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

Optimal. Leaf size=215 \[ \frac {4 \csc (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}+\frac {4 \sin (c+d x)}{3 a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {12 \sin (c+d x) \cos (c+d x)}{5 a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {44 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{5 a^2 d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

[Out]

-2*cot(d*x+c)/a^2/d/e^2/(e*csc(d*x+c))^(1/2)-2*cos(d*x+c)^2*cot(d*x+c)/a^2/d/e^2/(e*csc(d*x+c))^(1/2)+4*csc(d*
x+c)/a^2/d/e^2/(e*csc(d*x+c))^(1/2)+4/3*sin(d*x+c)/a^2/d/e^2/(e*csc(d*x+c))^(1/2)-12/5*cos(d*x+c)*sin(d*x+c)/a
^2/d/e^2/(e*csc(d*x+c))^(1/2)+44/5*(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))/a^2/d/e^2/(e*csc(d*x+c))^(1/2)/sin(d*x+c)^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3878, 3872, 2875, 2873, 2567, 2639, 2564, 14, 2569} \[ \frac {4 \csc (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}+\frac {4 \sin (c+d x)}{3 a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {12 \sin (c+d x) \cos (c+d x)}{5 a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {44 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{5 a^2 d e^2 \sqrt {\sin (c+d x)} \sqrt {e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cot[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (2*Cos[c + d*x]^2*Cot[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c +
d*x]]) + (4*Csc[c + d*x])/(a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (44*EllipticE[(c - Pi/2 + d*x)/2, 2])/(5*a^2*d*e
^2*Sqrt[e*Csc[c + d*x]]*Sqrt[Sin[c + d*x]]) + (4*Sin[c + d*x])/(3*a^2*d*e^2*Sqrt[e*Csc[c + d*x]]) - (12*Cos[c
+ d*x]*Sin[c + d*x])/(5*a^2*d*e^2*Sqrt[e*Csc[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 2569

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

Rule 3878

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*((g_.)*sec[(e_.) + (f_.)*(x_)])^(p_), x_Symbol] :> Dist[g^Int
Part[p]*(g*Sec[e + f*x])^FracPart[p]*Cos[e + f*x]^FracPart[p], Int[(a + b*Csc[e + f*x])^m/Cos[e + f*x]^p, x],
x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{(e \csc (c+d x))^{5/2} (a+a \sec (c+d x))^2} \, dx &=\frac {\int \frac {\sin ^{\frac {5}{2}}(c+d x)}{(a+a \sec (c+d x))^2} \, dx}{e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \frac {\cos ^2(c+d x) \sin ^{\frac {5}{2}}(c+d x)}{(-a-a \cos (c+d x))^2} \, dx}{e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \frac {\cos ^2(c+d x) (-a+a \cos (c+d x))^2}{\sin ^{\frac {3}{2}}(c+d x)} \, dx}{a^4 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \left (\frac {a^2 \cos ^2(c+d x)}{\sin ^{\frac {3}{2}}(c+d x)}-\frac {2 a^2 \cos ^3(c+d x)}{\sin ^{\frac {3}{2}}(c+d x)}+\frac {a^2 \cos ^4(c+d x)}{\sin ^{\frac {3}{2}}(c+d x)}\right ) \, dx}{a^4 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=\frac {\int \frac {\cos ^2(c+d x)}{\sin ^{\frac {3}{2}}(c+d x)} \, dx}{a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {\int \frac {\cos ^4(c+d x)}{\sin ^{\frac {3}{2}}(c+d x)} \, dx}{a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {2 \int \frac {\cos ^3(c+d x)}{\sin ^{\frac {3}{2}}(c+d x)} \, dx}{a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=-\frac {2 \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \int \sqrt {\sin (c+d x)} \, dx}{a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {6 \int \cos ^2(c+d x) \sqrt {\sin (c+d x)} \, dx}{a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {1-x^2}{x^{3/2}} \, dx,x,\sin (c+d x)\right )}{a^2 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=-\frac {2 \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {4 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{a^2 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {12 \cos (c+d x) \sin (c+d x)}{5 a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {12 \int \sqrt {\sin (c+d x)} \, dx}{5 a^2 e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}-\frac {2 \operatorname {Subst}\left (\int \left (\frac {1}{x^{3/2}}-\sqrt {x}\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}\\ &=-\frac {2 \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {2 \cos ^2(c+d x) \cot (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}+\frac {4 \csc (c+d x)}{a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {44 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{5 a^2 d e^2 \sqrt {e \csc (c+d x)} \sqrt {\sin (c+d x)}}+\frac {4 \sin (c+d x)}{3 a^2 d e^2 \sqrt {e \csc (c+d x)}}-\frac {12 \cos (c+d x) \sin (c+d x)}{5 a^2 d e^2 \sqrt {e \csc (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 2.14, size = 125, normalized size = 0.58 \[ \frac {88 \sqrt {1-e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};e^{2 i (c+d x)}\right ) (\cot (c+d x)+i)-123 \cot (c+d x)+\csc (c+d x) (-264 i \sin (c+d x)-20 \cos (2 (c+d x))+3 \cos (3 (c+d x))+140)}{30 a^2 d e^2 \sqrt {e \csc (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-123*Cot[c + d*x] + 88*Sqrt[1 - E^((2*I)*(c + d*x))]*(I + Cot[c + d*x])*Hypergeometric2F1[1/2, 3/4, 7/4, E^((
2*I)*(c + d*x))] + Csc[c + d*x]*(140 - 20*Cos[2*(c + d*x)] + 3*Cos[3*(c + d*x)] - (264*I)*Sin[c + d*x]))/(30*a
^2*d*e^2*Sqrt[e*Csc[c + d*x]])

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e \csc \left (d x + c\right )}}{a^{2} e^{3} \csc \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 2 \, a^{2} e^{3} \csc \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a^{2} e^{3} \csc \left (d x + c\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.30, size = 563, normalized size = 2.62 \[ \frac {\left (-66 \cos \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}+132 \cos \left (d x +c \right ) \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}+3 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-66 \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}+132 \sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (d x +c \right )-i+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}}-10 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+33 \cos \left (d x +c \right ) \sqrt {2}-26 \sqrt {2}\right ) \sqrt {2}}{15 a^{2} d \left (\frac {e}{\sin \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15/a^2/d*(-66*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*
x+c))^(1/2)*EllipticF(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+
c))^(1/2)+132*cos(d*x+c)*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+
c))^(1/2)*EllipticE(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)+3*2^(1/2)*cos(d*x+c)^3-66*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))
/sin(d*x+c))^(1/2)*EllipticF(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/s
in(d*x+c))^(1/2)+132*((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c))^
(1/2)*EllipticE(((I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-I*(-1+cos(d*x+c))/sin(d*x+c))^(1
/2)-10*cos(d*x+c)^2*2^(1/2)+33*cos(d*x+c)*2^(1/2)-26*2^(1/2))/(e/sin(d*x+c))^(5/2)/sin(d*x+c)^3*2^(1/2)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(e*csc(d*x+c))**(5/2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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