3.101 \(\int \cot ^2(e+f x) (a+a \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=151 \[ \frac{49 a^3 \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}+\frac{31 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{5 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{f}+\frac{7 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}-\frac{\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f} \]

[Out]

(-5*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/f + (49*a^3*Cos[e + f*x])/(15*f*Sqrt[a +
a*Sin[e + f*x]]) + (31*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f) + (7*a*Cos[e + f*x]*(a + a*Sin[e + f
*x])^(3/2))/(5*f) - (Cot[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/f

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Rubi [A]  time = 0.428696, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{49 a^3 \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}+\frac{31 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{5 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{f}+\frac{7 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}-\frac{\cot (e+f x) (a \sin (e+f x)+a)^{5/2}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(-5*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/Sqrt[a + a*Sin[e + f*x]]])/f + (49*a^3*Cos[e + f*x])/(15*f*Sqrt[a +
a*Sin[e + f*x]]) + (31*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f) + (7*a*Cos[e + f*x]*(a + a*Sin[e + f
*x])^(3/2))/(5*f) - (Cot[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/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))^{5/2} \, dx &=-\frac{\cot (e+f x) (a+a \sin (e+f x))^{5/2}}{f}+\frac{\int \csc (e+f x) \left (\frac{5 a}{2}-\frac{7}{2} a \sin (e+f x)\right ) (a+a \sin (e+f x))^{5/2} \, dx}{a}\\ &=\frac{7 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{5/2}}{f}+\frac{2 \int \csc (e+f x) (a+a \sin (e+f x))^{3/2} \left (\frac{25 a^2}{4}-\frac{31}{4} a^2 \sin (e+f x)\right ) \, dx}{5 a}\\ &=\frac{31 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}+\frac{7 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{5/2}}{f}+\frac{4 \int \csc (e+f x) \sqrt{a+a \sin (e+f x)} \left (\frac{75 a^3}{8}-\frac{49}{8} a^3 \sin (e+f x)\right ) \, dx}{15 a}\\ &=\frac{49 a^3 \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}+\frac{31 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}+\frac{7 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{5/2}}{f}+\frac{1}{2} \left (5 a^2\right ) \int \csc (e+f x) \sqrt{a+a \sin (e+f x)} \, dx\\ &=\frac{49 a^3 \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}+\frac{31 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}+\frac{7 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{5/2}}{f}-\frac{\left (5 a^3\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{5 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}+\frac{49 a^3 \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}+\frac{31 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}+\frac{7 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac{\cot (e+f x) (a+a \sin (e+f x))^{5/2}}{f}\\ \end{align*}

Mathematica [A]  time = 1.19308, size = 261, normalized size = 1.73 \[ -\frac{a^2 \csc ^4\left (\frac{1}{2} (e+f x)\right ) \sqrt{a (\sin (e+f x)+1)} \left (-125 \sin \left (\frac{1}{2} (e+f x)\right )-93 \sin \left (\frac{3}{2} (e+f x)\right )-25 \sin \left (\frac{5}{2} (e+f x)\right )+3 \sin \left (\frac{7}{2} (e+f x)\right )+125 \cos \left (\frac{1}{2} (e+f x)\right )-93 \cos \left (\frac{3}{2} (e+f x)\right )+25 \cos \left (\frac{5}{2} (e+f x)\right )+3 \cos \left (\frac{7}{2} (e+f x)\right )+150 \sin (e+f x) \log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-150 \sin (e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )\right )}{30 f \left (\cot \left (\frac{1}{2} (e+f x)\right )+1\right ) \left (\csc \left (\frac{1}{4} (e+f x)\right )-\sec \left (\frac{1}{4} (e+f x)\right )\right ) \left (\csc \left (\frac{1}{4} (e+f x)\right )+\sec \left (\frac{1}{4} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2*(a + a*Sin[e + f*x])^(5/2),x]

[Out]

-(a^2*Csc[(e + f*x)/2]^4*Sqrt[a*(1 + Sin[e + f*x])]*(125*Cos[(e + f*x)/2] - 93*Cos[(3*(e + f*x))/2] + 25*Cos[(
5*(e + f*x))/2] + 3*Cos[(7*(e + f*x))/2] - 125*Sin[(e + f*x)/2] + 150*Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)
/2]]*Sin[e + f*x] - 150*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sin[e + f*x] - 93*Sin[(3*(e + f*x))/2] -
25*Sin[(5*(e + f*x))/2] + 3*Sin[(7*(e + f*x))/2]))/(30*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 = 0.733, size = 162, normalized size = 1.1 \begin{align*}{\frac{1+\sin \left ( fx+e \right ) }{15\,\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 ( 90\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{5/2}-40\,{a}^{3/2} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}+6\,\sqrt{a} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{5/2}-75\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }}{\sqrt{a}}} \right ){a}^{3} \right ) -15\,\sqrt{a-a\sin \left ( fx+e \right ) }{a}^{5/2} \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))^(5/2),x)

[Out]

1/15*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(sin(f*x+e)*(90*(a-a*sin(f*x+e))^(1/2)*a^(5/2)-40*a^(3/2)*(a-a*
sin(f*x+e))^(3/2)+6*a^(1/2)*(a-a*sin(f*x+e))^(5/2)-75*arctanh((a-a*sin(f*x+e))^(1/2)/a^(1/2))*a^3)-15*(a-a*sin
(f*x+e))^(1/2)*a^(5/2))/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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*cot(f*x + e)^2, x)

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Fricas [B]  time = 1.57082, size = 934, normalized size = 6.19 \begin{align*} \frac{75 \,{\left (a^{2} \cos \left (f x + e\right )^{2} - a^{2} -{\left (a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sin \left (f x + e\right )\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 (6 \, a^{2} \cos \left (f x + e\right )^{4} + 28 \, a^{2} \cos \left (f x + e\right )^{3} - 40 \, a^{2} \cos \left (f x + e\right )^{2} - 13 \, a^{2} \cos \left (f x + e\right ) + 49 \, a^{2} +{\left (6 \, a^{2} \cos \left (f x + e\right )^{3} - 22 \, a^{2} \cos \left (f x + e\right )^{2} - 62 \, a^{2} \cos \left (f x + e\right ) - 49 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{60 \,{\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))^(5/2),x, algorithm="fricas")

[Out]

1/60*(75*(a^2*cos(f*x + e)^2 - a^2 - (a^2*cos(f*x + e) + a^2)*sin(f*x + e))*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)*sqrt(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*(6*a^2*cos(f*x + e)^4 + 28*a^
2*cos(f*x + e)^3 - 40*a^2*cos(f*x + e)^2 - 13*a^2*cos(f*x + e) + 49*a^2 + (6*a^2*cos(f*x + e)^3 - 22*a^2*cos(f
*x + e)^2 - 62*a^2*cos(f*x + e) - 49*a^2)*sin(f*x + e))*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))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 2.90585, size = 714, normalized size = 4.73 \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))^(5/2),x, algorithm="giac")

[Out]

1/30*(150*a^3*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) - 75*a^(5/2)*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) + 30*a^(7/2)*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) - (150*sqrt(2)*a^3*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)
) - 75*sqrt(2)*sqrt(-a)*a^(5/2)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 150*a^3*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sq
rt(-a)) - 75*sqrt(-a)*a^(5/2)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 83*sqrt(2)*sqrt(-a)*a^(5/2) + 181*sqrt(-a)*a^(5
/2))*sgn(tan(1/2*f*x + 1/2*e) + 1)/(sqrt(2)*sqrt(-a) + sqrt(-a)) + (127*a^5*sgn(tan(1/2*f*x + 1/2*e) + 1) + (2
05*a^5*sgn(tan(1/2*f*x + 1/2*e) + 1) - (160*a^5*sgn(tan(1/2*f*x + 1/2*e) + 1) - (45*a^5*sgn(tan(1/2*f*x + 1/2*
e) + 1) + (15*a^5*sgn(tan(1/2*f*x + 1/2*e) + 1)*tan(1/2*f*x + 1/2*e) - 112*a^5*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))*tan(1/2*f*x + 1/2*e)^2)/(a*tan(1/2*f*x + 1/2
*e)^2 + a)^(5/2))/f