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

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

[Out]

-5*a^(5/2)*arctanh(cos(f*x+e)*a^(1/2)/(a+a*sin(f*x+e))^(1/2))/f+7/5*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-cot(
f*x+e)*(a+a*sin(f*x+e))^(5/2)/f+49/15*a^3*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)+31/15*a^2*cos(f*x+e)*(a+a*sin(f*
x+e))^(1/2)/f

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Rubi [A]  time = 0.43, antiderivative size = 151, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 1.26, 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]

-1/30*(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]))/(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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fricas [B]  time = 0.45, size = 363, normalized size = 2.40 \[ \frac {75 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - a^{2} - {\left (a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} - 9 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) + 4 \, {\left (6 \, a^{2} \cos \left (f x + e\right )^{4} + 28 \, a^{2} \cos \left (f x + e\right )^{3} - 40 \, a^{2} \cos \left (f x + e\right )^{2} - 13 \, a^{2} \cos \left (f x + e\right ) + 49 \, a^{2} + {\left (6 \, a^{2} \cos \left (f x + e\right )^{3} - 22 \, a^{2} \cos \left (f x + e\right )^{2} - 62 \, a^{2} \cos \left (f x + e\right ) - 49 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{60 \, {\left (f \cos \left (f x + e\right )^{2} - {\left (f \cos \left (f x + e\right ) + f\right )} \sin \left (f x + e\right ) - f\right )}} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [A]  time = 0.73, size = 162, normalized size = 1.07 \[ \frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sin \left (f x +e \right ) \left (90 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {5}{2}}-40 a^{\frac {3}{2}} \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}}+6 \sqrt {a}\, \left (a -a \sin \left (f x +e \right )\right )^{\frac {5}{2}}-75 \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}}{\sqrt {a}}\right ) a^{3}\right )-15 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {5}{2}}\right )}{15 \sin \left (f x +e \right ) \sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

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*(sin(f*x+e)-1))^(1/2)*(sin(f*x+e)*(90*(a-a*sin(f*x+e))^(1/2)*a^(5/2)-40*a^(3/2)*(a-a*s
in(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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \cot \left (f x + e\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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