3.76 \(\int \csc ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=41 \[ \frac {2 d (d \tan (a+b x))^{3/2}}{3 b}-\frac {2 d^3}{b \sqrt {d \tan (a+b x)}} \]

[Out]

-2*d^3/b/(d*tan(b*x+a))^(1/2)+2/3*d*(d*tan(b*x+a))^(3/2)/b

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Rubi [A]  time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2591, 14} \[ \frac {2 d (d \tan (a+b x))^{3/2}}{3 b}-\frac {2 d^3}{b \sqrt {d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^4*(d*Tan[a + b*x])^(5/2),x]

[Out]

(-2*d^3)/(b*Sqrt[d*Tan[a + b*x]]) + (2*d*(d*Tan[a + b*x])^(3/2))/(3*b)

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 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \csc ^4(a+b x) (d \tan (a+b x))^{5/2} \, dx &=\frac {d \operatorname {Subst}\left (\int \frac {d^2+x^2}{x^{3/2}} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=\frac {d \operatorname {Subst}\left (\int \left (\frac {d^2}{x^{3/2}}+\sqrt {x}\right ) \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac {2 d^3}{b \sqrt {d \tan (a+b x)}}+\frac {2 d (d \tan (a+b x))^{3/2}}{3 b}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 32, normalized size = 0.78 \[ -\frac {2 d \left (3 \cot ^2(a+b x)-1\right ) (d \tan (a+b x))^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^4*(d*Tan[a + b*x])^(5/2),x]

[Out]

(-2*d*(-1 + 3*Cot[a + b*x]^2)*(d*Tan[a + b*x])^(3/2))/(3*b)

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fricas [A]  time = 0.59, size = 58, normalized size = 1.41 \[ -\frac {2 \, {\left (4 \, d^{2} \cos \left (b x + a\right )^{2} - d^{2}\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{3 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^4*(d*tan(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

-2/3*(4*d^2*cos(b*x + a)^2 - d^2)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*cos(b*x + a)*sin(b*x + a))

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giac [A]  time = 0.54, size = 42, normalized size = 1.02 \[ \frac {2}{3} \, d^{2} {\left (\frac {\sqrt {d \tan \left (b x + a\right )} \tan \left (b x + a\right )}{b} - \frac {3 \, d}{\sqrt {d \tan \left (b x + a\right )} b}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^4*(d*tan(b*x+a))^(5/2),x, algorithm="giac")

[Out]

2/3*d^2*(sqrt(d*tan(b*x + a))*tan(b*x + a)/b - 3*d/(sqrt(d*tan(b*x + a))*b))

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maple [A]  time = 0.50, size = 50, normalized size = 1.22 \[ -\frac {2 \left (4 \left (\cos ^{2}\left (b x +a \right )\right )-1\right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \cos \left (b x +a \right )}{3 b \sin \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^4*(d*tan(b*x+a))^(5/2),x)

[Out]

-2/3/b*(4*cos(b*x+a)^2-1)*(d*sin(b*x+a)/cos(b*x+a))^(5/2)*cos(b*x+a)/sin(b*x+a)^3

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maxima [A]  time = 0.57, size = 36, normalized size = 0.88 \[ -\frac {2 \, d^{3} {\left (\frac {3}{\sqrt {d \tan \left (b x + a\right )}} - \frac {\left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}}{d^{2}}\right )}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^4*(d*tan(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

-2/3*d^3*(3/sqrt(d*tan(b*x + a)) - (d*tan(b*x + a))^(3/2)/d^2)/b

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mupad [B]  time = 2.72, size = 64, normalized size = 1.56 \[ -\frac {4\,d^2\,\left (\sin \left (2\,a+2\,b\,x\right )+\sin \left (4\,a+4\,b\,x\right )\right )\,\sqrt {\frac {d\,\sin \left (2\,a+2\,b\,x\right )}{\cos \left (2\,a+2\,b\,x\right )+1}}}{3\,b\,{\sin \left (2\,a+2\,b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(5/2)/sin(a + b*x)^4,x)

[Out]

-(4*d^2*(sin(2*a + 2*b*x) + sin(4*a + 4*b*x))*((d*sin(2*a + 2*b*x))/(cos(2*a + 2*b*x) + 1))^(1/2))/(3*b*sin(2*
a + 2*b*x)^2)

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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(csc(b*x+a)**4*(d*tan(b*x+a))**(5/2),x)

[Out]

Timed out

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