3.118 \(\int \frac{\csc ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=183 \[ -\frac{2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 f (a+b)^4 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 f (a+b)^3 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{\cot ^5(e+f x)}{5 f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{2 (5 a+2 b) \cot ^3(e+f x)}{15 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}} \]

[Out]

-((15*a^2 - 10*a*b - b^2)*Cot[e + f*x])/(15*(a + b)^3*f*Sqrt[a + b + b*Tan[e + f*x]^2]) - (2*(5*a + 2*b)*Cot[e
+ f*x]^3)/(15*(a + b)^2*f*Sqrt[a + b + b*Tan[e + f*x]^2]) - Cot[e + f*x]^5/(5*(a + b)*f*Sqrt[a + b + b*Tan[e
+ f*x]^2]) - (2*b*(15*a^2 - 10*a*b - b^2)*Tan[e + f*x])/(15*(a + b)^4*f*Sqrt[a + b + b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.181575, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4132, 462, 453, 271, 191} \[ -\frac{2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 f (a+b)^4 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 f (a+b)^3 \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{\cot ^5(e+f x)}{5 f (a+b) \sqrt{a+b \tan ^2(e+f x)+b}}-\frac{2 (5 a+2 b) \cot ^3(e+f x)}{15 f (a+b)^2 \sqrt{a+b \tan ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-((15*a^2 - 10*a*b - b^2)*Cot[e + f*x])/(15*(a + b)^3*f*Sqrt[a + b + b*Tan[e + f*x]^2]) - (2*(5*a + 2*b)*Cot[e
+ f*x]^3)/(15*(a + b)^2*f*Sqrt[a + b + b*Tan[e + f*x]^2]) - Cot[e + f*x]^5/(5*(a + b)*f*Sqrt[a + b + b*Tan[e
+ f*x]^2]) - (2*b*(15*a^2 - 10*a*b - b^2)*Tan[e + f*x])/(15*(a + b)^4*f*Sqrt[a + b + b*Tan[e + f*x]^2])

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/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 271

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

Rule 191

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

Rubi steps

\begin{align*} \int \frac{\csc ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{2 (5 a+2 b)+5 (a+b) x^2}{x^4 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=-\frac{2 (5 a+2 b) \cot ^3(e+f x)}{15 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}+\frac{\left (15 a^2-10 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac{\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 (a+b)^3 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{2 (5 a+2 b) \cot ^3(e+f x)}{15 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\left (2 b \left (15 a^2-10 a b-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^3 f}\\ &=-\frac{\left (15 a^2-10 a b-b^2\right ) \cot (e+f x)}{15 (a+b)^3 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{2 (5 a+2 b) \cot ^3(e+f x)}{15 (a+b)^2 f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{\cot ^5(e+f x)}{5 (a+b) f \sqrt{a+b+b \tan ^2(e+f x)}}-\frac{2 b \left (15 a^2-10 a b-b^2\right ) \tan (e+f x)}{15 (a+b)^4 f \sqrt{a+b+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.986108, size = 126, normalized size = 0.69 \[ -\frac{\tan (e+f x) \sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (4 a \left (a^2-4 a b-5 b^2\right ) \csc ^2(e+f x)-8 a^2 (a-5 b)+3 (a+b)^3 \csc ^6(e+f x)+(a-5 b) (a+b)^2 \csc ^4(e+f x)\right )}{30 f (a+b)^4 \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(-8*a^2*(a - 5*b) + 4*a*(a^2 - 4*a*b - 5*b^2)*Csc[e + f*x]^2 + (a - 5*b)*(a +
b)^2*Csc[e + f*x]^4 + 3*(a + b)^3*Csc[e + f*x]^6)*Sec[e + f*x]^2*Tan[e + f*x])/(30*(a + b)^4*f*(a + b*Sec[e +
f*x]^2)^(3/2))

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Maple [A]  time = 0.413, size = 204, normalized size = 1.1 \begin{align*} -{\frac{ \left ( 8\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{3}-40\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}{a}^{2}b-20\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3}+104\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}b-20\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}a{b}^{2}+15\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{3}-85\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}b+49\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}a{b}^{2}+5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{b}^{3}+30\,{a}^{2}b-20\,a{b}^{2}-2\,{b}^{3} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{15\,f \left ( a+b \right ) ^{4} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5}} \left ({\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

-1/15/f/(a+b)^4/(b+a*cos(f*x+e)^2)^2*(8*cos(f*x+e)^6*a^3-40*cos(f*x+e)^6*a^2*b-20*cos(f*x+e)^4*a^3+104*cos(f*x
+e)^4*a^2*b-20*cos(f*x+e)^4*a*b^2+15*cos(f*x+e)^2*a^3-85*cos(f*x+e)^2*a^2*b+49*cos(f*x+e)^2*a*b^2+5*cos(f*x+e)
^2*b^3+30*a^2*b-20*a*b^2-2*b^3)*cos(f*x+e)^3*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)/sin(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*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.9212, size = 706, normalized size = 3.86 \begin{align*} -\frac{{\left (8 \,{\left (a^{3} - 5 \, a^{2} b\right )} \cos \left (f x + e\right )^{7} - 4 \,{\left (5 \, a^{3} - 26 \, a^{2} b + 5 \, a b^{2}\right )} \cos \left (f x + e\right )^{5} +{\left (15 \, a^{3} - 85 \, a^{2} b + 49 \, a b^{2} + 5 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \,{\left (15 \, a^{2} b - 10 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left ({\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{6} -{\left (2 \, a^{5} + 7 \, a^{4} b + 8 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 2 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{4} +{\left (a^{5} + 2 \, a^{4} b - 2 \, a^{3} b^{2} - 8 \, a^{2} b^{3} - 7 \, a b^{4} - 2 \, b^{5}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} 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*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(8*(a^3 - 5*a^2*b)*cos(f*x + e)^7 - 4*(5*a^3 - 26*a^2*b + 5*a*b^2)*cos(f*x + e)^5 + (15*a^3 - 85*a^2*b +
49*a*b^2 + 5*b^3)*cos(f*x + e)^3 + 2*(15*a^2*b - 10*a*b^2 - b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/co
s(f*x + e)^2)/(((a^5 + 4*a^4*b + 6*a^3*b^2 + 4*a^2*b^3 + a*b^4)*f*cos(f*x + e)^6 - (2*a^5 + 7*a^4*b + 8*a^3*b^
2 + 2*a^2*b^3 - 2*a*b^4 - b^5)*f*cos(f*x + e)^4 + (a^5 + 2*a^4*b - 2*a^3*b^2 - 8*a^2*b^3 - 7*a*b^4 - 2*b^5)*f*
cos(f*x + e)^2 + (a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*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*sec(f*x+e)**2)**(3/2),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}}{{\left (b \sec \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(csc(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(3/2), x)