3.39 \(\int \frac {\tan ^2(c+d x) (B \tan (c+d x)+C \tan ^2(c+d x))}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=250 \[ \frac {a (b B-a C) \tan ^2(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {a^2 \left (a^3 (-C)-3 a b^2 C+2 b^3 B\right )}{b^3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {\left (a^3 B+3 a^2 b C-3 a b^2 B-b^3 C\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {x \left (a^3 (-C)+3 a^2 b B+3 a b^2 C-b^3 B\right )}{\left (a^2+b^2\right )^3}+\frac {a \left (a^5 C+3 a^3 b^2 C+a^2 b^3 B+6 a b^4 C-3 b^5 B\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^3} \]

[Out]

-(3*B*a^2*b-B*b^3-C*a^3+3*C*a*b^2)*x/(a^2+b^2)^3+(B*a^3-3*B*a*b^2+3*C*a^2*b-C*b^3)*ln(cos(d*x+c))/(a^2+b^2)^3/
d+a*(B*a^2*b^3-3*B*b^5+C*a^5+3*C*a^3*b^2+6*C*a*b^4)*ln(a+b*tan(d*x+c))/b^3/(a^2+b^2)^3/d+1/2*a*(B*b-C*a)*tan(d
*x+c)^2/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^2-a^2*(2*B*b^3-C*a^3-3*C*a*b^2)/b^3/(a^2+b^2)^2/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.58, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3632, 3605, 3635, 3626, 3617, 31, 3475} \[ \frac {a (b B-a C) \tan ^2(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {a^2 \left (a^3 (-C)-3 a b^2 C+2 b^3 B\right )}{b^3 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {a \left (a^2 b^3 B+3 a^3 b^2 C+a^5 C+6 a b^4 C-3 b^5 B\right ) \log (a+b \tan (c+d x))}{b^3 d \left (a^2+b^2\right )^3}+\frac {\left (3 a^2 b C+a^3 B-3 a b^2 B-b^3 C\right ) \log (\cos (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {x \left (3 a^2 b B+a^3 (-C)+3 a b^2 C-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]^2*(B*Tan[c + d*x] + C*Tan[c + d*x]^2))/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((3*a^2*b*B - b^3*B - a^3*C + 3*a*b^2*C)*x)/(a^2 + b^2)^3) + ((a^3*B - 3*a*b^2*B + 3*a^2*b*C - b^3*C)*Log[Co
s[c + d*x]])/((a^2 + b^2)^3*d) + (a*(a^2*b^3*B - 3*b^5*B + a^5*C + 3*a^3*b^2*C + 6*a*b^4*C)*Log[a + b*Tan[c +
d*x]])/(b^3*(a^2 + b^2)^3*d) + (a*(b*B - a*C)*Tan[c + d*x]^2)/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^2) - (a^
2*(2*b^3*B - a^3*C - 3*a*b^2*C))/(b^3*(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 3605

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(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)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])

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 3632

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] :> Dist[1/b^2, Int[(a + b*Tan[e + f*x])
^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3635

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(c^2*C - B*c*d + A*d^2)*
(c + d*Tan[e + f*x])^(n + 1))/(d^2*f*(n + 1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x
])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d +
a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*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[c^2 + d^2, 0] && LtQ[n, -1]

Rubi steps

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

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Mathematica [C]  time = 4.92, size = 462, normalized size = 1.85 \[ \frac {\sec ^2(c+d x) (B+C \tan (c+d x)) (a \cos (c+d x)+b \sin (c+d x)) \left (-2 C \left (a^2+b^2\right )^3 \log (\cos (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^2+a^3 b^2 \left (a^2+b^2\right ) (b B-a C)-2 a b \left (a^2+b^2\right ) \left (a^3 C+4 a b^2 C-3 b^3 B\right ) \sin (c+d x) (a \cos (c+d x)+b \sin (c+d x))+2 b^3 (c+d x) \left (a^3 C-3 a^2 b B-3 a b^2 C+b^3 B\right ) (a \cos (c+d x)+b \sin (c+d x))^2+2 i a (c+d x) \left (a^5 C+3 a^3 b^2 C+a^2 b^3 B+6 a b^4 C-3 b^5 B\right ) (a \cos (c+d x)+b \sin (c+d x))^2+a \left (a^5 C+3 a^3 b^2 C+a^2 b^3 B+6 a b^4 C-3 b^5 B\right ) (a \cos (c+d x)+b \sin (c+d x))^2 \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )-2 i a \left (a^5 C+3 a^3 b^2 C+a^2 b^3 B+6 a b^4 C-3 b^5 B\right ) \tan ^{-1}(\tan (c+d x)) (a \cos (c+d x)+b \sin (c+d x))^2\right )}{2 b^3 d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))^3 (B \cos (c+d x)+C \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]^2*(B*Tan[c + d*x] + C*Tan[c + d*x]^2))/(a + b*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x])*(a^3*b^2*(a^2 + b^2)*(b*B - a*C) - 2*a*b*(a^2 + b^2)*(-3*b^3
*B + a^3*C + 4*a*b^2*C)*Sin[c + d*x]*(a*Cos[c + d*x] + b*Sin[c + d*x]) + 2*b^3*(-3*a^2*b*B + b^3*B + a^3*C - 3
*a*b^2*C)*(c + d*x)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 + (2*I)*a*(a^2*b^3*B - 3*b^5*B + a^5*C + 3*a^3*b^2*C +
6*a*b^4*C)*(c + d*x)*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 - (2*I)*a*(a^2*b^3*B - 3*b^5*B + a^5*C + 3*a^3*b^2*C
+ 6*a*b^4*C)*ArcTan[Tan[c + d*x]]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2 - 2*(a^2 + b^2)^3*C*Log[Cos[c + d*x]]*(
a*Cos[c + d*x] + b*Sin[c + d*x])^2 + a*(a^2*b^3*B - 3*b^5*B + a^5*C + 3*a^3*b^2*C + 6*a*b^4*C)*Log[(a*Cos[c +
d*x] + b*Sin[c + d*x])^2]*(a*Cos[c + d*x] + b*Sin[c + d*x])^2)*(B + C*Tan[c + d*x]))/(2*b^3*(a^2 + b^2)^3*d*(B
*Cos[c + d*x] + C*Sin[c + d*x])*(a + b*Tan[c + d*x])^3)

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fricas [B]  time = 1.46, size = 666, normalized size = 2.66 \[ \frac {C a^{6} b^{2} + B a^{5} b^{3} + 7 \, C a^{4} b^{4} - 5 \, B a^{3} b^{5} + 2 \, {\left (C a^{5} b^{3} - 3 \, B a^{4} b^{4} - 3 \, C a^{3} b^{5} + B a^{2} b^{6}\right )} d x - {\left (3 \, C a^{6} b^{2} - B a^{5} b^{3} + 9 \, C a^{4} b^{4} - 7 \, B a^{3} b^{5} - 2 \, {\left (C a^{3} b^{5} - 3 \, B a^{2} b^{6} - 3 \, C a b^{7} + B b^{8}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (C a^{8} + 3 \, C a^{6} b^{2} + B a^{5} b^{3} + 6 \, C a^{4} b^{4} - 3 \, B a^{3} b^{5} + {\left (C a^{6} b^{2} + 3 \, C a^{4} b^{4} + B a^{3} b^{5} + 6 \, C a^{2} b^{6} - 3 \, B a b^{7}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (C a^{7} b + 3 \, C a^{5} b^{3} + B a^{4} b^{4} + 6 \, C a^{3} b^{5} - 3 \, B a^{2} b^{6}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - {\left (C a^{8} + 3 \, C a^{6} b^{2} + 3 \, C a^{4} b^{4} + C a^{2} b^{6} + {\left (C a^{6} b^{2} + 3 \, C a^{4} b^{4} + 3 \, C a^{2} b^{6} + C b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (C a^{7} b + 3 \, C a^{5} b^{3} + 3 \, C a^{3} b^{5} + C a b^{7}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (C a^{7} b + 3 \, C a^{5} b^{3} - 3 \, B a^{4} b^{4} - 4 \, C a^{3} b^{5} + 3 \, B a^{2} b^{6} - 2 \, {\left (C a^{4} b^{4} - 3 \, B a^{3} b^{5} - 3 \, C a^{2} b^{6} + B a b^{7}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{5} + 3 \, a^{4} b^{7} + 3 \, a^{2} b^{9} + b^{11}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b^{4} + 3 \, a^{5} b^{6} + 3 \, a^{3} b^{8} + a b^{10}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} b^{3} + 3 \, a^{6} b^{5} + 3 \, a^{4} b^{7} + a^{2} b^{9}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 3.01, size = 458, normalized size = 1.83 \[ \frac {\frac {2 \, {\left (C a^{3} - 3 \, B a^{2} b - 3 \, C a b^{2} + B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (B a^{3} + 3 \, C a^{2} b - 3 \, B a b^{2} - C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (C a^{6} + 3 \, C a^{4} b^{2} + B a^{3} b^{3} + 6 \, C a^{2} b^{4} - 3 \, B a b^{5}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}} - \frac {3 \, C a^{6} b \tan \left (d x + c\right )^{2} + 9 \, C a^{4} b^{3} \tan \left (d x + c\right )^{2} + 3 \, B a^{3} b^{4} \tan \left (d x + c\right )^{2} + 18 \, C a^{2} b^{5} \tan \left (d x + c\right )^{2} - 9 \, B a b^{6} \tan \left (d x + c\right )^{2} + 2 \, C a^{7} \tan \left (d x + c\right ) + 2 \, B a^{6} b \tan \left (d x + c\right ) + 6 \, C a^{5} b^{2} \tan \left (d x + c\right ) + 14 \, B a^{4} b^{3} \tan \left (d x + c\right ) + 28 \, C a^{3} b^{4} \tan \left (d x + c\right ) - 12 \, B a^{2} b^{5} \tan \left (d x + c\right ) + B a^{7} - C a^{6} b + 9 \, B a^{5} b^{2} + 11 \, C a^{4} b^{3} - 4 \, B a^{3} b^{4}}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/2*(2*(C*a^3 - 3*B*a^2*b - 3*C*a*b^2 + B*b^3)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - (B*a^3 + 3*C*a^
2*b - 3*B*a*b^2 - C*b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(C*a^6 + 3*C*a^4*b^2
+ B*a^3*b^3 + 6*C*a^2*b^4 - 3*B*a*b^5)*log(abs(b*tan(d*x + c) + a))/(a^6*b^3 + 3*a^4*b^5 + 3*a^2*b^7 + b^9) -
(3*C*a^6*b*tan(d*x + c)^2 + 9*C*a^4*b^3*tan(d*x + c)^2 + 3*B*a^3*b^4*tan(d*x + c)^2 + 18*C*a^2*b^5*tan(d*x + c
)^2 - 9*B*a*b^6*tan(d*x + c)^2 + 2*C*a^7*tan(d*x + c) + 2*B*a^6*b*tan(d*x + c) + 6*C*a^5*b^2*tan(d*x + c) + 14
*B*a^4*b^3*tan(d*x + c) + 28*C*a^3*b^4*tan(d*x + c) - 12*B*a^2*b^5*tan(d*x + c) + B*a^7 - C*a^6*b + 9*B*a^5*b^
2 + 11*C*a^4*b^3 - 4*B*a^3*b^4)/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*(b*tan(d*x + c) + a)^2))/d

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maple [B]  time = 0.28, size = 566, normalized size = 2.26 \[ \frac {a^{3} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 a \,b^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {a^{6} \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right )^{3} b^{3}}+\frac {3 a^{4} \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right )^{3} b}+\frac {6 a^{2} b \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {a^{4} B}{d \,b^{2} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {3 a^{2} B}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 a^{5} C}{d \,b^{3} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {4 a^{3} C}{d b \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {a^{3} B}{2 d \,b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {a^{4} C}{2 d \,b^{3} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} B}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) C \,a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3} C}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {C \arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 C \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^3,x)

[Out]

1/d*a^3/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*B-3/d*a/(a^2+b^2)^3*b^2*ln(a+b*tan(d*x+c))*B+1/d*a^6/(a^2+b^2)^3/b^3*ln
(a+b*tan(d*x+c))*C+3/d*a^4/(a^2+b^2)^3/b*ln(a+b*tan(d*x+c))*C+6/d*a^2/(a^2+b^2)^3*b*ln(a+b*tan(d*x+c))*C-1/d*a
^4/b^2/(a^2+b^2)^2/(a+b*tan(d*x+c))*B-3/d*a^2/(a^2+b^2)^2/(a+b*tan(d*x+c))*B+2/d*a^5/b^3/(a^2+b^2)^2/(a+b*tan(
d*x+c))*C+4/d*a^3/b/(a^2+b^2)^2/(a+b*tan(d*x+c))*C+1/2/d*a^3/b^2/(a^2+b^2)/(a+b*tan(d*x+c))^2*B-1/2/d*a^4/b^3/
(a^2+b^2)/(a+b*tan(d*x+c))^2*C-1/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*a^3*B+3/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)
*B*a*b^2-3/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*C*a^2*b+1/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*b^3*C-3/d/(a^2+b^2)
^3*B*arctan(tan(d*x+c))*a^2*b+1/d/(a^2+b^2)^3*B*arctan(tan(d*x+c))*b^3+1/d/(a^2+b^2)^3*C*arctan(tan(d*x+c))*a^
3-3/d/(a^2+b^2)^3*C*arctan(tan(d*x+c))*a*b^2

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maxima [A]  time = 0.65, size = 366, normalized size = 1.46 \[ \frac {\frac {2 \, {\left (C a^{3} - 3 \, B a^{2} b - 3 \, C a b^{2} + B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (C a^{6} + 3 \, C a^{4} b^{2} + B a^{3} b^{3} + 6 \, C a^{2} b^{4} - 3 \, B a b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} b^{3} + 3 \, a^{4} b^{5} + 3 \, a^{2} b^{7} + b^{9}} - \frac {{\left (B a^{3} + 3 \, C a^{2} b - 3 \, B a b^{2} - C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {3 \, C a^{6} - B a^{5} b + 7 \, C a^{4} b^{2} - 5 \, B a^{3} b^{3} + 2 \, {\left (2 \, C a^{5} b - B a^{4} b^{2} + 4 \, C a^{3} b^{3} - 3 \, B a^{2} b^{4}\right )} \tan \left (d x + c\right )}{a^{6} b^{3} + 2 \, a^{4} b^{5} + a^{2} b^{7} + {\left (a^{4} b^{5} + 2 \, a^{2} b^{7} + b^{9}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{4} + 2 \, a^{3} b^{6} + a b^{8}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.32, size = 307, normalized size = 1.23 \[ \frac {\frac {3\,C\,a^6-B\,a^5\,b+7\,C\,a^4\,b^2-5\,B\,a^3\,b^3}{2\,b^3\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {a^2\,\mathrm {tan}\left (c+d\,x\right )\,\left (-2\,C\,a^3+B\,a^2\,b-4\,C\,a\,b^2+3\,B\,b^3\right )}{b^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}+3\,a^2\,b+a\,b^2\,3{}\mathrm {i}-b^3\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}\right )}+\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^5+3\,C\,a^3\,b^2+B\,a^2\,b^3+6\,C\,a\,b^4-3\,B\,b^5\right )}{b^3\,d\,{\left (a^2+b^2\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(c + d*x)^2*(B*tan(c + d*x) + C*tan(c + d*x)^2))/(a + b*tan(c + d*x))^3,x)

[Out]

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

[Out]

Exception raised: AttributeError

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