3.1.56 \(\int \csc ^2(a+b x) \sqrt {d \tan (a+b x)} \, dx\) [56]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {2671, 30} \begin {gather*} -\frac {2 d}{b \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sqrt[d*Tan[a + b*x]],x]

[Out]

(-2*d)/(b*Sqrt[d*Tan[a + b*x]])

Rule 30

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

Rule 2671

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]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

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

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Mathematica [A]
time = 0.08, size = 18, normalized size = 1.00 \begin {gather*} -\frac {2 d}{b \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sqrt[d*Tan[a + b*x]],x]

[Out]

(-2*d)/(b*Sqrt[d*Tan[a + b*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(37\) vs. \(2(16)=32\).
time = 0.36, size = 38, normalized size = 2.11

method	result	size
default	\(-\frac {2 \sqrt {\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}}\, \cos \left (b x +a \right )}{b \sin \left (b x +a \right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.34, size = 23, normalized size = 1.28 \begin {gather*} -\frac {2 \, \sqrt {d \tan \left (b x + a\right )}}{b \tan \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*(d*tan(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (16) = 32\).
time = 0.37, size = 37, normalized size = 2.06 \begin {gather*} -\frac {2 \, \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )}{b \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*(d*tan(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d \tan {\left (a + b x \right )}} \csc ^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**2*(d*tan(b*x+a))**(1/2),x)

[Out]

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

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Giac [A]
time = 0.71, size = 16, normalized size = 0.89 \begin {gather*} -\frac {2 \, d}{\sqrt {d \tan \left (b x + a\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*(d*tan(b*x+a))^(1/2),x, algorithm="giac")

[Out]

-2*d/(sqrt(d*tan(b*x + a))*b)

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Mupad [B]
time = 3.05, size = 48, normalized size = 2.67 \begin {gather*} -\frac {\sin \left (2\,a+2\,b\,x\right )\,\sqrt {\frac {d\,\sin \left (2\,a+2\,b\,x\right )}{\cos \left (2\,a+2\,b\,x\right )+1}}}{b\,{\sin \left (a+b\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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