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

Optimal. Leaf size=121 \[ \frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{f}+\frac{5 a \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 f}-\frac{\cot (e+f x) (a \sin (e+f x)+a)^{3/2}}{f} \]

[Out]

(-3*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/f + (11*a^2*Cos[e + f*x])/(3*f*Sqrt[a +
a*Sin[e + f*x]]) + (5*a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f) - (Cot[e + f*x]*(a + a*Sin[e + f*x])^(3/2
))/f

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Rubi [A]  time = 0.320568, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2716, 2976, 2981, 2773, 206} \[ \frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}-\frac{3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{f}+\frac{5 a \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{3 f}-\frac{\cot (e+f x) (a \sin (e+f x)+a)^{3/2}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/f + (11*a^2*Cos[e + f*x])/(3*f*Sqrt[a +
a*Sin[e + f*x]]) + (5*a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f) - (Cot[e + f*x]*(a + a*Sin[e + f*x])^(3/2
))/f

Rule 2716

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> -Simp[(a + b*Sin[e +
f*x])^m/(f*Tan[e + f*x]), x] + Dist[1/a, Int[((a + b*Sin[e + f*x])^m*(b*m - a*(m + 1)*Sin[e + f*x]))/Sin[e + f
*x], x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] &&  !LtQ[m, -1]

Rule 2976

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&
NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \cot ^2(e+f x) (a+a \sin (e+f x))^{3/2} \, dx &=-\frac{\cot (e+f x) (a+a \sin (e+f x))^{3/2}}{f}+\frac{\int \csc (e+f x) \left (\frac{3 a}{2}-\frac{5}{2} a \sin (e+f x)\right ) (a+a \sin (e+f x))^{3/2} \, dx}{a}\\ &=\frac{5 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{3/2}}{f}+\frac{2 \int \csc (e+f x) \sqrt{a+a \sin (e+f x)} \left (\frac{9 a^2}{4}-\frac{11}{4} a^2 \sin (e+f x)\right ) \, dx}{3 a}\\ &=\frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{5 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{3/2}}{f}+\frac{1}{2} (3 a) \int \csc (e+f x) \sqrt{a+a \sin (e+f x)} \, dx\\ &=\frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{5 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{3/2}}{f}-\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}\\ &=-\frac{3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}+\frac{11 a^2 \cos (e+f x)}{3 f \sqrt{a+a \sin (e+f x)}}+\frac{5 a \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{3 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{3/2}}{f}\\ \end{align*}

Mathematica [A]  time = 0.759996, size = 233, normalized size = 1.93 \[ -\frac{a \csc ^4\left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sin (e+f x)+1)} \left (-14 \sin \left (\frac{1}{2} (e+f x)\right )-9 \sin \left (\frac{3}{2} (e+f x)\right )-\sin \left (\frac{5}{2} (e+f x)\right )+14 \cos \left (\frac{1}{2} (e+f x)\right )-9 \cos \left (\frac{3}{2} (e+f x)\right )+\cos \left (\frac{5}{2} (e+f x)\right )+9 \sin (e+f x) \log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-9 \sin (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )\right )}{3 f \left (\cot \left (\frac{1}{2} (e+f x)\right )+1\right ) \left (\csc \left (\frac{1}{4} (e+f x)\right )-\sec \left (\frac{1}{4} (e+f x)\right )\right ) \left (\csc \left (\frac{1}{4} (e+f x)\right )+\sec \left (\frac{1}{4} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Csc[(e + f*x)/2]^4*Sqrt[a*(1 + Sin[e + f*x])]*(14*Cos[(e + f*x)/2] - 9*Cos[(3*(e + f*x))/2] + Cos[(5*(e +
f*x))/2] - 14*Sin[(e + f*x)/2] + 9*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[e + f*x] - 9*Log[1 - Cos[(
e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] - 9*Sin[(3*(e + f*x))/2] - Sin[(5*(e + f*x))/2]))/(3*f*(1 + Cot[(
e + f*x)/2])*(Csc[(e + f*x)/4] - Sec[(e + f*x)/4])*(Csc[(e + f*x)/4] + Sec[(e + f*x)/4]))

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Maple [A]  time = 0.642, size = 144, normalized size = 1.2 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{3\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) f}\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( \sin \left ( fx+e \right ) \left ( -12\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2}+2\,\sqrt{a} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}+9\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }}{\sqrt{a}}} \right ){a}^{2} \right ) +3\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{3/2} \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(sin(f*x+e)*(-12*(a-a*sin(f*x+e))^(1/2)*a^(3/2)+2*a^(1/2)*(a-a*
sin(f*x+e))^(3/2)+9*arctanh((a-a*sin(f*x+e))^(1/2)/a^(1/2))*a^2)+3*(a-a*sin(f*x+e))^(1/2)*a^(3/2))/sin(f*x+e)/
a^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.84636, size = 833, normalized size = 6.88 \begin{align*} \frac{9 \,{\left (a \cos \left (f x + e\right )^{2} -{\left (a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) - a\right )} \sqrt{a} \log \left (\frac{a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a} - 9 \, a \cos \left (f x + e\right ) +{\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) + 4 \,{\left (2 \, a \cos \left (f x + e\right )^{3} - 8 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) -{\left (2 \, a \cos \left (f x + e\right )^{2} + 10 \, a \cos \left (f x + e\right ) + 11 \, a\right )} \sin \left (f x + e\right ) + 11 \, a\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{12 \,{\left (f \cos \left (f x + e\right )^{2} -{\left (f \cos \left (f x + e\right ) + f\right )} \sin \left (f x + e\right ) - f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(9*(a*cos(f*x + e)^2 - (a*cos(f*x + e) + a)*sin(f*x + e) - a)*sqrt(a)*log((a*cos(f*x + e)^3 - 7*a*cos(f*x
+ e)^2 - 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)*sin(f*x + e) - 2*cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*s
qrt(a) - 9*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 8*a*cos(f*x + e) - a)*sin(f*x + e) - a)/(cos(f*x + e)^3 + cos(
f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e) - 1)) + 4*(2*a*cos(f*x + e)^3 - 8*a*cos(f*x + e)
^2 + a*cos(f*x + e) - (2*a*cos(f*x + e)^2 + 10*a*cos(f*x + e) + 11*a)*sin(f*x + e) + 11*a)*sqrt(a*sin(f*x + e)
+ a))/(f*cos(f*x + e)^2 - (f*cos(f*x + e) + f)*sin(f*x + e) - 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)**2*(a+a*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 2.63422, size = 671, normalized size = 5.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(18*a^2*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))*sgn(tan(1/2*
f*x + 1/2*e) + 1)/sqrt(-a) - 9*a^(3/2)*log(abs(-sqrt(a)*tan(1/2*f*x + 1/2*e) + sqrt(a*tan(1/2*f*x + 1/2*e)^2 +
a)))*sgn(tan(1/2*f*x + 1/2*e) + 1) + 6*a^(5/2)*sgn(tan(1/2*f*x + 1/2*e) + 1)/((sqrt(a)*tan(1/2*f*x + 1/2*e) -
sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - a) - (18*sqrt(2)*a^2*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - 9
*sqrt(2)*sqrt(-a)*a^(3/2)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 18*a^2*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a))
- 9*sqrt(-a)*a^(3/2)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 19*sqrt(2)*sqrt(-a)*a^(3/2) + 41*sqrt(-a)*a^(3/2))*sgn(
tan(1/2*f*x + 1/2*e) + 1)/(sqrt(2)*sqrt(-a) + sqrt(-a)) + (23*a^3*sgn(tan(1/2*f*x + 1/2*e) + 1) - (12*a^3*sgn(
tan(1/2*f*x + 1/2*e) + 1) - (18*a^3*sgn(tan(1/2*f*x + 1/2*e) + 1) + (3*a^3*sgn(tan(1/2*f*x + 1/2*e) + 1)*tan(1
/2*f*x + 1/2*e) - 20*a^3*sgn(tan(1/2*f*x + 1/2*e) + 1))*tan(1/2*f*x + 1/2*e))*tan(1/2*f*x + 1/2*e))*tan(1/2*f*
x + 1/2*e))/(a*tan(1/2*f*x + 1/2*e)^2 + a)^(3/2))/f