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

Optimal. Leaf size=189 \[ -\frac {a^2 (A b-a B)}{2 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {\left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

[Out]

-(A*a^3-3*A*a*b^2+3*B*a^2*b-B*b^3)*x/(a^2+b^2)^3-(3*A*a^2*b-A*b^3-B*a^3+3*B*a*b^2)*ln(a*cos(d*x+c)+b*sin(d*x+c
))/(a^2+b^2)^3/d-1/2*a^2*(A*b-B*a)/b^2/(a^2+b^2)/d/(a+b*tan(d*x+c))^2+a*(2*A*b^3-a*(a^2+3*b^2)*B)/b^2/(a^2+b^2
)^2/d/(a+b*tan(d*x+c))

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Rubi [A]  time = 0.37, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3604, 3628, 3531, 3530} \[ -\frac {a^2 (A b-a B)}{2 b^2 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {a \left (2 A b^3-a B \left (a^2+3 b^2\right )\right )}{b^2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {\left (3 a^2 A b+a^3 (-B)+3 a b^2 B-A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}-\frac {x \left (a^3 A+3 a^2 b B-3 a A b^2-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^3*A - 3*a*A*b^2 + 3*a^2*b*B - b^3*B)*x)/(a^2 + b^2)^3) - ((3*a^2*A*b - A*b^3 - a^3*B + 3*a*b^2*B)*Log[a*
Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)^3*d) - (a^2*(A*b - a*B))/(2*b^2*(a^2 + b^2)*d*(a + b*Tan[c + d*x]
)^2) + (a*(2*A*b^3 - a*(a^2 + 3*b^2)*B))/(b^2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x]))

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rule 3604

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

Rule 3628

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

Rubi steps

\begin {align*} \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=-\frac {a^2 (A b-a B)}{2 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {-a (A b-a B)+b (A b-a B) \tan (c+d x)+\left (a^2+b^2\right ) B \tan ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac {a^2 (A b-a B)}{2 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {-b \left (a^2 A-A b^2+2 a b B\right )+b \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{b \left (a^2+b^2\right )^2}\\ &=-\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x}{\left (a^2+b^2\right )^3}-\frac {a^2 (A b-a B)}{2 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\left (a^2+b^2\right )^3}\\ &=-\frac {\left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x}{\left (a^2+b^2\right )^3}-\frac {\left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac {a^2 (A b-a B)}{2 b^2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a \left (2 A b^3-a \left (a^2+3 b^2\right ) B\right )}{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 = 5.39, size = 288, normalized size = 1.52 \[ \frac {(A b-a B) \left (\frac {b \left (\frac {\left (a^2+b^2\right ) \left (5 a^2+4 a b \tan (c+d x)+b^2\right )}{(a+b \tan (c+d x))^2}+\left (2 b^2-6 a^2\right ) \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^3}+\frac {i \log (-\tan (c+d x)+i)}{(a+i b)^3}-\frac {\log (\tan (c+d x)+i)}{(b+i a)^3}\right )+B \left (\frac {2 b \left (\frac {a^2+b^2}{a+b \tan (c+d x)}-2 a \log (a+b \tan (c+d x))\right )}{\left (a^2+b^2\right )^2}+\frac {i \log (-\tan (c+d x)+i)}{(a+i b)^2}-\frac {i \log (\tan (c+d x)+i)}{(a-i b)^2}\right )-\frac {a B+A b}{b (a+b \tan (c+d x))^2}-\frac {2 B \tan (c+d x)}{(a+b \tan (c+d x))^2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((A*b + a*B)/(b*(a + b*Tan[c + d*x])^2)) - (2*B*Tan[c + d*x])/(a + b*Tan[c + d*x])^2 + B*((I*Log[I - Tan[c +
d*x]])/(a + I*b)^2 - (I*Log[I + Tan[c + d*x]])/(a - I*b)^2 + (2*b*(-2*a*Log[a + b*Tan[c + d*x]] + (a^2 + b^2)
/(a + b*Tan[c + d*x])))/(a^2 + b^2)^2) + (A*b - a*B)*((I*Log[I - Tan[c + d*x]])/(a + I*b)^3 - Log[I + Tan[c +
d*x]]/(I*a + b)^3 + (b*((-6*a^2 + 2*b^2)*Log[a + b*Tan[c + d*x]] + ((a^2 + b^2)*(5*a^2 + b^2 + 4*a*b*Tan[c + d
*x]))/(a + b*Tan[c + d*x])^2))/(a^2 + b^2)^3))/(2*b*d)

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fricas [B]  time = 0.59, size = 478, normalized size = 2.53 \[ \frac {B a^{5} - 3 \, A a^{4} b - 5 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3} - 2 \, {\left (A a^{5} + 3 \, B a^{4} b - 3 \, A a^{3} b^{2} - B a^{2} b^{3}\right )} d x + {\left (B a^{5} + A a^{4} b + 7 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3} - 2 \, {\left (A a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4} - B b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (B a^{5} - 3 \, A a^{4} b - 3 \, B a^{3} b^{2} + A a^{2} b^{3} + {\left (B a^{3} b^{2} - 3 \, A a^{2} b^{3} - 3 \, B a b^{4} + A b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{4} b - 3 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + A a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (A a^{5} + 3 \, B a^{4} b - 3 \, A a^{3} b^{2} - 3 \, B a^{2} b^{3} + 2 \, A a b^{4} - 2 \, {\left (A a^{4} b + 3 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} - B a b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.03, size = 410, normalized size = 2.17 \[ -\frac {\frac {2 \, {\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} + \frac {3 \, B a^{3} b^{4} \tan \left (d x + c\right )^{2} - 9 \, A a^{2} b^{5} \tan \left (d x + c\right )^{2} - 9 \, B a b^{6} \tan \left (d x + c\right )^{2} + 3 \, A b^{7} \tan \left (d x + c\right )^{2} + 2 \, B a^{6} b \tan \left (d x + c\right ) + 14 \, B a^{4} b^{3} \tan \left (d x + c\right ) - 22 \, A a^{3} b^{4} \tan \left (d x + c\right ) - 12 \, B a^{2} b^{5} \tan \left (d x + c\right ) + 2 \, A a b^{6} \tan \left (d x + c\right ) + B a^{7} + A a^{6} b + 9 \, B a^{5} b^{2} - 11 \, A a^{4} b^{3} - 4 \, B a^{3} b^{4}}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(A*a^3 + 3*B*a^2*b - 3*A*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*A*a
^2*b - 3*B*a*b^2 + A*b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(B*a^3*b - 3*A*a^2*b
^2 - 3*B*a*b^3 + A*b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) + (3*B*a^3*b^4*tan(
d*x + c)^2 - 9*A*a^2*b^5*tan(d*x + c)^2 - 9*B*a*b^6*tan(d*x + c)^2 + 3*A*b^7*tan(d*x + c)^2 + 2*B*a^6*b*tan(d*
x + c) + 14*B*a^4*b^3*tan(d*x + c) - 22*A*a^3*b^4*tan(d*x + c) - 12*B*a^2*b^5*tan(d*x + c) + 2*A*a*b^6*tan(d*x
+ c) + B*a^7 + A*a^6*b + 9*B*a^5*b^2 - 11*A*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 = 0.29, size = 495, normalized size = 2.62 \[ -\frac {a^{2} A}{2 d b \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {a^{3} B}{2 d \,b^{2} \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {3 b \,a^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) A}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) A \,b^{3}}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {a^{3} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 \ln \left (a +b \tan \left (d x +c \right )\right ) B a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {2 a b A}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{4} B}{d \,b^{2} \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {3 a^{2} B}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) A \,b^{3}}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} B}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {A \arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 A \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*a^2/b/(a^2+b^2)/(a+b*tan(d*x+c))^2*A+1/2/d*a^3/b^2/(a^2+b^2)/(a+b*tan(d*x+c))^2*B-3/d*b*a^2/(a^2+b^2)^3
*ln(a+b*tan(d*x+c))*A+1/d/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*A*b^3+1/d*a^3/(a^2+b^2)^3*ln(a+b*tan(d*x+c))*B-3/d/(a
^2+b^2)^3*ln(a+b*tan(d*x+c))*B*a*b^2+2/d*a/(a^2+b^2)^2*b/(a+b*tan(d*x+c))*A-1/d/b^2*a^4/(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+3/2/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*A*a^2*b-1/2/d/(a^2+b^2)^3
*ln(1+tan(d*x+c)^2)*A*b^3-1/2/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^2-1/d/(a^2+b^2)^3*A*arctan(tan(d*x+c))*a^3+3/d/(a^2+b^2)^3*A*arctan(tan(d*x+c))*a*b^2-3/d/(a^2+b^2)^3*B*arct
an(tan(d*x+c))*a^2*b+1/d/(a^2+b^2)^3*B*arctan(tan(d*x+c))*b^3

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maxima [A]  time = 0.88, size = 333, normalized size = 1.76 \[ -\frac {\frac {2 \, {\left (A a^{3} + 3 \, B a^{2} b - 3 \, A a b^{2} - B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {2 \, {\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {{\left (B a^{3} - 3 \, A a^{2} b - 3 \, B a b^{2} + A b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {B a^{5} + A a^{4} b + 5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} + 2 \, {\left (B a^{4} b + 3 \, B a^{2} b^{3} - 2 \, A a b^{4}\right )} \tan \left (d x + c\right )}{a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.67, size = 280, normalized size = 1.48 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^3-3\,A\,a^2\,b-3\,B\,a\,b^2+A\,b^3\right )}{d\,{\left (a^2+b^2\right )}^3}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}-\frac {\frac {a\,\left (B\,a^4+A\,a^3\,b+5\,B\,a^2\,b^2-3\,A\,a\,b^3\right )}{2\,b^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^4+3\,B\,a^2\,b^2-2\,A\,a\,b^3\right )}{b\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)

[Out]

Exception raised: AttributeError

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