \sqrt{e \sin (c+d x)}} \]

[Out]

(4*e^3)/(9*a^2*d*(e*Sin[c + d*x])^(9/2)) - (2*e^3*Cos[c + d*x])/(9*a^2*d*(e*Sin[c + d*x])^(9/2)) - (2*e^3*Cos[
c + d*x]^3)/(9*a^2*d*(e*Sin[c + d*x])^(9/2)) - (4*e)/(5*a^2*d*(e*Sin[c + d*x])^(5/2)) + (16*e*Cos[c + d*x])/(4
5*a^2*d*(e*Sin[c + d*x])^(5/2)) - (4*Cos[c + d*x])/(15*a^2*d*e*Sqrt[e*Sin[c + d*x]]) - (4*EllipticE[(c - Pi/2
+ d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(15*a^2*d*e^2*Sqrt[Sin[c + d*x]])

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Rubi [A]  time = 0.6638, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3872, 2875, 2873, 2567, 2636, 2640, 2639, 2564, 14} \[ \frac{4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{4 E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{15 a^2 d e^2 \sqrt{\sin (c+d x)}}-\frac{4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac{16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac{4 \cos (c+d x)}{15 a^2 d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*e^3)/(9*a^2*d*(e*Sin[c + d*x])^(9/2)) - (2*e^3*Cos[c + d*x])/(9*a^2*d*(e*Sin[c + d*x])^(9/2)) - (2*e^3*Cos[
c + d*x]^3)/(9*a^2*d*(e*Sin[c + d*x])^(9/2)) - (4*e)/(5*a^2*d*(e*Sin[c + d*x])^(5/2)) + (16*e*Cos[c + d*x])/(4
5*a^2*d*(e*Sin[c + d*x])^(5/2)) - (4*Cos[c + d*x])/(15*a^2*d*e*Sqrt[e*Sin[c + d*x]]) - (4*EllipticE[(c - Pi/2
+ d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(15*a^2*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 2875

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

Rule 2873

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

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]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx &=\int \frac{\cos ^2(c+d x)}{(-a-a \cos (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx\\ &=\frac{e^4 \int \frac{\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{(e \sin (c+d x))^{11/2}} \, dx}{a^4}\\ &=\frac{e^4 \int \left (\frac{a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{11/2}}-\frac{2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{11/2}}+\frac{a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{11/2}}\right ) \, dx}{a^4}\\ &=\frac{e^4 \int \frac{\cos ^2(c+d x)}{(e \sin (c+d x))^{11/2}} \, dx}{a^2}+\frac{e^4 \int \frac{\cos ^4(c+d x)}{(e \sin (c+d x))^{11/2}} \, dx}{a^2}-\frac{\left (2 e^4\right ) \int \frac{\cos ^3(c+d x)}{(e \sin (c+d x))^{11/2}} \, dx}{a^2}\\ &=-\frac{2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{\left (2 e^2\right ) \int \frac{1}{(e \sin (c+d x))^{7/2}} \, dx}{9 a^2}-\frac{\left (2 e^2\right ) \int \frac{\cos ^2(c+d x)}{(e \sin (c+d x))^{7/2}} \, dx}{3 a^2}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{1-\frac{x^2}{e^2}}{x^{11/2}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=-\frac{2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}+\frac{16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac{2 \int \frac{1}{(e \sin (c+d x))^{3/2}} \, dx}{15 a^2}+\frac{4 \int \frac{1}{(e \sin (c+d x))^{3/2}} \, dx}{15 a^2}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^{11/2}}-\frac{1}{e^2 x^{7/2}}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=\frac{4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac{16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac{4 \cos (c+d x)}{15 a^2 d e \sqrt{e \sin (c+d x)}}+\frac{2 \int \sqrt{e \sin (c+d x)} \, dx}{15 a^2 e^2}-\frac{4 \int \sqrt{e \sin (c+d x)} \, dx}{15 a^2 e^2}\\ &=\frac{4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac{16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac{4 \cos (c+d x)}{15 a^2 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}{15 a^2 e^2 \sqrt{\sin (c+d x)}}-\frac{\left (4 \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{15 a^2 e^2 \sqrt{\sin (c+d x)}}\\ &=\frac{4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac{4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac{16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac{4 \cos (c+d x)}{15 a^2 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)}}{15 a^2 d e^2 \sqrt{\sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 1.42102, size = 163, normalized size = 0.73 \[ \frac{\sec ^4\left (\frac{1}{2} (c+d x)\right ) (\cos (c+d x)+i \sin (c+d x)) \left (e^{-2 i (c+d x)} \sqrt{1-e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )^4 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},e^{2 i (c+d x)}\right )+16 i \sin (c+d x)+13 i \sin (2 (c+d x))-40 \cos (c+d x)-19 \cos (2 (c+d x))-31\right )}{180 a^2 d e \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]^4*(Cos[c + d*x] + I*Sin[c + d*x])*(-31 - 40*Cos[c + d*x] - 19*Cos[2*(c + d*x)] + ((1 + E^(I*
(c + d*x)))^4*Sqrt[1 - E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, E^((2*I)*(c + d*x))])/E^((2*I)*(c
+ d*x)) + (16*I)*Sin[c + d*x] + (13*I)*Sin[2*(c + d*x)]))/(180*a^2*d*e*Sqrt[e*Sin[c + d*x]])

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Maple [A]  time = 1.753, size = 213, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{4\,{e}^{3} \left ( 9\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-4 \right ) }{45\,{a}^{2}} \left ( e\sin \left ( dx+c \right ) \right ) ^{-{\frac{9}{2}}}}+{\frac{2}{45\,e{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{5}\cos \left ( dx+c \right ) } \left ( 6\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{11/2}{\it EllipticE} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) -3\,\sqrt{-\sin \left ( dx+c \right ) +1}\sqrt{2+2\,\sin \left ( dx+c \right ) } \left ( \sin \left ( dx+c \right ) \right ) ^{11/2}{\it EllipticF} \left ( \sqrt{-\sin \left ( dx+c \right ) +1},1/2\,\sqrt{2} \right ) +6\, \left ( \sin \left ( dx+c \right ) \right ) ^{7}-19\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}+23\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}-10\,\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))^2/(e*sin(d*x+c))^(3/2),x)

[Out]

(4/45*e^3/a^2/(e*sin(d*x+c))^(9/2)*(9*cos(d*x+c)^2-4)+2/45/e*(6*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*s
in(d*x+c)^(11/2)*EllipticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-3*(-sin(d*x+c)+1)^(1/2)*(2+2*sin(d*x+c))^(1/2)*s
in(d*x+c)^(11/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))+6*sin(d*x+c)^7-19*sin(d*x+c)^5+23*sin(d*x+c)^3-1
0*sin(d*x+c))/a^2/sin(d*x+c)^5/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))^2/(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^{2} e^{2} \cos \left (d x + c\right )^{2} - a^{2} e^{2} +{\left (a^{2} e^{2} \cos \left (d x + c\right )^{2} - a^{2} e^{2}\right )} \sec \left (d x + c\right )^{2} + 2 \,{\left (a^{2} e^{2} \cos \left (d x + c\right )^{2} - a^{2} 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))^2/(e*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

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