3.478 \(\int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=239 \[ -\frac {a^2 \tan ^3(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {a^2 \left (3 a^2+7 b^2\right ) \tan ^2(c+d x)}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3}+\frac {\left (3 a^4+6 a^2 b^2+b^4\right ) \tan (c+d x)}{b^3 d \left (a^2+b^2\right )^2}-\frac {a^3 \left (3 a^4+9 a^2 b^2+10 b^4\right ) \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )^3} \]

[Out]

b*(3*a^2-b^2)*x/(a^2+b^2)^3-a*(a^2-3*b^2)*ln(cos(d*x+c))/(a^2+b^2)^3/d-a^3*(3*a^4+9*a^2*b^2+10*b^4)*ln(a+b*tan
(d*x+c))/b^4/(a^2+b^2)^3/d+(3*a^4+6*a^2*b^2+b^4)*tan(d*x+c)/b^3/(a^2+b^2)^2/d-1/2*a^2*tan(d*x+c)^3/b/(a^2+b^2)
/d/(a+b*tan(d*x+c))^2-1/2*a^2*(3*a^2+7*b^2)*tan(d*x+c)^2/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.56, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3565, 3645, 3647, 3626, 3617, 31, 3475} \[ -\frac {a^2 \left (3 a^2+7 b^2\right ) \tan ^2(c+d x)}{2 b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {a^2 \tan ^3(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {\left (6 a^2 b^2+3 a^4+b^4\right ) \tan (c+d x)}{b^3 d \left (a^2+b^2\right )^2}-\frac {a^3 \left (9 a^2 b^2+3 a^4+10 b^4\right ) \log (a+b \tan (c+d x))}{b^4 d \left (a^2+b^2\right )^3}-\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {b x \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + b*Tan[c + d*x])^3,x]

[Out]

(b*(3*a^2 - b^2)*x)/(a^2 + b^2)^3 - (a*(a^2 - 3*b^2)*Log[Cos[c + d*x]])/((a^2 + b^2)^3*d) - (a^3*(3*a^4 + 9*a^
2*b^2 + 10*b^4)*Log[a + b*Tan[c + d*x]])/(b^4*(a^2 + b^2)^3*d) + ((3*a^4 + 6*a^2*b^2 + b^4)*Tan[c + d*x])/(b^3
*(a^2 + b^2)^2*d) - (a^2*Tan[c + d*x]^3)/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (a^2*(3*a^2 + 7*b^2)*Tan
[c + d*x]^2)/(2*b^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3617

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A/(b*f), Subst[Int[(a + x)^m, x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A, C]

Rule 3626

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/((a_.) + (b_.)*tan[(e_.) + (f_.)*
(x_)]), x_Symbol] :> Simp[((a*A + b*B - a*C)*x)/(a^2 + b^2), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2), I
nt[(1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x]), x], x] - Dist[(A*b - a*B - b*C)/(a^2 + b^2), Int[Tan[e + f*x], x
], x]) /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && NeQ[a^2 + b^2, 0] && NeQ[A*b - a
*B - b*C, 0]

Rule 3645

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {\tan ^5(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=-\frac {a^2 \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {\tan ^2(c+d x) \left (3 a^2-2 a b \tan (c+d x)+\left (3 a^2+2 b^2\right ) \tan ^2(c+d x)\right )}{(a+b \tan (c+d x))^2} \, dx}{2 b \left (a^2+b^2\right )}\\ &=-\frac {a^2 \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (3 a^2+7 b^2\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {\tan (c+d x) \left (2 a^2 \left (3 a^2+7 b^2\right )-4 a b^3 \tan (c+d x)+2 \left (3 a^4+6 a^2 b^2+b^4\right ) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{2 b^2 \left (a^2+b^2\right )^2}\\ &=\frac {\left (3 a^4+6 a^2 b^2+b^4\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (3 a^2+7 b^2\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {-2 a \left (3 a^4+6 a^2 b^2+b^4\right )+2 b^3 \left (a^2-b^2\right ) \tan (c+d x)-6 a \left (a^2+b^2\right )^2 \tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{2 b^3 \left (a^2+b^2\right )^2}\\ &=\frac {b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac {\left (3 a^4+6 a^2 b^2+b^4\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (3 a^2+7 b^2\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\left (a \left (a^2-3 b^2\right )\right ) \int \tan (c+d x) \, dx}{\left (a^2+b^2\right )^3}-\frac {\left (a^3 \left (3 a^4+9 a^2 b^2+10 b^4\right )\right ) \int \frac {1+\tan ^2(c+d x)}{a+b \tan (c+d x)} \, dx}{b^3 \left (a^2+b^2\right )^3}\\ &=\frac {b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {\left (3 a^4+6 a^2 b^2+b^4\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (3 a^2+7 b^2\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (a^3 \left (3 a^4+9 a^2 b^2+10 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \tan (c+d x)\right )}{b^4 \left (a^2+b^2\right )^3 d}\\ &=\frac {b \left (3 a^2-b^2\right ) x}{\left (a^2+b^2\right )^3}-\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {a^3 \left (3 a^4+9 a^2 b^2+10 b^4\right ) \log (a+b \tan (c+d x))}{b^4 \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^4+6 a^2 b^2+b^4\right ) \tan (c+d x)}{b^3 \left (a^2+b^2\right )^2 d}-\frac {a^2 \tan ^3(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a^2 \left (3 a^2+7 b^2\right ) \tan ^2(c+d x)}{2 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 6.53, size = 694, normalized size = 2.90 \[ \frac {a^5 \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))}{2 b^2 d (a-i b)^2 (a+i b)^2 (a+b \tan (c+d x))^3}+\frac {b \left (3 a^2-b^2\right ) (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{d (a-i b)^3 (a+i b)^3 (a+b \tan (c+d x))^3}+\frac {\sec ^3(c+d x) \left (2 a^5 \sin (c+d x)+5 a^3 b^2 \sin (c+d x)\right ) (a \cos (c+d x)+b \sin (c+d x))^2}{b^3 d (a-i b)^2 (a+i b)^2 (a+b \tan (c+d x))^3}-\frac {i \left (-3 a^7-9 a^5 b^2-10 a^3 b^4\right ) \tan ^{-1}(\tan (c+d x)) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{b^4 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^3}+\frac {\left (-3 a^7-9 a^5 b^2-10 a^3 b^4\right ) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )}{2 b^4 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^3}+\frac {3 a \sec ^3(c+d x) \log (\cos (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^3}{b^4 d (a+b \tan (c+d x))^3}+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{b^3 d (a+b \tan (c+d x))^3}-\frac {i \left (3 a^{12} b^3-3 i a^{11} b^4+15 a^{10} b^5-15 i a^9 b^6+31 a^8 b^7-31 i a^7 b^8+29 a^6 b^9-29 i a^5 b^{10}+10 a^4 b^{11}-10 i a^3 b^{12}\right ) (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3}{b^7 d (a-i b)^6 (a+i b)^5 (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + b*Tan[c + d*x])^3,x]

[Out]

(a^5*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x]))/(2*(a - I*b)^2*(a + I*b)^2*b^2*d*(a + b*Tan[c + d*x])^3
) + (b*(3*a^2 - b^2)*(c + d*x)*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/((a - I*b)^3*(a + I*b)^3*d*
(a + b*Tan[c + d*x])^3) - (I*(3*a^12*b^3 - (3*I)*a^11*b^4 + 15*a^10*b^5 - (15*I)*a^9*b^6 + 31*a^8*b^7 - (31*I)
*a^7*b^8 + 29*a^6*b^9 - (29*I)*a^5*b^10 + 10*a^4*b^11 - (10*I)*a^3*b^12)*(c + d*x)*Sec[c + d*x]^3*(a*Cos[c + d
*x] + b*Sin[c + d*x])^3)/((a - I*b)^6*(a + I*b)^5*b^7*d*(a + b*Tan[c + d*x])^3) - (I*(-3*a^7 - 9*a^5*b^2 - 10*
a^3*b^4)*ArcTan[Tan[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(b^4*(a^2 + b^2)^3*d*(a + b*
Tan[c + d*x])^3) + (3*a*Log[Cos[c + d*x]]*Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3)/(b^4*d*(a + b*Ta
n[c + d*x])^3) + ((-3*a^7 - 9*a^5*b^2 - 10*a^3*b^4)*Log[(a*Cos[c + d*x] + b*Sin[c + d*x])^2]*Sec[c + d*x]^3*(a
*Cos[c + d*x] + b*Sin[c + d*x])^3)/(2*b^4*(a^2 + b^2)^3*d*(a + b*Tan[c + d*x])^3) + (Sec[c + d*x]^3*(a*Cos[c +
d*x] + b*Sin[c + d*x])^2*(2*a^5*Sin[c + d*x] + 5*a^3*b^2*Sin[c + d*x]))/((a - I*b)^2*(a + I*b)^2*b^3*d*(a + b
*Tan[c + d*x])^3) + (Sec[c + d*x]^3*(a*Cos[c + d*x] + b*Sin[c + d*x])^3*Tan[c + d*x])/(b^3*d*(a + b*Tan[c + d*
x])^3)

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fricas [B]  time = 0.61, size = 549, normalized size = 2.30 \[ -\frac {3 \, a^{7} b^{2} + 9 \, a^{5} b^{4} - 2 \, {\left (a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{3} - 2 \, {\left (3 \, a^{4} b^{5} - a^{2} b^{7}\right )} d x - {\left (9 \, a^{7} b^{2} + 23 \, a^{5} b^{4} + 12 \, a^{3} b^{6} + 4 \, a b^{8} + 2 \, {\left (3 \, a^{2} b^{7} - b^{9}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (3 \, a^{9} + 9 \, a^{7} b^{2} + 10 \, a^{5} b^{4} + {\left (3 \, a^{7} b^{2} + 9 \, a^{5} b^{4} + 10 \, a^{3} b^{6}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{8} b + 9 \, a^{6} b^{3} + 10 \, a^{4} b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 3 \, {\left (a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6} + {\left (a^{7} b^{2} + 3 \, a^{5} b^{4} + 3 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{8} b + 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} + a^{2} b^{7}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (3 \, a^{8} b + 6 \, a^{6} b^{3} - 2 \, a^{4} b^{5} + a^{2} b^{7} + 2 \, {\left (3 \, a^{3} b^{6} - a b^{8}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{6} + 3 \, a^{4} b^{8} + 3 \, a^{2} b^{10} + b^{12}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b^{5} + 3 \, a^{5} b^{7} + 3 \, a^{3} b^{9} + a b^{11}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} b^{4} + 3 \, a^{6} b^{6} + 3 \, a^{4} b^{8} + a^{2} b^{10}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(3*a^7*b^2 + 9*a^5*b^4 - 2*(a^6*b^3 + 3*a^4*b^5 + 3*a^2*b^7 + b^9)*tan(d*x + c)^3 - 2*(3*a^4*b^5 - a^2*b^
7)*d*x - (9*a^7*b^2 + 23*a^5*b^4 + 12*a^3*b^6 + 4*a*b^8 + 2*(3*a^2*b^7 - b^9)*d*x)*tan(d*x + c)^2 + (3*a^9 + 9
*a^7*b^2 + 10*a^5*b^4 + (3*a^7*b^2 + 9*a^5*b^4 + 10*a^3*b^6)*tan(d*x + c)^2 + 2*(3*a^8*b + 9*a^6*b^3 + 10*a^4*
b^5)*tan(d*x + c))*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) - 3*(a^9 + 3*a^7*
b^2 + 3*a^5*b^4 + a^3*b^6 + (a^7*b^2 + 3*a^5*b^4 + 3*a^3*b^6 + a*b^8)*tan(d*x + c)^2 + 2*(a^8*b + 3*a^6*b^3 +
3*a^4*b^5 + a^2*b^7)*tan(d*x + c))*log(1/(tan(d*x + c)^2 + 1)) - 2*(3*a^8*b + 6*a^6*b^3 - 2*a^4*b^5 + a^2*b^7
+ 2*(3*a^3*b^6 - a*b^8)*d*x)*tan(d*x + c))/((a^6*b^6 + 3*a^4*b^8 + 3*a^2*b^10 + b^12)*d*tan(d*x + c)^2 + 2*(a^
7*b^5 + 3*a^5*b^7 + 3*a^3*b^9 + a*b^11)*d*tan(d*x + c) + (a^8*b^4 + 3*a^6*b^6 + 3*a^4*b^8 + a^2*b^10)*d)

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giac [A]  time = 7.94, size = 325, normalized size = 1.36 \[ \frac {\frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (3 \, a^{7} + 9 \, a^{5} b^{2} + 10 \, a^{3} b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}} + \frac {9 \, a^{7} b^{2} \tan \left (d x + c\right )^{2} + 27 \, a^{5} b^{4} \tan \left (d x + c\right )^{2} + 30 \, a^{3} b^{6} \tan \left (d x + c\right )^{2} + 12 \, a^{8} b \tan \left (d x + c\right ) + 38 \, a^{6} b^{3} \tan \left (d x + c\right ) + 50 \, a^{4} b^{5} \tan \left (d x + c\right ) + 4 \, a^{9} + 13 \, a^{7} b^{2} + 21 \, a^{5} b^{4}}{{\left (a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}} + \frac {2 \, \tan \left (d x + c\right )}{b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1)
/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(3*a^7 + 9*a^5*b^2 + 10*a^3*b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b^
4 + 3*a^4*b^6 + 3*a^2*b^8 + b^10) + (9*a^7*b^2*tan(d*x + c)^2 + 27*a^5*b^4*tan(d*x + c)^2 + 30*a^3*b^6*tan(d*x
+ c)^2 + 12*a^8*b*tan(d*x + c) + 38*a^6*b^3*tan(d*x + c) + 50*a^4*b^5*tan(d*x + c) + 4*a^9 + 13*a^7*b^2 + 21*
a^5*b^4)/((a^6*b^4 + 3*a^4*b^6 + 3*a^2*b^8 + b^10)*(b*tan(d*x + c) + a)^2) + 2*tan(d*x + c)/b^3)/d

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maple [A]  time = 0.21, size = 307, normalized size = 1.28 \[ \frac {\tan \left (d x +c \right )}{d \,b^{3}}-\frac {3 a^{6}}{d \,b^{4} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {5 a^{4}}{d \,b^{2} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{5}}{2 d \,b^{4} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {3 a^{7} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,b^{4} \left (a^{2}+b^{2}\right )^{3}}-\frac {9 a^{5} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,b^{2} \left (a^{2}+b^{2}\right )^{3}}-\frac {10 a^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3}}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{2} a}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {\arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+b*tan(d*x+c))^3,x)

[Out]

1/d/b^3*tan(d*x+c)-3/d/b^4*a^6/(a^2+b^2)^2/(a+b*tan(d*x+c))-5/d/b^2*a^4/(a^2+b^2)^2/(a+b*tan(d*x+c))+1/2/d/b^4
*a^5/(a^2+b^2)/(a+b*tan(d*x+c))^2-3/d/b^4*a^7/(a^2+b^2)^3*ln(a+b*tan(d*x+c))-9/d/b^2*a^5/(a^2+b^2)^3*ln(a+b*ta
n(d*x+c))-10/d*a^3/(a^2+b^2)^3*ln(a+b*tan(d*x+c))+1/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*a^3-3/2/d/(a^2+b^2)^3*l
n(1+tan(d*x+c)^2)*b^2*a+3/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^2*b-1/d/(a^2+b^2)^3*arctan(tan(d*x+c))*b^3

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maxima [A]  time = 0.70, size = 293, normalized size = 1.23 \[ \frac {\frac {2 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (3 \, a^{7} + 9 \, a^{5} b^{2} + 10 \, a^{3} b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} b^{4} + 3 \, a^{4} b^{6} + 3 \, a^{2} b^{8} + b^{10}} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {5 \, a^{7} + 9 \, a^{5} b^{2} + 2 \, {\left (3 \, a^{6} b + 5 \, a^{4} b^{3}\right )} \tan \left (d x + c\right )}{a^{6} b^{4} + 2 \, a^{4} b^{6} + a^{2} b^{8} + {\left (a^{4} b^{6} + 2 \, a^{2} b^{8} + b^{10}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{5} + 2 \, a^{3} b^{7} + a b^{9}\right )} \tan \left (d x + c\right )} + \frac {2 \, \tan \left (d x + c\right )}{b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*(2*(3*a^2*b - b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(3*a^7 + 9*a^5*b^2 + 10*a^3*b^4)*log(
b*tan(d*x + c) + a)/(a^6*b^4 + 3*a^4*b^6 + 3*a^2*b^8 + b^10) + (a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1)/(a^6 +
3*a^4*b^2 + 3*a^2*b^4 + b^6) - (5*a^7 + 9*a^5*b^2 + 2*(3*a^6*b + 5*a^4*b^3)*tan(d*x + c))/(a^6*b^4 + 2*a^4*b^6
+ a^2*b^8 + (a^4*b^6 + 2*a^2*b^8 + b^10)*tan(d*x + c)^2 + 2*(a^5*b^5 + 2*a^3*b^7 + a*b^9)*tan(d*x + c)) + 2*t
an(d*x + c)/b^3)/d

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mupad [B]  time = 4.25, size = 263, normalized size = 1.10 \[ \frac {\mathrm {tan}\left (c+d\,x\right )}{b^3\,d}-\frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^6+5\,a^4\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {5\,a^7+9\,a^5\,b^2}{2\,b\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,b^3+2\,a\,b^4\,\mathrm {tan}\left (c+d\,x\right )+b^5\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}\right )}-\frac {a^3\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (3\,a^4+9\,a^2\,b^2+10\,b^4\right )}{b^4\,d\,{\left (a^2+b^2\right )}^3}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-a^3\,1{}\mathrm {i}+3\,a^2\,b+a\,b^2\,3{}\mathrm {i}-b^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5/(a + b*tan(c + d*x))^3,x)

[Out]

tan(c + d*x)/(b^3*d) - ((tan(c + d*x)*(3*a^6 + 5*a^4*b^2))/(a^4 + b^4 + 2*a^2*b^2) + (5*a^7 + 9*a^5*b^2)/(2*b*
(a^4 + b^4 + 2*a^2*b^2)))/(d*(a^2*b^3 + b^5*tan(c + d*x)^2 + 2*a*b^4*tan(c + d*x))) - (log(tan(c + d*x) - 1i)*
1i)/(2*d*(a*b^2*3i + 3*a^2*b - a^3*1i - b^3)) - log(tan(c + d*x) + 1i)/(2*d*(3*a*b^2 + a^2*b*3i - a^3 - b^3*1i
)) - (a^3*log(a + b*tan(c + d*x))*(3*a^4 + 10*b^4 + 9*a^2*b^2))/(b^4*d*(a^2 + b^2)^3)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: AttributeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**5/(a+b*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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