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

Optimal. Leaf size=231 \[ -\frac {(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac {\sqrt {b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{11/2} f}-\frac {b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2} \]

[Out]

-1/5*(5*a^2-30*a*b+27*b^2)*cot(f*x+e)/a^5/f-1/15*(10*a-9*b)*cot(f*x+e)^3/a^4/f-1/8*(15*a^2-70*a*b+63*b^2)*arct
an(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/a^(11/2)/f-1/5*cot(f*x+e)^5/a/f/(a+b*tan(f*x+e)^2)^2-1/20*b*(5*a^2-10*a
*b+9*b^2)*tan(f*x+e)/a^4/f/(a+b*tan(f*x+e)^2)^2-1/40*b*(35*a^2-110*a*b+99*b^2)*tan(f*x+e)/a^5/f/(a+b*tan(f*x+e
)^2)

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3663, 462, 456, 1259, 1261, 205} \[ -\frac {\sqrt {b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{11/2} f}-\frac {b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac {(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[b]*(15*a^2 - 70*a*b + 63*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(11/2)*f) - ((5*a^2 - 30*a*b
+ 27*b^2)*Cot[e + f*x])/(5*a^5*f) - ((10*a - 9*b)*Cot[e + f*x]^3)/(15*a^4*f) - Cot[e + f*x]^5/(5*a*f*(a + b*T
an[e + f*x]^2)^2) - (b*(5*a^2 - 10*a*b + 9*b^2)*Tan[e + f*x])/(20*a^4*f*(a + b*Tan[e + f*x]^2)^2) - (b*(35*a^2
- 110*a*b + 99*b^2)*Tan[e + f*x])/(40*a^5*f*(a + b*Tan[e + f*x]^2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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]

Rubi steps

\begin {align*} \int \frac {\csc ^6(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^6 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {10 a-9 b+5 a x^2}{x^4 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \operatorname {Subst}\left (\int \frac {4 \left (\frac {9}{a}-\frac {10}{b}\right )+4 \left (\frac {10}{a}-\frac {5}{b}-\frac {9 b}{a^2}\right ) x^2+\frac {3 \left (5 a^2-10 a b+9 b^2\right ) x^4}{a^3}}{x^4 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{20 a f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-8 a (10 a-9 b) b-8 b \left (5 a^2-20 a b+18 b^2\right ) x^2+\frac {b^2 \left (35 a^2-110 a b+99 b^2\right ) x^4}{a}}{x^4 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{40 a^4 b f}\\ &=-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\operatorname {Subst}\left (\int \left (-\frac {8 (10 a-9 b) b}{x^4}-\frac {8 b \left (5 a^2-30 a b+27 b^2\right )}{a x^2}+\frac {5 b^2 \left (15 a^2-70 a b+63 b^2\right )}{a \left (a+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{40 a^4 b f}\\ &=-\frac {\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac {(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}-\frac {\left (b \left (15 a^2-70 a b+63 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^5 f}\\ &=-\frac {\sqrt {b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 a^{11/2} f}-\frac {\left (5 a^2-30 a b+27 b^2\right ) \cot (e+f x)}{5 a^5 f}-\frac {(10 a-9 b) \cot ^3(e+f x)}{15 a^4 f}-\frac {\cot ^5(e+f x)}{5 a f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (5 a^2-10 a b+9 b^2\right ) \tan (e+f x)}{20 a^4 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b \left (35 a^2-110 a b+99 b^2\right ) \tan (e+f x)}{40 a^5 f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.68, size = 346, normalized size = 1.50 \[ \frac {-960 \sqrt {b} \left (15 a^2-70 a b+63 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )-\frac {2 \sqrt {a} \cot (e+f x) \csc ^4(e+f x) \left (-128 a^4 \cos (6 (e+f x))+64 a^4 \cos (8 (e+f x))+1600 a^4+1788 a^3 b \cos (6 (e+f x))-863 a^3 b \cos (8 (e+f x))-165 a^3 b-8800 a^2 b^2 \cos (6 (e+f x))+2479 a^2 b^2 \cos (8 (e+f x))+637 a^2 b^2+4 \left (416 a^4-447 a^3 b-1400 a^2 b^2+13125 a b^3-13230 b^4\right ) \cos (2 (e+f x))-4 \left (32 a^4-257 a^3 b-2821 a^2 b^2+8925 a b^3-6615 b^4\right ) \cos (4 (e+f x))+14700 a b^3 \cos (6 (e+f x))-2625 a b^3 \cos (8 (e+f x))-28875 a b^3-7560 b^4 \cos (6 (e+f x))+945 b^4 \cos (8 (e+f x))+33075 b^4\right )}{((a-b) \cos (2 (e+f x))+a+b)^2}}{7680 a^{11/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-960*Sqrt[b]*(15*a^2 - 70*a*b + 63*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - (2*Sqrt[a]*(1600*a^4 - 165*a
^3*b + 637*a^2*b^2 - 28875*a*b^3 + 33075*b^4 + 4*(416*a^4 - 447*a^3*b - 1400*a^2*b^2 + 13125*a*b^3 - 13230*b^4
)*Cos[2*(e + f*x)] - 4*(32*a^4 - 257*a^3*b - 2821*a^2*b^2 + 8925*a*b^3 - 6615*b^4)*Cos[4*(e + f*x)] - 128*a^4*
Cos[6*(e + f*x)] + 1788*a^3*b*Cos[6*(e + f*x)] - 8800*a^2*b^2*Cos[6*(e + f*x)] + 14700*a*b^3*Cos[6*(e + f*x)]
- 7560*b^4*Cos[6*(e + f*x)] + 64*a^4*Cos[8*(e + f*x)] - 863*a^3*b*Cos[8*(e + f*x)] + 2479*a^2*b^2*Cos[8*(e + f
*x)] - 2625*a*b^3*Cos[8*(e + f*x)] + 945*b^4*Cos[8*(e + f*x)])*Cot[e + f*x]*Csc[e + f*x]^4)/(a + b + (a - b)*C
os[2*(e + f*x)])^2)/(7680*a^(11/2)*f)

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fricas [B]  time = 0.58, size = 1199, normalized size = 5.19 \[ \text {result too large to display} \]

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,x, algorithm="fricas")

[Out]

[-1/480*(4*(64*a^4 - 863*a^3*b + 2479*a^2*b^2 - 2625*a*b^3 + 945*b^4)*cos(f*x + e)^9 - 4*(160*a^4 - 2173*a^3*b
+ 7158*a^2*b^2 - 8925*a*b^3 + 3780*b^4)*cos(f*x + e)^7 + 4*(120*a^4 - 1685*a^3*b + 7104*a^2*b^2 - 11025*a*b^3
+ 5670*b^4)*cos(f*x + e)^5 + 20*(75*a^3*b - 530*a^2*b^2 + 1155*a*b^3 - 756*b^4)*cos(f*x + e)^3 - 15*((15*a^4
- 100*a^3*b + 218*a^2*b^2 - 196*a*b^3 + 63*b^4)*cos(f*x + e)^8 - 2*(15*a^4 - 115*a^3*b + 303*a^2*b^2 - 329*a*b
^3 + 126*b^4)*cos(f*x + e)^6 + (15*a^4 - 160*a^3*b + 573*a^2*b^2 - 798*a*b^3 + 378*b^4)*cos(f*x + e)^4 + 15*a^
2*b^2 - 70*a*b^3 + 63*b^4 + 2*(15*a^3*b - 100*a^2*b^2 + 203*a*b^3 - 126*b^4)*cos(f*x + e)^2)*sqrt(-b/a)*log(((
a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 + 4*((a^2 + a*b)*cos(f*x + e)^3 - a*b*cos(f
*x + e))*sqrt(-b/a)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b
^2))*sin(f*x + e) + 60*(15*a^2*b^2 - 70*a*b^3 + 63*b^4)*cos(f*x + e))/(((a^7 - 2*a^6*b + a^5*b^2)*f*cos(f*x +
e)^8 + a^5*b^2*f - 2*(a^7 - 3*a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 - 6*a^6*b + 6*a^5*b^2)*f*cos(f*x + e)
^4 + 2*(a^6*b - 2*a^5*b^2)*f*cos(f*x + e)^2)*sin(f*x + e)), -1/240*(2*(64*a^4 - 863*a^3*b + 2479*a^2*b^2 - 262
5*a*b^3 + 945*b^4)*cos(f*x + e)^9 - 2*(160*a^4 - 2173*a^3*b + 7158*a^2*b^2 - 8925*a*b^3 + 3780*b^4)*cos(f*x +
e)^7 + 2*(120*a^4 - 1685*a^3*b + 7104*a^2*b^2 - 11025*a*b^3 + 5670*b^4)*cos(f*x + e)^5 + 10*(75*a^3*b - 530*a^
2*b^2 + 1155*a*b^3 - 756*b^4)*cos(f*x + e)^3 - 15*((15*a^4 - 100*a^3*b + 218*a^2*b^2 - 196*a*b^3 + 63*b^4)*cos
(f*x + e)^8 - 2*(15*a^4 - 115*a^3*b + 303*a^2*b^2 - 329*a*b^3 + 126*b^4)*cos(f*x + e)^6 + (15*a^4 - 160*a^3*b
+ 573*a^2*b^2 - 798*a*b^3 + 378*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 - 70*a*b^3 + 63*b^4 + 2*(15*a^3*b - 100*a^2*b
^2 + 203*a*b^3 - 126*b^4)*cos(f*x + e)^2)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(b*cos(f
*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*(15*a^2*b^2 - 70*a*b^3 + 63*b^4)*cos(f*x + e))/(((a^7 - 2*a^6*b + a^5
*b^2)*f*cos(f*x + e)^8 + a^5*b^2*f - 2*(a^7 - 3*a^6*b + 2*a^5*b^2)*f*cos(f*x + e)^6 + (a^7 - 6*a^6*b + 6*a^5*b
^2)*f*cos(f*x + e)^4 + 2*(a^6*b - 2*a^5*b^2)*f*cos(f*x + e)^2)*sin(f*x + e))]

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giac [A]  time = 4.13, size = 263, normalized size = 1.14 \[ -\frac {\frac {15 \, {\left (15 \, a^{2} b - 70 \, a b^{2} + 63 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{\sqrt {a b} a^{5}} + \frac {15 \, {\left (7 \, a^{2} b^{2} \tan \left (f x + e\right )^{3} - 22 \, a b^{3} \tan \left (f x + e\right )^{3} + 15 \, b^{4} \tan \left (f x + e\right )^{3} + 9 \, a^{3} b \tan \left (f x + e\right ) - 26 \, a^{2} b^{2} \tan \left (f x + e\right ) + 17 \, a b^{3} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2} a^{5}} + \frac {8 \, {\left (15 \, a^{2} \tan \left (f x + e\right )^{4} - 90 \, a b \tan \left (f x + e\right )^{4} + 90 \, b^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} - 15 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}\right )}}{a^{5} \tan \left (f x + e\right )^{5}}}{120 \, f} \]

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,x, algorithm="giac")

[Out]

-1/120*(15*(15*a^2*b - 70*a*b^2 + 63*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*
b)))/(sqrt(a*b)*a^5) + 15*(7*a^2*b^2*tan(f*x + e)^3 - 22*a*b^3*tan(f*x + e)^3 + 15*b^4*tan(f*x + e)^3 + 9*a^3*
b*tan(f*x + e) - 26*a^2*b^2*tan(f*x + e) + 17*a*b^3*tan(f*x + e))/((b*tan(f*x + e)^2 + a)^2*a^5) + 8*(15*a^2*t
an(f*x + e)^4 - 90*a*b*tan(f*x + e)^4 + 90*b^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 - 15*a*b*tan(f*x + e)^2
+ 3*a^2)/(a^5*tan(f*x + e)^5))/f

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maple [A]  time = 0.73, size = 380, normalized size = 1.65 \[ -\frac {7 b^{2} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {11 b^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{4 f \,a^{4} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b^{4} \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{5} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {9 b \tan \left (f x +e \right )}{8 f \,a^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}+\frac {13 b^{2} \tan \left (f x +e \right )}{4 f \,a^{3} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {17 b^{3} \tan \left (f x +e \right )}{8 f \,a^{4} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{2}}-\frac {15 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 f \,a^{3} \sqrt {a b}}+\frac {35 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{4 f \,a^{4} \sqrt {a b}}-\frac {63 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{8 f \,a^{5} \sqrt {a b}}-\frac {1}{5 f \,a^{3} \tan \left (f x +e \right )^{5}}+\frac {b}{f \,a^{4} \tan \left (f x +e \right )^{3}}-\frac {2}{3 f \,a^{3} \tan \left (f x +e \right )^{3}}-\frac {1}{f \,a^{3} \tan \left (f x +e \right )}+\frac {6 b}{f \,a^{4} \tan \left (f x +e \right )}-\frac {6 b^{2}}{f \,a^{5} \tan \left (f x +e \right )} \]

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,x)

[Out]

-7/8/f/a^3*b^2/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3+11/4/f/a^4*b^3/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-15/8/f*b^4/a
^5/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-9/8/f/a^2*b/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)+13/4/f/a^3*b^2/(a+b*tan(f*x+e
)^2)^2*tan(f*x+e)-17/8/f*b^3/a^4/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)-15/8/f/a^3*b/(a*b)^(1/2)*arctan(tan(f*x+e)*b/
(a*b)^(1/2))+35/4/f/a^4*b^2/(a*b)^(1/2)*arctan(tan(f*x+e)*b/(a*b)^(1/2))-63/8/f*b^3/a^5/(a*b)^(1/2)*arctan(tan
(f*x+e)*b/(a*b)^(1/2))-1/5/f/a^3/tan(f*x+e)^5+1/f/a^4/tan(f*x+e)^3*b-2/3/f/a^3/tan(f*x+e)^3-1/f/a^3/tan(f*x+e)
+6/f/a^4/tan(f*x+e)*b-6/f/a^5/tan(f*x+e)*b^2

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maxima [A]  time = 0.73, size = 212, normalized size = 0.92 \[ -\frac {\frac {15 \, {\left (15 \, a^{2} b^{2} - 70 \, a b^{3} + 63 \, b^{4}\right )} \tan \left (f x + e\right )^{8} + 25 \, {\left (15 \, a^{3} b - 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \tan \left (f x + e\right )^{6} + 8 \, {\left (15 \, a^{4} - 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \tan \left (f x + e\right )^{4} + 24 \, a^{4} + 8 \, {\left (10 \, a^{4} - 9 \, a^{3} b\right )} \tan \left (f x + e\right )^{2}}{a^{5} b^{2} \tan \left (f x + e\right )^{9} + 2 \, a^{6} b \tan \left (f x + e\right )^{7} + a^{7} \tan \left (f x + e\right )^{5}} + \frac {15 \, {\left (15 \, a^{2} b - 70 \, a b^{2} + 63 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}}}{120 \, f} \]

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,x, algorithm="maxima")

[Out]

-1/120*((15*(15*a^2*b^2 - 70*a*b^3 + 63*b^4)*tan(f*x + e)^8 + 25*(15*a^3*b - 70*a^2*b^2 + 63*a*b^3)*tan(f*x +
e)^6 + 8*(15*a^4 - 70*a^3*b + 63*a^2*b^2)*tan(f*x + e)^4 + 24*a^4 + 8*(10*a^4 - 9*a^3*b)*tan(f*x + e)^2)/(a^5*
b^2*tan(f*x + e)^9 + 2*a^6*b*tan(f*x + e)^7 + a^7*tan(f*x + e)^5) + 15*(15*a^2*b - 70*a*b^2 + 63*b^3)*arctan(b
*tan(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^5))/f

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mupad [B]  time = 13.31, size = 199, normalized size = 0.86 \[ -\frac {\frac {1}{5\,a}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{15\,a^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,a-9\,b\right )}{15\,a^2}+\frac {5\,b\,{\mathrm {tan}\left (e+f\,x\right )}^6\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{24\,a^4}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^8\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{8\,a^5}}{f\,\left (a^2\,{\mathrm {tan}\left (e+f\,x\right )}^5+2\,a\,b\,{\mathrm {tan}\left (e+f\,x\right )}^7+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a}}\right )\,\left (15\,a^2-70\,a\,b+63\,b^2\right )}{8\,a^{11/2}\,f} \]

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),x)

[Out]

- (1/(5*a) + (tan(e + f*x)^4*(15*a^2 - 70*a*b + 63*b^2))/(15*a^3) + (tan(e + f*x)^2*(10*a - 9*b))/(15*a^2) + (
5*b*tan(e + f*x)^6*(15*a^2 - 70*a*b + 63*b^2))/(24*a^4) + (b^2*tan(e + f*x)^8*(15*a^2 - 70*a*b + 63*b^2))/(8*a
^5))/(f*(a^2*tan(e + f*x)^5 + b^2*tan(e + f*x)^9 + 2*a*b*tan(e + f*x)^7)) - (b^(1/2)*atan((b^(1/2)*tan(e + f*x
))/a^(1/2))*(15*a^2 - 70*a*b + 63*b^2))/(8*a^(11/2)*f)

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

[Out]

Timed out

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