3.502 \(\int \cot (e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=78 \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{f}+\frac{a \sqrt{a+b \sin ^2(e+f x)}}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f} \]

[Out]

-((a^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/f) + (a*Sqrt[a + b*Sin[e + f*x]^2])/f + (a + b*Sin[e +
f*x]^2)^(3/2)/(3*f)

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Rubi [A]  time = 0.0791743, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3194, 50, 63, 208} \[ -\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{f}+\frac{a \sqrt{a+b \sin ^2(e+f x)}}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-((a^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/f) + (a*Sqrt[a + b*Sin[e + f*x]^2])/f + (a + b*Sin[e +
f*x]^2)^(3/2)/(3*f)

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{a \sqrt{a+b \sin ^2(e+f x)}}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{a \sqrt{a+b \sin ^2(e+f x)}}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{b f}\\ &=-\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{f}+\frac{a \sqrt{a+b \sin ^2(e+f x)}}{f}+\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.127664, size = 69, normalized size = 0.88 \[ \frac{\sqrt{a+b \sin ^2(e+f x)} \left (4 a+b \sin ^2(e+f x)\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(-3*a^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + Sqrt[a + b*Sin[e + f*x]^2]*(4*a + b*Sin[e + f*x]^2))
/(3*f)

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Maple [A]  time = 1.359, size = 91, normalized size = 1.2 \begin{align*}{\frac{b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3\,f}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}+{\frac{4\,a}{3\,f}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}-{\frac{1}{f}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/3/f*b*sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)+4/3*a*(a+b*sin(f*x+e)^2)^(1/2)/f-1/f*a^(3/2)*ln((2*a+2*a^(1/2)*(
a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.37492, size = 439, normalized size = 5.63 \begin{align*} \left [\frac{3 \, a^{\frac{3}{2}} \log \left (\frac{2 \,{\left (b \cos \left (f x + e\right )^{2} + 2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a} - 2 \, a - b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \,{\left (b \cos \left (f x + e\right )^{2} - 4 \, a - b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f}, \frac{3 \, \sqrt{-a} a \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a}}{a}\right ) -{\left (b \cos \left (f x + e\right )^{2} - 4 \, a - b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/6*(3*a^(3/2)*log(2*(b*cos(f*x + e)^2 + 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a) - 2*a - b)/(cos(f*x + e)^2
- 1)) - 2*(b*cos(f*x + e)^2 - 4*a - b)*sqrt(-b*cos(f*x + e)^2 + a + b))/f, 1/3*(3*sqrt(-a)*a*arctan(sqrt(-b*c
os(f*x + e)^2 + a + b)*sqrt(-a)/a) - (b*cos(f*x + e)^2 - 4*a - b)*sqrt(-b*cos(f*x + e)^2 + a + b))/f]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11539, size = 96, normalized size = 1.23 \begin{align*} \frac{\frac{3 \, a^{2} \arctan \left (\frac{\sqrt{b \sin \left (f x + e\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} +{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} + 3 \, \sqrt{b \sin \left (f x + e\right )^{2} + a} a}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

1/3*(3*a^2*arctan(sqrt(b*sin(f*x + e)^2 + a)/sqrt(-a))/sqrt(-a) + (b*sin(f*x + e)^2 + a)^(3/2) + 3*sqrt(b*sin(
f*x + e)^2 + a)*a)/f