3.2.39 \(\int \frac {\csc ^6(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\) [139]

Optimal. Leaf size=171 \[ -\frac {\left (15 a^2-40 a b+24 b^2\right ) \cot (e+f x)}{15 a^3 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\cot ^5(e+f x)}{5 a f \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 b \left (15 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{15 a^4 f \sqrt {a+b \tan ^2(e+f x)}} \]

[Out]

-1/15*(15*a^2-40*a*b+24*b^2)*cot(f*x+e)/a^3/f/(a+b*tan(f*x+e)^2)^(1/2)-2/15*(5*a-3*b)*cot(f*x+e)^3/a^2/f/(a+b*
tan(f*x+e)^2)^(1/2)-1/5*cot(f*x+e)^5/a/f/(a+b*tan(f*x+e)^2)^(1/2)-2/15*b*(15*a^2-40*a*b+24*b^2)*tan(f*x+e)/a^4
/f/(a+b*tan(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3744, 473, 464, 277, 197} \begin {gather*} -\frac {2 (5 a-3 b) \cot ^3(e+f x)}{15 a^2 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {2 b \left (15 a^2-40 a b+24 b^2\right ) \tan (e+f x)}{15 a^4 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\left (15 a^2-40 a b+24 b^2\right ) \cot (e+f x)}{15 a^3 f \sqrt {a+b \tan ^2(e+f x)}}-\frac {\cot ^5(e+f x)}{5 a f \sqrt {a+b \tan ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*((15*a^2 - 40*a*b + 24*b^2)*Cot[e + f*x])/(a^3*f*Sqrt[a + b*Tan[e + f*x]^2]) - (2*(5*a - 3*b)*Cot[e + f*
x]^3)/(15*a^2*f*Sqrt[a + b*Tan[e + f*x]^2]) - Cot[e + f*x]^5/(5*a*f*Sqrt[a + b*Tan[e + f*x]^2]) - (2*b*(15*a^2
- 40*a*b + 24*b^2)*Tan[e + f*x])/(15*a^4*f*Sqrt[a + b*Tan[e + f*x]^2])

Rule 197

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]

Rule 277

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 464

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 473

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 3744

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]

Rubi steps

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

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Mathematica [A]
time = 1.50, size = 135, normalized size = 0.79 \begin {gather*} -\frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (\cot (e+f x) \left (8 a^2-41 a b+33 b^2+a (4 a-9 b) \csc ^2(e+f x)+3 a^2 \csc ^4(e+f x)\right )+\frac {15 (a-b)^2 b \sin (2 (e+f x))}{a+b+(a-b) \cos (2 (e+f x))}\right )}{15 \sqrt {2} a^4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(Cot[e + f*x]*(8*a^2 - 41*a*b + 33*b^2 + a*(4*a
- 9*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4) + (15*(a - b)^2*b*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e +
f*x)])))/(Sqrt[2]*a^4*f)

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Maple [A]
time = 0.42, size = 264, normalized size = 1.54

method	result	size
default	\(-\frac {\left (8 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3}-64 \left (\cos ^{6}\left (f x +e \right )\right ) a^{2} b +104 \left (\cos ^{6}\left (f x +e \right )\right ) a \,b^{2}-48 \left (\cos ^{6}\left (f x +e \right )\right ) b^{3}-20 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3}+164 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2} b -288 \left (\cos ^{4}\left (f x +e \right )\right ) a \,b^{2}+144 \left (\cos ^{4}\left (f x +e \right )\right ) b^{3}+15 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3}-130 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b +264 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{2}-144 \left (\cos ^{2}\left (f x +e \right )\right ) b^{3}+30 a^{2} b -80 a \,b^{2}+48 b^{3}\right ) \left (\cos ^{3}\left (f x +e \right )\right ) \left (\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}\right )^{\frac {3}{2}}}{15 f \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2} \sin \left (f x +e \right )^{5} a^{4}}\)	\(264\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/f/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*(8*cos(f*x+e)^6*a^3-64*cos(f*x+e)^6*a^2*b+104*cos(f*x+e)^6*a*b^2-4
8*cos(f*x+e)^6*b^3-20*cos(f*x+e)^4*a^3+164*cos(f*x+e)^4*a^2*b-288*cos(f*x+e)^4*a*b^2+144*cos(f*x+e)^4*b^3+15*c
os(f*x+e)^2*a^3-130*cos(f*x+e)^2*a^2*b+264*cos(f*x+e)^2*a*b^2-144*cos(f*x+e)^2*b^3+30*a^2*b-80*a*b^2+48*b^3)*c
os(f*x+e)^3*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)/sin(f*x+e)^5/a^4

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Maxima [A]
time = 0.29, size = 273, normalized size = 1.60 \begin {gather*} -\frac {\frac {30 \, b \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2}} - \frac {80 \, b^{2} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3}} + \frac {48 \, b^{3} \tan \left (f x + e\right )}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{4}} + \frac {15}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )} - \frac {40 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2} \tan \left (f x + e\right )} + \frac {24 \, b^{2}}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{3} \tan \left (f x + e\right )} + \frac {10}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )^{3}} - \frac {6 \, b}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a^{2} \tan \left (f x + e\right )^{3}} + \frac {3}{\sqrt {b \tan \left (f x + e\right )^{2} + a} a \tan \left (f x + e\right )^{5}}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(30*b*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^2) - 80*b^2*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^
3) + 48*b^3*tan(f*x + e)/(sqrt(b*tan(f*x + e)^2 + a)*a^4) + 15/(sqrt(b*tan(f*x + e)^2 + a)*a*tan(f*x + e)) - 4
0*b/(sqrt(b*tan(f*x + e)^2 + a)*a^2*tan(f*x + e)) + 24*b^2/(sqrt(b*tan(f*x + e)^2 + a)*a^3*tan(f*x + e)) + 10/
(sqrt(b*tan(f*x + e)^2 + a)*a*tan(f*x + e)^3) - 6*b/(sqrt(b*tan(f*x + e)^2 + a)*a^2*tan(f*x + e)^3) + 3/(sqrt(
b*tan(f*x + e)^2 + a)*a*tan(f*x + e)^5))/f

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Fricas [A]
time = 86.68, size = 242, normalized size = 1.42 \begin {gather*} -\frac {{\left (8 \, {\left (a^{3} - 8 \, a^{2} b + 13 \, a b^{2} - 6 \, b^{3}\right )} \cos \left (f x + e\right )^{7} - 4 \, {\left (5 \, a^{3} - 41 \, a^{2} b + 72 \, a b^{2} - 36 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{3} - 130 \, a^{2} b + 264 \, a b^{2} - 144 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 2 \, {\left (15 \, a^{2} b - 40 \, a b^{2} + 24 \, b^{3}\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 ({\left (a^{5} - a^{4} b\right )} f \cos \left (f x + e\right )^{6} + a^{4} b f - {\left (2 \, a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{5} - 3 \, a^{4} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/15*(8*(a^3 - 8*a^2*b + 13*a*b^2 - 6*b^3)*cos(f*x + e)^7 - 4*(5*a^3 - 41*a^2*b + 72*a*b^2 - 36*b^3)*cos(f*x
+ e)^5 + (15*a^3 - 130*a^2*b + 264*a*b^2 - 144*b^3)*cos(f*x + e)^3 + 2*(15*a^2*b - 40*a*b^2 + 24*b^3)*cos(f*x
+ e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(((a^5 - a^4*b)*f*cos(f*x + e)^6 + a^4*b*f - (2*a^5 -
3*a^4*b)*f*cos(f*x + e)^4 + (a^5 - 3*a^4*b)*f*cos(f*x + e)^2)*sin(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{6}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(e + f*x)**6/(a + b*tan(e + f*x)**2)**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^6/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^6*(a + b*tan(e + f*x)^2)^(3/2)),x)

[Out]

\text{Hanged}

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