3.127 \(\int \frac{\csc ^6(e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=123 \[ -\frac{\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac{2 (5 a-2 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f} \]

[Out]

-((15*a^2 - 20*a*b + 8*b^2)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*f) - (2*(5*a - 2*b)*Cot[e + f*x]^
3*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^2*f) - (Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a*f)

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Rubi [A]  time = 0.136504, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3663, 462, 453, 264} \[ -\frac{\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac{2 (5 a-2 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-((15*a^2 - 20*a*b + 8*b^2)*Cot[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^3*f) - (2*(5*a - 2*b)*Cot[e + f*x]^
3*Sqrt[a + b*Tan[e + f*x]^2])/(15*a^2*f) - (Cot[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2])/(5*a*f)

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 453

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\csc ^6(e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{2 (5 a-2 b)+5 a x^2}{x^4 \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac{2 (5 a-2 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}+\frac{\left (15 a^2-20 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac{\left (15 a^2-20 a b+8 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^3 f}-\frac{2 (5 a-2 b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{15 a^2 f}-\frac{\cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 1.81809, size = 90, normalized size = 0.73 \[ -\frac{\cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+4 a (a-b) \csc ^2(e+f x)+8 (a-b)^2\right ) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}{15 \sqrt{2} a^3 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-(Cot[e + f*x]*(8*(a - b)^2 + 4*a*(a - b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4)*Sqrt[(a + b + (a - b)*Cos[2*(
e + f*x)])*Sec[e + f*x]^2])/(15*Sqrt[2]*a^3*f)

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Maple [A]  time = 0.214, size = 148, normalized size = 1.2 \begin{align*} -{\frac{ \left ( 8\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}ab+8\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{2}-20\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}+36\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab-16\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{2}+15\,{a}^{2}-20\,ab+8\,{b}^{2} \right ) \cos \left ( fx+e \right ) }{15\,f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}}\sqrt{{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

-1/15/f/a^3*(8*cos(f*x+e)^4*a^2-16*cos(f*x+e)^4*a*b+8*cos(f*x+e)^4*b^2-20*cos(f*x+e)^2*a^2+36*cos(f*x+e)^2*a*b
-16*cos(f*x+e)^2*b^2+15*a^2-20*a*b+8*b^2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*cos(f*x+e)/si
n(f*x+e)^5

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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(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.82206, size = 339, normalized size = 2.76 \begin{align*} -\frac{{\left (8 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \,{\left (5 \, a^{2} - 9 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} - 20 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{4} - 2 \, a^{3} f \cos \left (f x + e\right )^{2} + a^{3} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/15*(8*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 4*(5*a^2 - 9*a*b + 4*b^2)*cos(f*x + e)^3 + (15*a^2 - 20*a*b + 8*
b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a^3*f*cos(f*x + e)^4 - 2*a^3*f*cos(f*x
+ e)^2 + a^3*f)*sin(f*x + e))

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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(csc(f*x+e)**6/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{6}}{\sqrt{b \tan \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^6/sqrt(b*tan(f*x + e)^2 + a), x)