3.2.13 \(\int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\) [113]
Optimal. Leaf size=140 \[ \frac {2 \csc (a+b x)}{15 b d (d \tan (a+b x))^{3/2}}-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}+\frac {4 \cos (a+b x)}{15 b d^2 \sqrt {d \tan (a+b x)}}+\frac {4 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{15 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}} \]
[Out]
4/15*cos(b*x+a)/b/d^2/(d*tan(b*x+a))^(1/2)-4/15*(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)+2/15*csc(b*x+a)/b/d/(d*tan(b*x
+a))^(3/2)-2/9*csc(b*x+a)^3/b/d/(d*tan(b*x+a))^(3/2)
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Rubi [A]
time = 0.13, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2677, 2679,
2681, 2650, 2652, 2719} \begin {gather*} \frac {4 \cos (a+b x)}{15 b d^2 \sqrt {d \tan (a+b x)}}+\frac {4 \sin (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{15 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}+\frac {2 \csc (a+b x)}{15 b d (d \tan (a+b x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Csc[a + b*x]^3/(d*Tan[a + b*x])^(5/2),x]
[Out]
(2*Csc[a + b*x])/(15*b*d*(d*Tan[a + b*x])^(3/2)) - (2*Csc[a + b*x]^3)/(9*b*d*(d*Tan[a + b*x])^(3/2)) + (4*Cos[
a + b*x])/(15*b*d^2*Sqrt[d*Tan[a + b*x]]) + (4*EllipticE[a - Pi/4 + b*x, 2]*Sin[a + b*x])/(15*b*d^2*Sqrt[Sin[2
*a + 2*b*x]]*Sqrt[d*Tan[a + b*x]])
Rule 2650
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 2652
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 2677
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 2679
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[b*(a*Sin[e +
f*x])^(m + 2)*((b*Tan[e + f*x])^(n - 1)/(a^2*f*(m + n + 1))), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& IntegersQ[2*m, 2*n]
Rule 2681
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 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]
Rubi steps
\begin {align*} \int \frac {\csc ^3(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}-\frac {\int \frac {\csc ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx}{3 d^2}\\ &=\frac {2 \csc (a+b x)}{15 b d (d \tan (a+b x))^{3/2}}-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}-\frac {2 \int \frac {\csc (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx}{15 d^2}\\ &=\frac {2 \csc (a+b x)}{15 b d (d \tan (a+b x))^{3/2}}-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}-\frac {\left (2 \sqrt {\sin (a+b x)}\right ) \int \frac {\sqrt {\cos (a+b x)}}{\sin ^{\frac {3}{2}}(a+b x)} \, dx}{15 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {2 \csc (a+b x)}{15 b d (d \tan (a+b x))^{3/2}}-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}+\frac {4 \cos (a+b x)}{15 b d^2 \sqrt {d \tan (a+b x)}}+\frac {\left (4 \sqrt {\sin (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{15 d^2 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {2 \csc (a+b x)}{15 b d (d \tan (a+b x))^{3/2}}-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}+\frac {4 \cos (a+b x)}{15 b d^2 \sqrt {d \tan (a+b x)}}+\frac {(4 \sin (a+b x)) \int \sqrt {\sin (2 a+2 b x)} \, dx}{15 d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ &=\frac {2 \csc (a+b x)}{15 b d (d \tan (a+b x))^{3/2}}-\frac {2 \csc ^3(a+b x)}{9 b d (d \tan (a+b x))^{3/2}}+\frac {4 \cos (a+b x)}{15 b d^2 \sqrt {d \tan (a+b x)}}+\frac {4 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sin (a+b x)}{15 b d^2 \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in
optimal.
time = 0.90, size = 116, normalized size = 0.83 \begin {gather*} \frac {2 \left (4 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right ) \sec ^2(a+b x)+\left (-6+3 \csc ^2(a+b x)+8 \csc ^4(a+b x)-5 \csc ^6(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sin (a+b x) \sqrt {d \tan (a+b x)}}{45 b d^3 \sqrt {\sec ^2(a+b x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Csc[a + b*x]^3/(d*Tan[a + b*x])^(5/2),x]
[Out]
(2*(4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x]^2 + (-6 + 3*Csc[a + b*x]^2 + 8*Csc[a + b*
x]^4 - 5*Csc[a + b*x]^6)*Sqrt[Sec[a + b*x]^2])*Sin[a + b*x]*Sqrt[d*Tan[a + b*x]])/(45*b*d^3*Sqrt[Sec[a + b*x]^
2])
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1454\) vs.
\(2(147)=294\).
time = 0.40, size = 1455, normalized size = 10.39
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result |
size |
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\(\text {Expression too large to display}\) |
\(1455\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(b*x+a)^3/(d*tan(b*x+a))^(5/2),x,method=_RETURNVERBOSE)
[Out]
1/45/b*(6*((1-cos(b*x+a)+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)*((-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))*cos(b*x+a)^5-12*(
(1-cos(b*x+a)+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)*((-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))*cos(b*x+a)^5+6*((1-cos(b*x+a
)+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)*((-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))*cos(b*x+a)^4-12*((1-cos(b*x+a)+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)*((-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))*cos(b*x+a)^4-12*((1-cos(b*x+a)+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)*((-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))*cos(b*x+a)^3+24*((1-cos(b*x+a)+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)*((-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))*cos(b*x+a)^3-12*((1-cos(b*x+a)+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)*((-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))*cos(b*x+a)^2+24*((1-cos(b*x+a)+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)*((-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))*cos(b*x+a)^2+6*2^(1/2)*cos(b*x+a)^5+6*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))*((-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)*((cos(b*
x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-12*cos(b*x+a)*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*
2^(1/2))*((-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)*((cos(b*x+a)-1+sin(b*
x+a))/sin(b*x+a))^(1/2)-3*cos(b*x+a)^4*2^(1/2)+6*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^
(1/2))*((-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)*((cos(b*x+a)-1+sin(b*x+
a))/sin(b*x+a))^(1/2)-12*EllipticE(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-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)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-12
*cos(b*x+a)^3*2^(1/2)-2*cos(b*x+a)^2*2^(1/2)+6*cos(b*x+a)*2^(1/2))/cos(b*x+a)^3/sin(b*x+a)^2/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)^3/(d*tan(b*x+a))^(5/2),x, algorithm="maxima")
[Out]
integrate(csc(b*x + a)^3/(d*tan(b*x + a))^(5/2), x)
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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)^3/(d*tan(b*x+a))^(5/2),x, algorithm="fricas")
[Out]
Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{3}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)**3/(d*tan(b*x+a))**(5/2),x)
[Out]
Integral(csc(a + b*x)**3/(d*tan(a + b*x))**(5/2), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)^3/(d*tan(b*x+a))^(5/2),x, algorithm="giac")
[Out]
integrate(csc(b*x + a)^3/(d*tan(b*x + a))^(5/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\sin \left (a+b\,x\right )}^3\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(sin(a + b*x)^3*(d*tan(a + b*x))^(5/2)),x)
[Out]
int(1/(sin(a + b*x)^3*(d*tan(a + b*x))^(5/2)), x)
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