3.1.75 \(\int \csc ^2(a+b x) (d \tan (a+b x))^{5/2} \, dx\) [75]

Optimal. Leaf size=20 \[ \frac {2 d (d \tan (a+b x))^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 20, 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 (d \tan (a+b x))^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(d*Tan[a + b*x])^(3/2))/(3*b)

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

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Mathematica [A]
time = 0.08, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 d (d \tan (a+b x))^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(d*Tan[a + b*x])^(3/2))/(3*b)

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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 = 0.32, size = 38, normalized size = 1.90

method	result	size
default	\(\frac {2 \cos \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}}}{3 b \sin \left (b x +a \right )}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 23, normalized size = 1.15 \begin {gather*} \frac {2 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}}}{3 \, 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))^(5/2),x, algorithm="maxima")

[Out]

2/3*(d*tan(b*x + a))^(5/2)/(b*tan(b*x + a))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (16) = 32\).
time = 0.40, size = 40, normalized size = 2.00 \begin {gather*} \frac {2 \, d^{2} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \sin \left (b x + a\right )}{3 \, b \cos \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))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.61, size = 24, normalized size = 1.20 \begin {gather*} \frac {2 \, \sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )}{3 \, 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))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.48, size = 56, normalized size = 2.80 \begin {gather*} \frac {2\,d^2\,\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}}}{3\,b\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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