\(\int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx\) [132]

Optimal result

Integrand size = 25, antiderivative size = 224 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\frac {4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac {16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 \cos (c+d x)}{15 a^2 d e \sqrt {e \sin (c+d x)}}-\frac {4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 a^2 d e^2 \sqrt {\sin (c+d x)}} \]

[Out]

4/9*e^3/a^2/d/(e*sin(d*x+c))^(9/2)-2/9*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(9/2)-2/9*e^3*cos(d*x+c)^3/a^2/d/(e
*sin(d*x+c))^(9/2)-4/5*e/a^2/d/(e*sin(d*x+c))^(5/2)+16/45*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(5/2)-4/15*cos(d*x
+c)/a^2/d/e/(e*sin(d*x+c))^(1/2)+4/15*(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^2/d/e^2/sin(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3957, 2954, 2952, 2647, 2716, 2721, 2719, 2644, 14} \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\frac {4 e^3}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {2 e^3 \cos ^3(c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {2 e^3 \cos (c+d x)}{9 a^2 d (e \sin (c+d x))^{9/2}}-\frac {4 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{15 a^2 d e^2 \sqrt {\sin (c+d x)}}-\frac {4 e}{5 a^2 d (e \sin (c+d x))^{5/2}}+\frac {16 e \cos (c+d x)}{45 a^2 d (e \sin (c+d x))^{5/2}}-\frac {4 \cos (c+d x)}{15 a^2 d e \sqrt {e \sin (c+d x)}} \]

[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 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]]

Rule 2644

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 2647

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 2716

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

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

Rule 2952

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 2954

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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*} \text {integral}& = \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 ) \text {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 ) \text {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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.99 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) (\cos (c+d x)+i \sin (c+d x)) \left (-31-40 \cos (c+d x)-19 \cos (2 (c+d x))+e^{-2 i (c+d x)} \left (1+e^{i (c+d x)}\right )^4 \sqrt {1-e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{2 i (c+d x)}\right )+16 i \sin (c+d x)+13 i \sin (2 (c+d x))\right )}{180 a^2 d e \sqrt {e \sin (c+d x)}} \]

[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]])

Maple [A] (verified)

Time = 6.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {\frac {4 e^{3} \left (9 \cos \left (d x +c \right )^{2}-4\right )}{45 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {9}{2}}}+\frac {\frac {4 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {11}{2}} \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{15}-\frac {2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sin \left (d x +c \right )^{\frac {11}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{15}+\frac {4 \sin \left (d x +c \right )^{7}}{15}-\frac {38 \sin \left (d x +c \right )^{5}}{45}+\frac {46 \sin \left (d x +c \right )^{3}}{45}-\frac {4 \sin \left (d x +c \right )}{9}}{e \,a^{2} \sin \left (d x +c \right )^{5} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(213\)

[In]

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

[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*sin(d*x+c)+2)^(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*sin(d*x+c)+2)^(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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} \sqrt {-i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} \sqrt {i \, e} \sin \left (d x + c\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (6 \, \cos \left (d x + c\right )^{3} + 12 \, \cos \left (d x + c\right )^{2} + 19 \, \cos \left (d x + c\right ) + 8\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{45 \, {\left (a^{2} d e^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} d e^{2} \cos \left (d x + c\right ) + a^{2} d e^{2}\right )} \sin \left (d x + c\right )} \]

[In]

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

[Out]

-2/45*(3*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*sqrt(-I*e)*sin(d*x + c)*weierstrass
Zeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt
(2)*cos(d*x + c) - I*sqrt(2))*sqrt(I*e)*sin(d*x + c)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x +
c) - I*sin(d*x + c))) + (6*cos(d*x + c)^3 + 12*cos(d*x + c)^2 + 19*cos(d*x + c) + 8)*sqrt(e*sin(d*x + c)))/((
a^2*d*e^2*cos(d*x + c)^2 + 2*a^2*d*e^2*cos(d*x + c) + a^2*d*e^2)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

[In]

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

[Out]

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