3.526 \(\int \frac{\cot ^5(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=167 \[ \frac{8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^4(e+f x)}{4 a f \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

-((8*a^2 + 24*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(8*a^(7/2)*f) + (8*a^2 + 24*a*b + 15*
b^2)/(8*a^3*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((8*a + 5*b)*Csc[e + f*x]^2)/(8*a^2*f*Sqrt[a + b*Sin[e + f*x]^2])
- Csc[e + f*x]^4/(4*a*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.162834, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3194, 89, 78, 51, 63, 208} \[ \frac{8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^4(e+f x)}{4 a f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-((8*a^2 + 24*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a]])/(8*a^(7/2)*f) + (8*a^2 + 24*a*b + 15*
b^2)/(8*a^3*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((8*a + 5*b)*Csc[e + f*x]^2)/(8*a^2*f*Sqrt[a + b*Sin[e + f*x]^2])
- Csc[e + f*x]^4/(4*a*f*Sqrt[a + b*Sin[e + f*x]^2])

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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 \frac{\cot ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^3 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{\csc ^4(e+f x)}{4 a f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (-8 a-5 b)+2 a x}{x^2 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 a f}\\ &=\frac{(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^4(e+f x)}{4 a f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2+24 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^2 f}\\ &=\frac{8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^4(e+f x)}{4 a f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2+24 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 a^3 f}\\ &=\frac{8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^4(e+f x)}{4 a f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2+24 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{8 a^3 b f}\\ &=-\frac{\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{8 a^2+24 a b+15 b^2}{8 a^3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{(8 a+5 b) \csc ^2(e+f x)}{8 a^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\csc ^4(e+f x)}{4 a f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.340294, size = 94, normalized size = 0.56 \[ \frac{\left (8 a^2+24 a b+15 b^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \sin ^2(e+f x)}{a}+1\right )+a \csc ^2(e+f x) \left (-2 a \csc ^2(e+f x)+8 a+5 b\right )}{8 a^3 f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(a*Csc[e + f*x]^2*(8*a + 5*b - 2*a*Csc[e + f*x]^2) + (8*a^2 + 24*a*b + 15*b^2)*Hypergeometric2F1[-1/2, 1, 1/2,
1 + (b*Sin[e + f*x]^2)/a])/(8*a^3*f*Sqrt[a + b*Sin[e + f*x]^2])

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Maple [A]  time = 1.669, size = 288, normalized size = 1.7 \begin{align*} -{\frac{1}{4\,af \left ( \sin \left ( fx+e \right ) \right ) ^{4}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{5\,b}{8\,{a}^{2}f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}}+{\frac{15\,{b}^{2}}{8\,f{a}^{3}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{15\,{b}^{2}}{8\,f}\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 ){a}^{-{\frac{7}{2}}}}+{\frac{1}{af}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}}-{\frac{1}{f}\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 ){a}^{-{\frac{3}{2}}}}+{\frac{1}{af \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}}+3\,{\frac{b}{{a}^{2}f\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}-3\,{\frac{b}{f{a}^{5/2}}\ln \left ({\frac{2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}{\sin \left ( fx+e \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

-1/4/f/a/sin(f*x+e)^4/(a+b*sin(f*x+e)^2)^(1/2)+5/8/f/a^2*b/sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2)+15/8/f/a^3*b^
2/(a+b*sin(f*x+e)^2)^(1/2)-15/8/f/a^(7/2)*b^2*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)^(1/2))/sin(f*x+e))+1/a/f/(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))+1/f/a/sin(f*x+e)^2/
(a+b*sin(f*x+e)^2)^(1/2)+3/f/a^2*b/(a+b*sin(f*x+e)^2)^(1/2)-3/f/a^(5/2)*b*ln((2*a+2*a^(1/2)*(a+b*sin(f*x+e)^2)
^(1/2))/sin(f*x+e))

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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)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.91502, size = 1539, normalized size = 9.22 \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)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(((8*a^2*b + 24*a*b^2 + 15*b^3)*cos(f*x + e)^6 - (8*a^3 + 48*a^2*b + 87*a*b^2 + 45*b^3)*cos(f*x + e)^4 -
8*a^3 - 32*a^2*b - 39*a*b^2 - 15*b^3 + (16*a^3 + 72*a^2*b + 102*a*b^2 + 45*b^3)*cos(f*x + e)^2)*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*((8*a^3 +
24*a^2*b + 15*a*b^2)*cos(f*x + e)^4 + 14*a^3 + 29*a^2*b + 15*a*b^2 - (24*a^3 + 53*a^2*b + 30*a*b^2)*cos(f*x +
e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/(a^4*b*f*cos(f*x + e)^6 - (a^5 + 3*a^4*b)*f*cos(f*x + e)^4 + (2*a^5 + 3
*a^4*b)*f*cos(f*x + e)^2 - (a^5 + a^4*b)*f), 1/8*(((8*a^2*b + 24*a*b^2 + 15*b^3)*cos(f*x + e)^6 - (8*a^3 + 48*
a^2*b + 87*a*b^2 + 45*b^3)*cos(f*x + e)^4 - 8*a^3 - 32*a^2*b - 39*a*b^2 - 15*b^3 + (16*a^3 + 72*a^2*b + 102*a*
b^2 + 45*b^3)*cos(f*x + e)^2)*sqrt(-a)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a)/a) - ((8*a^3 + 24*a^2*b
+ 15*a*b^2)*cos(f*x + e)^4 + 14*a^3 + 29*a^2*b + 15*a*b^2 - (24*a^3 + 53*a^2*b + 30*a*b^2)*cos(f*x + e)^2)*sq
rt(-b*cos(f*x + e)^2 + a + b))/(a^4*b*f*cos(f*x + e)^6 - (a^5 + 3*a^4*b)*f*cos(f*x + e)^4 + (2*a^5 + 3*a^4*b)*
f*cos(f*x + e)^2 - (a^5 + a^4*b)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{5}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral(cot(e + f*x)**5/(a + b*sin(e + f*x)**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{5}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^5/(b*sin(f*x + e)^2 + a)^(3/2), x)