3.112 \(\int \frac {\csc (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=110 \[ \frac {6 \cos (a+b x)}{5 b d^2 \sqrt {d \tan (a+b x)}}+\frac {6 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \csc (a+b x)}{5 b d (d \tan (a+b x))^{3/2}} \]

[Out]

6/5*cos(b*x+a)/b/d^2/(d*tan(b*x+a))^(1/2)-6/5*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/
4*Pi+b*x),2^(1/2))*sin(b*x+a)/b/d^2/sin(2*b*x+2*a)^(1/2)/(d*tan(b*x+a))^(1/2)-2/5*csc(b*x+a)/b/d/(d*tan(b*x+a)
)^(3/2)

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Rubi [A]  time = 0.14, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2597, 2601, 2570, 2572, 2639} \[ \frac {6 \cos (a+b x)}{5 b d^2 \sqrt {d \tan (a+b x)}}+\frac {6 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{5 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \csc (a+b x)}{5 b d (d \tan (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/(d*Tan[a + b*x])^(5/2),x]

[Out]

(-2*Csc[a + b*x])/(5*b*d*(d*Tan[a + b*x])^(3/2)) + (6*Cos[a + b*x])/(5*b*d^2*Sqrt[d*Tan[a + b*x]]) + (6*Ellipt
icE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(5*b*d^2*Sqrt[Sin[2*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])

Rule 2570

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

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
x]

Rule 2597

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sin[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n + 1)), x] - Dist[(n + 1)/(b^2*(m + n + 1)), Int[(a*Sin[e + f*x])
^m*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

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]

Rubi steps

\begin {align*} \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=-\frac {2 \csc (a+b x)}{5 b d (d \tan (a+b x))^{3/2}}-\frac {3 \int \frac {\csc (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx}{5 d^2}\\ &=-\frac {2 \csc (a+b x)}{5 b d (d \tan (a+b x))^{3/2}}-\frac {\left (3 \sqrt {\sin (a+b x)}\right ) \int \frac {\sqrt {\cos (a+b x)}}{\sin ^{\frac {3}{2}}(a+b x)} \, dx}{5 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {2 \csc (a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {6 \cos (a+b x)}{5 b d^2 \sqrt {d \tan (a+b x)}}+\frac {\left (6 \sqrt {\sin (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{5 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {2 \csc (a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {6 \cos (a+b x)}{5 b d^2 \sqrt {d \tan (a+b x)}}+\frac {(6 \sin (a+b x)) \int \sqrt {\sin (2 a+2 b x)} \, dx}{5 d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {2 \csc (a+b x)}{5 b d (d \tan (a+b x))^{3/2}}+\frac {6 \cos (a+b x)}{5 b d^2 \sqrt {d \tan (a+b x)}}+\frac {6 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{5 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ \end {align*}

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Mathematica [C]  time = 1.79, size = 105, normalized size = 0.95 \[ \frac {2 \sin (a+b x) \sqrt {d \tan (a+b x)} \left (2 \sec ^2(a+b x) \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right )-\left (\csc ^4(a+b x)-4 \csc ^2(a+b x)+3\right ) \sqrt {\sec ^2(a+b x)}\right )}{5 b d^3 \sqrt {\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/(d*Tan[a + b*x])^(5/2),x]

[Out]

(2*(2*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x]^2 - (3 - 4*Csc[a + b*x]^2 + Csc[a + b*x]^
4)*Sqrt[Sec[a + b*x]^2])*Sin[a + b*x]*Sqrt[d*Tan[a + b*x]])/(5*b*d^3*Sqrt[Sec[a + b*x]^2])

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \tan \left (b x + a\right )} \csc \left (b x + a\right )}{d^{3} \tan \left (b x + a\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*csc(b*x + a)/(d^3*tan(b*x + a)^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/(d*tan(b*x + a))^(5/2), x)

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maple [B]  time = 0.53, size = 965, normalized size = 8.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)/(d*tan(b*x+a))^(5/2),x)

[Out]

1/5/b*(6*cos(b*x+a)^3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/
2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3*co
s(b*x+a)^3*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos
(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+6*cos(b*x+a)^2*
((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/si
n(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3*cos(b*x+a)^2*((1-cos(b*x
+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(
1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-6*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*
((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*cos(b*x+a)*Elliptic
E(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+3*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+
a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*cos(b*x+a)*EllipticF(((1-cos(b*x
+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3*cos(b*x+a)^3*2^(1/2)-6*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-
1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*EllipticE(((1-cos(b*x+
a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+3*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))
/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a
))^(1/2),1/2*2^(1/2))-cos(b*x+a)^2*2^(1/2)+3*cos(b*x+a)*2^(1/2))/cos(b*x+a)^3/(d*sin(b*x+a)/cos(b*x+a))^(5/2)*
2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)/(d*tan(b*x + a))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)*(d*tan(a + b*x))^(5/2)),x)

[Out]

int(1/(sin(a + b*x)*(d*tan(a + b*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc {\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))**(5/2),x)

[Out]

Integral(csc(a + b*x)/(d*tan(a + b*x))**(5/2), x)

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