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]
(-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]])
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Rubi [A] time = 0.136172, antiderivative size = 110, normalized size of antiderivative = 1.,
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 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 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 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*}
Mathematica [C] time = 1.76854, 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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Maple [B] time = 0.164, size = 980, normalized size = 8.9 \begin{align*} \text{result too large to display} \end{align*}
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*2^(1/2)*(3*cos(b*x+a)^3*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*(-(cos(b*x
+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1)/sin(b*x+a))^(
1/2)-6*cos(b*x+a)^3*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*(-(cos(b*x+a)-1-sin(b
*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)+3*cos(
b*x+a)^2*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*(-(cos(b*x+a)-1-sin(b*x+a))/sin(
b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)-6*cos(b*x+a)^2*El
lipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/
2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)-3*cos(b*x+a)*EllipticF((-(co
s(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x
+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+6*cos(b*x+a)*EllipticE((-(cos(b*x+a)-1-si
n(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+
a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+3*cos(b*x+a)^3*2^(1/2)-3*EllipticF((-(cos(b*x+a)-1-sin
(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a
))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)+6*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/
2),1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1
-sin(b*x+a))/sin(b*x+a))^(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)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}
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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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(csc(b*x+a)/(d*tan(b*x+a))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )}{\left (d \tan \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
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)