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

Optimal. Leaf size=179 \[ \frac {a (b B-a C)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {a^2 B+2 a b C-b^2 B}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {\left (a^3 B+3 a^2 b C-3 a b^2 B-b^3 C\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {x \left (a^3 (-C)+3 a^2 b B+3 a b^2 C-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

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

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Rubi [A]  time = 0.25, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3628, 3529, 3531, 3530} \[ \frac {a (b B-a C)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {a^2 B+2 a b C-b^2 B}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {\left (3 a^2 b C+a^3 B-3 a b^2 B-b^3 C\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac {x \left (3 a^2 b B+a^3 (-C)+3 a b^2 C-b^3 B\right )}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

Rule 3529

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

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 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 {B \tan (c+d x)+C \tan ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\frac {a (b B-a C)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {\int \frac {b B-a C+(a B+b C) \tan (c+d x)}{(a+b \tan (c+d x))^2} \, dx}{a^2+b^2}\\ &=\frac {a (b B-a C)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^2 B-b^2 B+2 a b C}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}+\frac {\int \frac {2 a b B-a^2 C+b^2 C+\left (a^2 B-b^2 B+2 a b C\right ) \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\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)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^2 B-b^2 B+2 a b C}{\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 \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{\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 (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac {a (b B-a C)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}+\frac {a^2 B-b^2 B+2 a b C}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 4.07, size = 188, normalized size = 1.05 \[ \frac {\frac {a (b B-a C)}{b \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {2 \left (a^2 B+2 a b C-b^2 B\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac {2 \left (a^3 B+3 a^2 b C-3 a b^2 B-b^3 C\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3}+\frac {(B+i C) \log (-\tan (c+d x)+i)}{(a+i b)^3}+\frac {(B-i C) \log (\tan (c+d x)+i)}{(a-i b)^3}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((B + I*C)*Log[I - Tan[c + d*x]])/(a + I*b)^3 + ((B - I*C)*Log[I + Tan[c + d*x]])/(a - I*b)^3 - (2*(a^3*B - 3
*a*b^2*B + 3*a^2*b*C - b^3*C)*Log[a + b*Tan[c + d*x]])/(a^2 + b^2)^3 + (a*(b*B - a*C))/(b*(a^2 + b^2)*(a + b*T
an[c + d*x])^2) + (2*(a^2*B - b^2*B + 2*a*b*C))/((a^2 + b^2)^2*(a + b*Tan[c + d*x])))/(2*d)

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fricas [B]  time = 0.90, size = 488, normalized size = 2.73 \[ -\frac {3 \, C a^{4} b - 5 \, B a^{3} b^{2} - 3 \, C a^{2} b^{3} + B a b^{4} + 2 \, {\left (C a^{5} - 3 \, B a^{4} b - 3 \, C a^{3} b^{2} + B a^{2} b^{3}\right )} d x - {\left (C a^{4} b - 3 \, B a^{3} b^{2} - 5 \, C a^{2} b^{3} + 3 \, B a b^{4} - 2 \, {\left (C a^{3} b^{2} - 3 \, B a^{2} b^{3} - 3 \, C a b^{4} + B b^{5}\right )} d x\right )} \tan \left (d x + c\right )^{2} + {\left (B a^{5} + 3 \, C a^{4} b - 3 \, B a^{3} b^{2} - C a^{2} b^{3} + {\left (B a^{3} b^{2} + 3 \, C a^{2} b^{3} - 3 \, B a b^{4} - C b^{5}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{4} b + 3 \, C a^{3} b^{2} - 3 \, B a^{2} b^{3} - C a b^{4}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - 2 \, {\left (C a^{5} - 2 \, B a^{4} b - 3 \, C a^{3} b^{2} + 3 \, B a^{2} b^{3} + 2 \, C a b^{4} - B b^{5} - 2 \, {\left (C a^{4} b - 3 \, B a^{3} b^{2} - 3 \, C a^{2} b^{3} + B a b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \tan \left (d x + c\right ) + {\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6}\right )} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(3*C*a^4*b - 5*B*a^3*b^2 - 3*C*a^2*b^3 + B*a*b^4 + 2*(C*a^5 - 3*B*a^4*b - 3*C*a^3*b^2 + B*a^2*b^3)*d*x -
(C*a^4*b - 3*B*a^3*b^2 - 5*C*a^2*b^3 + 3*B*a*b^4 - 2*(C*a^3*b^2 - 3*B*a^2*b^3 - 3*C*a*b^4 + B*b^5)*d*x)*tan(d*
x + c)^2 + (B*a^5 + 3*C*a^4*b - 3*B*a^3*b^2 - C*a^2*b^3 + (B*a^3*b^2 + 3*C*a^2*b^3 - 3*B*a*b^4 - C*b^5)*tan(d*
x + c)^2 + 2*(B*a^4*b + 3*C*a^3*b^2 - 3*B*a^2*b^3 - C*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*(C*a^5 - 2*B*a^4*b - 3*C*a^3*b^2 + 3*B*a^2*b^3 + 2*C*a*b^4 - B*b^5
- 2*(C*a^4*b - 3*B*a^3*b^2 - 3*C*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 = 2.65, size = 410, normalized size = 2.29 \[ -\frac {\frac {2 \, {\left (C a^{3} - 3 \, B a^{2} b - 3 \, C a b^{2} + B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (B a^{3} + 3 \, C a^{2} b - 3 \, B a b^{2} - C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (B a^{3} b + 3 \, C a^{2} b^{2} - 3 \, B a b^{3} - C b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac {3 \, B a^{3} b^{3} \tan \left (d x + c\right )^{2} + 9 \, C a^{2} b^{4} \tan \left (d x + c\right )^{2} - 9 \, B a b^{5} \tan \left (d x + c\right )^{2} - 3 \, C b^{6} \tan \left (d x + c\right )^{2} + 8 \, B a^{4} b^{2} \tan \left (d x + c\right ) + 22 \, C a^{3} b^{3} \tan \left (d x + c\right ) - 18 \, B a^{2} b^{4} \tan \left (d x + c\right ) - 2 \, C a b^{5} \tan \left (d x + c\right ) - 2 \, B b^{6} \tan \left (d x + c\right ) - C a^{6} + 6 \, B a^{5} b + 11 \, C a^{4} b^{2} - 7 \, B a^{3} b^{3} - B a b^{5}}{{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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*(B*a^3*b + 3*C*a^2*b
^2 - 3*B*a*b^3 - C*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^3*tan(
d*x + c)^2 + 9*C*a^2*b^4*tan(d*x + c)^2 - 9*B*a*b^5*tan(d*x + c)^2 - 3*C*b^6*tan(d*x + c)^2 + 8*B*a^4*b^2*tan(
d*x + c) + 22*C*a^3*b^3*tan(d*x + c) - 18*B*a^2*b^4*tan(d*x + c) - 2*C*a*b^5*tan(d*x + c) - 2*B*b^6*tan(d*x +
c) - C*a^6 + 6*B*a^5*b + 11*C*a^4*b^2 - 7*B*a^3*b^3 - B*a*b^5)/((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*(b*tan(d
*x + c) + a)^2))/d

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maple [B]  time = 0.29, size = 488, normalized size = 2.73 \[ \frac {a B}{2 d \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (d x +c \right )\right )^{2}}-\frac {a^{2} C}{2 d \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {a^{2} B}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {b^{2} B}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}+\frac {2 C a b}{d \left (a^{2}+b^{2}\right )^{2} \left (a +b \tan \left (d x +c \right )\right )}-\frac {a^{3} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 a \,b^{2} \ln \left (a +b \tan \left (d x +c \right )\right ) B}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 a^{2} b \ln \left (a +b \tan \left (d x +c \right )\right ) C}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (a +b \tan \left (d x +c \right )\right ) b^{3} C}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) a^{3} B}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a \,b^{2}}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) C \,a^{2} b}{2 d \left (a^{2}+b^{2}\right )^{3}}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3} C}{2 d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 B \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d \left (a^{2}+b^{2}\right )^{3}}-\frac {C \arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d \left (a^{2}+b^{2}\right )^{3}}+\frac {3 C \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d \left (a^{2}+b^{2}\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*a/(a^2+b^2)/(a+b*tan(d*x+c))^2*B-1/2/d*a^2/(a^2+b^2)/b/(a+b*tan(d*x+c))^2*C+1/d*a^2/(a^2+b^2)^2/(a+b*tan
(d*x+c))*B-1/d/(a^2+b^2)^2/(a+b*tan(d*x+c))*b^2*B+2/d/(a^2+b^2)^2/(a+b*tan(d*x+c))*C*a*b-1/d*a^3/(a^2+b^2)^3*l
n(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-3/d*a^2/(a^2+b^2)^3*b*ln(a+b*tan(d*x+c))*C+1/d/
(a^2+b^2)^3*ln(a+b*tan(d*x+c))*b^3*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 = 0.64, size = 330, normalized size = 1.84 \[ -\frac {\frac {2 \, {\left (C a^{3} - 3 \, B a^{2} b - 3 \, C a b^{2} + B b^{3}\right )} {\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {2 \, {\left (B a^{3} + 3 \, C a^{2} b - 3 \, B a b^{2} - C b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (B a^{3} + 3 \, C a^{2} b - 3 \, B a b^{2} - C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {C a^{4} - 3 \, B a^{3} b - 3 \, C a^{2} b^{2} + B a b^{3} - 2 \, {\left (B a^{2} b^{2} + 2 \, C a b^{3} - B b^{4}\right )} \tan \left (d x + c\right )}{a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5} + {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 9.28, size = 282, normalized size = 1.58 \[ \frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^2\,b+2\,C\,a\,b^2-B\,b^3\right )}{a^4+2\,a^2\,b^2+b^4}-\frac {C\,a^4-3\,B\,a^3\,b-3\,C\,a^2\,b^2+B\,a\,b^3}{2\,b\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )+b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )}-\frac {\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (\frac {B\,a+3\,C\,b}{{\left (a^2+b^2\right )}^2}-\frac {4\,b^2\,\left (B\,a+C\,b\right )}{{\left (a^2+b^2\right )}^3}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3\,1{}\mathrm {i}+3\,a^2\,b+a\,b^2\,3{}\mathrm {i}-b^3\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )}{2\,d\,\left (-a^3+a^2\,b\,3{}\mathrm {i}+3\,a\,b^2-b^3\,1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((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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