3.133 \(\int \frac {1}{(a+a \sec (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=224 \[ \frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}+\frac {4 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{231 a^2 d e^2 \sqrt {e \sin (c+d x)}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}} \]

[Out]

4/11*e^3/a^2/d/(e*sin(d*x+c))^(11/2)-2/11*e^3*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(11/2)-2/11*e^3*cos(d*x+c)^3/a^2
/d/(e*sin(d*x+c))^(11/2)-4/7*e/a^2/d/(e*sin(d*x+c))^(7/2)+16/77*e*cos(d*x+c)/a^2/d/(e*sin(d*x+c))^(7/2)-4/231*
cos(d*x+c)/a^2/d/e/(e*sin(d*x+c))^(3/2)-4/231*(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)*El
lipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*sin(d*x+c)^(1/2)/a^2/d/e^2/(e*sin(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.67, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3872, 2875, 2873, 2567, 2636, 2642, 2641, 2564, 14} \[ \frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}+\frac {4 \sqrt {\sin (c+d x)} F\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{231 a^2 d e^2 \sqrt {e \sin (c+d x)}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[c + d*x])^2*(e*Sin[c + d*x])^(5/2)),x]

[Out]

(4*e^3)/(11*a^2*d*(e*Sin[c + d*x])^(11/2)) - (2*e^3*Cos[c + d*x])/(11*a^2*d*(e*Sin[c + d*x])^(11/2)) - (2*e^3*
Cos[c + d*x]^3)/(11*a^2*d*(e*Sin[c + d*x])^(11/2)) - (4*e)/(7*a^2*d*(e*Sin[c + d*x])^(7/2)) + (16*e*Cos[c + d*
x])/(77*a^2*d*(e*Sin[c + d*x])^(7/2)) - (4*Cos[c + d*x])/(231*a^2*d*e*(e*Sin[c + d*x])^(3/2)) + (4*EllipticF[(
c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(231*a^2*d*e^2*Sqrt[e*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 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 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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 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 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))^2 (e \sin (c+d x))^{5/2}} \, dx &=\int \frac {\cos ^2(c+d x)}{(-a-a \cos (c+d x))^2 (e \sin (c+d x))^{5/2}} \, dx\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{(e \sin (c+d x))^{13/2}} \, dx}{a^4}\\ &=\frac {e^4 \int \left (\frac {a^2 \cos ^2(c+d x)}{(e \sin (c+d x))^{13/2}}-\frac {2 a^2 \cos ^3(c+d x)}{(e \sin (c+d x))^{13/2}}+\frac {a^2 \cos ^4(c+d x)}{(e \sin (c+d x))^{13/2}}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{13/2}} \, dx}{a^2}+\frac {e^4 \int \frac {\cos ^4(c+d x)}{(e \sin (c+d x))^{13/2}} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \frac {\cos ^3(c+d x)}{(e \sin (c+d x))^{13/2}} \, dx}{a^2}\\ &=-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {\left (2 e^2\right ) \int \frac {1}{(e \sin (c+d x))^{9/2}} \, dx}{11 a^2}-\frac {\left (6 e^2\right ) \int \frac {\cos ^2(c+d x)}{(e \sin (c+d x))^{9/2}} \, dx}{11 a^2}-\frac {\left (2 e^3\right ) \operatorname {Subst}\left (\int \frac {1-\frac {x^2}{e^2}}{x^{13/2}} \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {10 \int \frac {1}{(e \sin (c+d x))^{5/2}} \, dx}{77 a^2}+\frac {12 \int \frac {1}{(e \sin (c+d x))^{5/2}} \, dx}{77 a^2}-\frac {\left (2 e^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{x^{13/2}}-\frac {1}{e^2 x^{9/2}}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=\frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}-\frac {10 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{231 a^2 e^2}+\frac {4 \int \frac {1}{\sqrt {e \sin (c+d x)}} \, dx}{77 a^2 e^2}\\ &=\frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}-\frac {\left (10 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{231 a^2 e^2 \sqrt {e \sin (c+d x)}}+\frac {\left (4 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\sin (c+d x)}} \, dx}{77 a^2 e^2 \sqrt {e \sin (c+d x)}}\\ &=\frac {4 e^3}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos (c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {2 e^3 \cos ^3(c+d x)}{11 a^2 d (e \sin (c+d x))^{11/2}}-\frac {4 e}{7 a^2 d (e \sin (c+d x))^{7/2}}+\frac {16 e \cos (c+d x)}{77 a^2 d (e \sin (c+d x))^{7/2}}-\frac {4 \cos (c+d x)}{231 a^2 d e (e \sin (c+d x))^{3/2}}+\frac {4 F\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {\sin (c+d x)}}{231 a^2 d e^2 \sqrt {e \sin (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.99, size = 113, normalized size = 0.50 \[ -\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (97 \cos (c+d x)+4 \cos (2 (c+d x))+\cos (3 (c+d x))+\sin ^{\frac {11}{2}}(c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) F\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )+52\right )}{1848 a^2 d e^2 \sqrt {e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sec[c + d*x])^2*(e*Sin[c + d*x])^(5/2)),x]

[Out]

-1/1848*(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]^5*(52 + 97*Cos[c + d*x] + 4*Cos[2*(c + d*x)] + Cos[3*(c + d*x)] + C
sc[(c + d*x)/2]^4*EllipticF[(-2*c + Pi - 2*d*x)/4, 2]*Sin[c + d*x]^(11/2)))/(a^2*d*e^2*Sqrt[e*Sin[c + d*x]])

________________________________________________________________________________________

fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {e \sin \left (d x + c\right )}}{{\left (a^{2} e^{3} \cos \left (d x + c\right )^{2} - a^{2} e^{3} + {\left (a^{2} e^{3} \cos \left (d x + c\right )^{2} - a^{2} e^{3}\right )} \sec \left (d x + c\right )^{2} + 2 \, {\left (a^{2} e^{3} \cos \left (d x + c\right )^{2} - a^{2} e^{3}\right )} \sec \left (d x + c\right )\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(e*sin(d*x + c))/((a^2*e^3*cos(d*x + c)^2 - a^2*e^3 + (a^2*e^3*cos(d*x + c)^2 - a^2*e^3)*sec(d*x
+ c)^2 + 2*(a^2*e^3*cos(d*x + c)^2 - a^2*e^3)*sec(d*x + c))*sin(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \sin \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((a*sec(d*x + c) + a)^2*(e*sin(d*x + c))^(5/2)), x)

________________________________________________________________________________________

maple [A]  time = 4.84, size = 160, normalized size = 0.71 \[ \frac {\frac {4 e^{3} \left (11 \left (\cos ^{2}\left (d x +c \right )\right )-4\right )}{77 a^{2} \left (e \sin \left (d x +c \right )\right )^{\frac {11}{2}}}-\frac {2 \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sin ^{\frac {13}{2}}\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{7}\left (d x +c \right )\right )+47 \left (\sin ^{5}\left (d x +c \right )\right )-87 \left (\sin ^{3}\left (d x +c \right )\right )+42 \sin \left (d x +c \right )\right )}{231 e^{2} a^{2} \sin \left (d x +c \right )^{6} \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(d*x+c))^2/(e*sin(d*x+c))^(5/2),x)

[Out]

(4/77*e^3/a^2/(e*sin(d*x+c))^(11/2)*(11*cos(d*x+c)^2-4)-2/231/e^2*((-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2
)*sin(d*x+c)^(13/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^7+47*sin(d*x+c)^5-87*sin(d*x+c)^
3+42*sin(d*x+c))/a^2/sin(d*x+c)^6/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d

________________________________________________________________________________________

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))^2/(e*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2}{a^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^{5/2}\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((e*sin(c + d*x))^(5/2)*(a + a/cos(c + d*x))^2),x)

[Out]

int(cos(c + d*x)^2/(a^2*(e*sin(c + d*x))^(5/2)*(cos(c + d*x) + 1)^2), x)

________________________________________________________________________________________

sympy [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))**2/(e*sin(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________