3.6.2 \(\int \cot (e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\) [502]

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

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Rubi [A]
time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, 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 = {3273, 52, 65, 214} \begin {gather*} -\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 {gather*}

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 52

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 65

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 214

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

Rule 3273

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]

Rubi steps

\begin {align*} \int \cot (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac {\text {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 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.09, size = 69, normalized size = 0.88 \begin {gather*} \frac {-3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a+b \sin ^2(e+f x)} \left (4 a+b \sin ^2(e+f x)\right )}{3 f} \end {gather*}

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 = 9.76, size = 86, normalized size = 1.10

method	result	size
default	\(\frac {\frac {b \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}+\frac {4 a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{3}-a^{\frac {3}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{\sin \left (f x +e \right )}\right )}{f}\)	\(86\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.31, size = 64, normalized size = 0.82 \begin {gather*} -\frac {3 \, a^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | \sin \left (f x + e\right ) \right |}}\right ) - {\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 {gather*}

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]

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

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Fricas [A]
time = 0.87, size = 175, normalized size = 2.24 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot {\left (e + f x \right )}\, dx \end {gather*}

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]

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

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Giac [A]
time = 0.43, size = 71, normalized size = 0.91 \begin {gather*} \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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {cot}\left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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