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

Optimal. Leaf size=140 \[ -\frac{(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac{(2 a-3 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}+\frac{\sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 f}-\frac{\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f} \]

[Out]

(Sqrt[a]*(2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*f) - ((2*a - 3*b)*Sqrt[a + b*Sin[e + f*x]
^2])/(2*f) - ((2*a - 3*b)*(a + b*Sin[e + f*x]^2)^(3/2))/(6*a*f) - (Csc[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2)
)/(2*a*f)

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Rubi [A]  time = 0.127071, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3194, 78, 50, 63, 208} \[ -\frac{(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac{(2 a-3 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}+\frac{\sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 f}-\frac{\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*(2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(2*f) - ((2*a - 3*b)*Sqrt[a + b*Sin[e + f*x]
^2])/(2*f) - ((2*a - 3*b)*(a + b*Sin[e + f*x]^2)^(3/2))/(6*a*f) - (Csc[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2)
)/(2*a*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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1-x) (a+b x)^{3/2}}{x^2} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac{(2 a-3 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=-\frac{(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac{\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac{(2 a-3 b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\sin ^2(e+f x)\right )}{4 f}\\ &=-\frac{(2 a-3 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}-\frac{(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac{\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac{(a (2 a-3 b)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 f}\\ &=-\frac{(2 a-3 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}-\frac{(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac{\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}-\frac{(a (2 a-3 b)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{2 b f}\\ &=\frac{\sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{2 f}-\frac{(2 a-3 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}-\frac{(2 a-3 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 a f}-\frac{\csc ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 a f}\\ \end{align*}

Mathematica [A]  time = 0.481449, size = 90, normalized size = 0.64 \[ \frac{3 \sqrt{a} (2 a-3 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )+\sqrt{a+b \sin ^2(e+f x)} \left (-3 a \csc ^2(e+f x)-8 a+b \cos (2 (e+f x))+5 b\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[a]*(2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]] + (-8*a + 5*b + b*Cos[2*(e + f*x)] - 3*a*Cs
c[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])/(6*f)

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Maple [A]  time = 1.432, size = 179, normalized size = 1.3 \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{b}{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 ) }-{\frac{3\,b}{2\,f}\sqrt{a}\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 ) }-{\frac{a}{2\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.40473, size = 699, normalized size = 4.99 \begin{align*} \left [-\frac{3 \,{\left ({\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + 3 \, b\right )} \sqrt{a} \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 (2 \, b \cos \left (f x + e\right )^{4} - 2 \,{\left (4 \, a - b\right )} \cos \left (f x + e\right )^{2} + 11 \, a - 4 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{12 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )}}, -\frac{3 \,{\left ({\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + 3 \, b\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a}}{a}\right ) -{\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \,{\left (4 \, a - b\right )} \cos \left (f x + e\right )^{2} + 11 \, a - 4 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{6 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*((2*a - 3*b)*cos(f*x + e)^2 - 2*a + 3*b)*sqrt(a)*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*(2*b*cos(f*x + e)^4 - 2*(4*a - b)*cos(f*x + e)^2 + 11*a
- 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2 - f), -1/6*(3*((2*a - 3*b)*cos(f*x + e)^2 - 2*a + 3*
b)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - (2*b*cos(f*x + e)^4 - 2*(4*a - b)*cos(f*x + e
)^2 + 11*a - 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2 - 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)**3*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \cot \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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