3.66 \(\int \csc ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=18 \[ \frac {2 d \sqrt {d \tan (a+b x)}}{b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2591, 30} \[ \frac {2 d \sqrt {d \tan (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*(d*Tan[a + b*x])^(3/2),x]

[Out]

(2*d*Sqrt[d*Tan[a + b*x]])/b

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 ^2(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {d \operatorname {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=\frac {2 d \sqrt {d \tan (a+b x)}}{b}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 1.00 \[ \frac {2 d \sqrt {d \tan (a+b x)}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*(d*Tan[a + b*x])^(3/2),x]

[Out]

(2*d*Sqrt[d*Tan[a + b*x]])/b

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fricas [A]  time = 0.43, size = 24, normalized size = 1.33 \[ \frac {2 \, d \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

2*d*sqrt(d*sin(b*x + a)/cos(b*x + a))/b

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giac [A]  time = 1.37, size = 16, normalized size = 0.89 \[ \frac {2 \, \sqrt {d \tan \left (b x + a\right )} d}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

2*sqrt(d*tan(b*x + a))*d/b

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maple [B]  time = 0.50, size = 58, normalized size = 3.22 \[ \frac {2 \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \cos \left (b x +a \right ) \left (-1+\cos \left (b x +a \right )\right )^{2} \left (\cos \left (b x +a \right )+1\right )^{2}}{b \sin \left (b x +a \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*(d*tan(b*x+a))^(3/2),x)

[Out]

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

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maxima [A]  time = 0.68, size = 23, normalized size = 1.28 \[ \frac {2 \, \left (d \tan \left (b x + a\right )\right )^{\frac {3}{2}}}{b \tan \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.77, size = 43, normalized size = 2.39 \[ \frac {2\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(3/2)/sin(a + b*x)^2,x)

[Out]

(2*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/b

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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)**2*(d*tan(b*x+a))**(3/2),x)

[Out]

Timed out

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