3.17 \(\int (a+b \tan (c+d x))^3 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=165 \[ \frac {b \left (a^2 B-2 a b C-b^2 B\right ) \tan (c+d x)}{d}-\frac {\left (a^3 B-3 a^2 b C-3 a b^2 B+b^3 C\right ) \log (\cos (c+d x))}{d}-x \left (a^3 C+3 a^2 b B-3 a b^2 C-b^3 B\right )+\frac {(a B-b C) (a+b \tan (c+d x))^2}{2 d}+\frac {B (a+b \tan (c+d x))^3}{3 d}+\frac {C (a+b \tan (c+d x))^4}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3630, 3528, 3525, 3475} \[ \frac {b \left (a^2 B-2 a b C-b^2 B\right ) \tan (c+d x)}{d}-\frac {\left (-3 a^2 b C+a^3 B-3 a b^2 B+b^3 C\right ) \log (\cos (c+d x))}{d}-x \left (3 a^2 b B+a^3 C-3 a b^2 C-b^3 B\right )+\frac {(a B-b C) (a+b \tan (c+d x))^2}{2 d}+\frac {B (a+b \tan (c+d x))^3}{3 d}+\frac {C (a+b \tan (c+d x))^4}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a^2*b*B - b^3*B + a^3*C - 3*a*b^2*C)*x) - ((a^3*B - 3*a*b^2*B - 3*a^2*b*C + b^3*C)*Log[Cos[c + d*x]])/d +
(b*(a^2*B - b^2*B - 2*a*b*C)*Tan[c + d*x])/d + ((a*B - b*C)*(a + b*Tan[c + d*x])^2)/(2*d) + (B*(a + b*Tan[c +
d*x])^3)/(3*d) + (C*(a + b*Tan[c + d*x])^4)/(4*b*d)

Rule 3475

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

Rule 3525

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

Rule 3528

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

Rule 3630

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

Rubi steps

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

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Mathematica [C]  time = 1.63, size = 209, normalized size = 1.27 \[ \frac {-12 b^2 B \left (b^2-6 a^2\right ) \tan (c+d x)+24 a b^3 B \tan ^2(c+d x)-6 (a B+b C) \left (6 a b^2 \tan (c+d x)+(-b+i a)^3 \log (-\tan (c+d x)+i)-(b+i a)^3 \log (\tan (c+d x)+i)+b^3 \tan ^2(c+d x)\right )+6 i B (a-i b)^4 \log (\tan (c+d x)+i)-6 i B (a+i b)^4 \log (-\tan (c+d x)+i)+3 C (a+b \tan (c+d x))^4+4 b^4 B \tan ^3(c+d x)}{12 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*I)*(a + I*b)^4*B*Log[I - Tan[c + d*x]] + (6*I)*(a - I*b)^4*B*Log[I + Tan[c + d*x]] - 12*b^2*(-6*a^2 + b^2
)*B*Tan[c + d*x] + 24*a*b^3*B*Tan[c + d*x]^2 + 4*b^4*B*Tan[c + d*x]^3 + 3*C*(a + b*Tan[c + d*x])^4 - 6*(a*B +
b*C)*((I*a - b)^3*Log[I - Tan[c + d*x]] - (I*a + b)^3*Log[I + Tan[c + d*x]] + 6*a*b^2*Tan[c + d*x] + b^3*Tan[c
+ d*x]^2))/(12*b*d)

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fricas [A]  time = 0.59, size = 178, normalized size = 1.08 \[ \frac {3 \, C b^{3} \tan \left (d x + c\right )^{4} + 4 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )^{3} - 12 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} d x + 6 \, {\left (3 \, C a^{2} b + 3 \, B a b^{2} - C b^{3}\right )} \tan \left (d x + c\right )^{2} - 6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 12 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*C*b^3*tan(d*x + c)^4 + 4*(3*C*a*b^2 + B*b^3)*tan(d*x + c)^3 - 12*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^
3)*d*x + 6*(3*C*a^2*b + 3*B*a*b^2 - C*b^3)*tan(d*x + c)^2 - 6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*log(1/(t
an(d*x + c)^2 + 1)) + 12*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*tan(d*x + c))/d

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giac [B]  time = 12.65, size = 2870, normalized size = 17.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*C*a^3*d*x*tan(d*x)^4*tan(c)^4 + 36*B*a^2*b*d*x*tan(d*x)^4*tan(c)^4 - 36*C*a*b^2*d*x*tan(d*x)^4*tan(c
)^4 - 12*B*b^3*d*x*tan(d*x)^4*tan(c)^4 + 6*B*a^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2
*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 - 18*C*a^2*b*log(4*(tan(d*
x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1)
)*tan(d*x)^4*tan(c)^4 - 18*B*a*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + ta
n(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 6*C*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2
*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^4*tan(
c)^4 - 48*C*a^3*d*x*tan(d*x)^3*tan(c)^3 - 144*B*a^2*b*d*x*tan(d*x)^3*tan(c)^3 + 144*C*a*b^2*d*x*tan(d*x)^3*tan
(c)^3 + 48*B*b^3*d*x*tan(d*x)^3*tan(c)^3 - 18*C*a^2*b*tan(d*x)^4*tan(c)^4 - 18*B*a*b^2*tan(d*x)^4*tan(c)^4 + 9
*C*b^3*tan(d*x)^4*tan(c)^4 - 24*B*a^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 +
tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 72*C*a^2*b*log(4*(tan(d*x)^4*tan(c)
^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^
3*tan(c)^3 + 72*B*a*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 -
2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 - 24*C*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^
3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^3*tan(c)^3 + 12*
C*a^3*tan(d*x)^4*tan(c)^3 + 36*B*a^2*b*tan(d*x)^4*tan(c)^3 - 36*C*a*b^2*tan(d*x)^4*tan(c)^3 - 12*B*b^3*tan(d*x
)^4*tan(c)^3 + 12*C*a^3*tan(d*x)^3*tan(c)^4 + 36*B*a^2*b*tan(d*x)^3*tan(c)^4 - 36*C*a*b^2*tan(d*x)^3*tan(c)^4
- 12*B*b^3*tan(d*x)^3*tan(c)^4 + 72*C*a^3*d*x*tan(d*x)^2*tan(c)^2 + 216*B*a^2*b*d*x*tan(d*x)^2*tan(c)^2 - 216*
C*a*b^2*d*x*tan(d*x)^2*tan(c)^2 - 72*B*b^3*d*x*tan(d*x)^2*tan(c)^2 - 18*C*a^2*b*tan(d*x)^4*tan(c)^2 - 18*B*a*b
^2*tan(d*x)^4*tan(c)^2 + 6*C*b^3*tan(d*x)^4*tan(c)^2 + 36*C*a^2*b*tan(d*x)^3*tan(c)^3 + 36*B*a*b^2*tan(d*x)^3*
tan(c)^3 - 24*C*b^3*tan(d*x)^3*tan(c)^3 - 18*C*a^2*b*tan(d*x)^2*tan(c)^4 - 18*B*a*b^2*tan(d*x)^2*tan(c)^4 + 6*
C*b^3*tan(d*x)^2*tan(c)^4 + 12*C*a*b^2*tan(d*x)^4*tan(c) + 4*B*b^3*tan(d*x)^4*tan(c) + 36*B*a^3*log(4*(tan(d*x
)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))
*tan(d*x)^2*tan(c)^2 - 108*C*a^2*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + ta
n(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 108*B*a*b^2*log(4*(tan(d*x)^4*tan(c)^2
- 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*
tan(c)^2 + 36*C*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*ta
n(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 - 36*C*a^3*tan(d*x)^3*tan(c)^2 - 108*B*a^2*b*tan(d*x)^3
*tan(c)^2 + 144*C*a*b^2*tan(d*x)^3*tan(c)^2 + 48*B*b^3*tan(d*x)^3*tan(c)^2 - 36*C*a^3*tan(d*x)^2*tan(c)^3 - 10
8*B*a^2*b*tan(d*x)^2*tan(c)^3 + 144*C*a*b^2*tan(d*x)^2*tan(c)^3 + 48*B*b^3*tan(d*x)^2*tan(c)^3 + 12*C*a*b^2*ta
n(d*x)*tan(c)^4 + 4*B*b^3*tan(d*x)*tan(c)^4 - 3*C*b^3*tan(d*x)^4 - 48*C*a^3*d*x*tan(d*x)*tan(c) - 144*B*a^2*b*
d*x*tan(d*x)*tan(c) + 144*C*a*b^2*d*x*tan(d*x)*tan(c) + 48*B*b^3*d*x*tan(d*x)*tan(c) + 36*C*a^2*b*tan(d*x)^3*t
an(c) + 36*B*a*b^2*tan(d*x)^3*tan(c) - 24*C*b^3*tan(d*x)^3*tan(c) - 36*C*a^2*b*tan(d*x)^2*tan(c)^2 - 36*B*a*b^
2*tan(d*x)^2*tan(c)^2 + 12*C*b^3*tan(d*x)^2*tan(c)^2 + 36*C*a^2*b*tan(d*x)*tan(c)^3 + 36*B*a*b^2*tan(d*x)*tan(
c)^3 - 24*C*b^3*tan(d*x)*tan(c)^3 - 3*C*b^3*tan(c)^4 - 12*C*a*b^2*tan(d*x)^3 - 4*B*b^3*tan(d*x)^3 - 24*B*a^3*l
og(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(t
an(c)^2 + 1))*tan(d*x)*tan(c) + 72*C*a^2*b*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c
)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)*tan(c) + 72*B*a*b^2*log(4*(tan(d*x)^4*tan(c
)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)
*tan(c) - 24*C*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan
(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)*tan(c) + 36*C*a^3*tan(d*x)^2*tan(c) + 108*B*a^2*b*tan(d*x)^2*tan(c)
- 144*C*a*b^2*tan(d*x)^2*tan(c) - 48*B*b^3*tan(d*x)^2*tan(c) + 36*C*a^3*tan(d*x)*tan(c)^2 + 108*B*a^2*b*tan(d
*x)*tan(c)^2 - 144*C*a*b^2*tan(d*x)*tan(c)^2 - 48*B*b^3*tan(d*x)*tan(c)^2 - 12*C*a*b^2*tan(c)^3 - 4*B*b^3*tan(
c)^3 + 12*C*a^3*d*x + 36*B*a^2*b*d*x - 36*C*a*b^2*d*x - 12*B*b^3*d*x - 18*C*a^2*b*tan(d*x)^2 - 18*B*a*b^2*tan(
d*x)^2 + 6*C*b^3*tan(d*x)^2 + 36*C*a^2*b*tan(d*x)*tan(c) + 36*B*a*b^2*tan(d*x)*tan(c) - 24*C*b^3*tan(d*x)*tan(
c) - 18*C*a^2*b*tan(c)^2 - 18*B*a*b^2*tan(c)^2 + 6*C*b^3*tan(c)^2 + 6*B*a^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan
(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1)) - 18*C*a^2*b*log(4*
(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)
^2 + 1)) - 18*B*a*b^2*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*
tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1)) + 6*C*b^3*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*t
an(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1)) - 12*C*a^3*tan(d*x) - 36*B*a^2*b*tan(d*x) + 36*C
*a*b^2*tan(d*x) + 12*B*b^3*tan(d*x) - 12*C*a^3*tan(c) - 36*B*a^2*b*tan(c) + 36*C*a*b^2*tan(c) + 12*B*b^3*tan(c
) - 18*C*a^2*b - 18*B*a*b^2 + 9*C*b^3)/(d*tan(d*x)^4*tan(c)^4 - 4*d*tan(d*x)^3*tan(c)^3 + 6*d*tan(d*x)^2*tan(c
)^2 - 4*d*tan(d*x)*tan(c) + d)

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maple [A]  time = 0.03, size = 314, normalized size = 1.90 \[ \frac {b^{3} C \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {B \left (\tan ^{3}\left (d x +c \right )\right ) b^{3}}{3 d}+\frac {C \left (\tan ^{3}\left (d x +c \right )\right ) a \,b^{2}}{d}+\frac {3 B \left (\tan ^{2}\left (d x +c \right )\right ) a \,b^{2}}{2 d}+\frac {3 C \left (\tan ^{2}\left (d x +c \right )\right ) a^{2} b}{2 d}-\frac {b^{3} C \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {3 B \tan \left (d x +c \right ) a^{2} b}{d}-\frac {B \tan \left (d x +c \right ) b^{3}}{d}+\frac {C \tan \left (d x +c \right ) a^{3}}{d}-\frac {3 C a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {a^{3} B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) B a \,b^{2}}{2 d}-\frac {3 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) C \,a^{2} b}{2 d}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right ) b^{3} C}{2 d}-\frac {3 B \arctan \left (\tan \left (d x +c \right )\right ) a^{2} b}{d}+\frac {B \arctan \left (\tan \left (d x +c \right )\right ) b^{3}}{d}-\frac {C \arctan \left (\tan \left (d x +c \right )\right ) a^{3}}{d}+\frac {3 C \arctan \left (\tan \left (d x +c \right )\right ) a \,b^{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.52, size = 179, normalized size = 1.08 \[ \frac {3 \, C b^{3} \tan \left (d x + c\right )^{4} + 4 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \tan \left (d x + c\right )^{3} + 6 \, {\left (3 \, C a^{2} b + 3 \, B a b^{2} - C b^{3}\right )} \tan \left (d x + c\right )^{2} - 12 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} {\left (d x + c\right )} + 6 \, {\left (B a^{3} - 3 \, C a^{2} b - 3 \, B a b^{2} + C b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (C a^{3} + 3 \, B a^{2} b - 3 \, C a b^{2} - B b^{3}\right )} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*C*b^3*tan(d*x + c)^4 + 4*(3*C*a*b^2 + B*b^3)*tan(d*x + c)^3 + 6*(3*C*a^2*b + 3*B*a*b^2 - C*b^3)*tan(d*
x + c)^2 - 12*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*(d*x + c) + 6*(B*a^3 - 3*C*a^2*b - 3*B*a*b^2 + C*b^3)*lo
g(tan(d*x + c)^2 + 1) + 12*(C*a^3 + 3*B*a^2*b - 3*C*a*b^2 - B*b^3)*tan(d*x + c))/d

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mupad [B]  time = 8.83, size = 181, normalized size = 1.10 \[ x\,\left (-C\,a^3-3\,B\,a^2\,b+3\,C\,a\,b^2+B\,b^3\right )-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (\frac {C\,b^3}{2}-\frac {3\,a\,b\,\left (B\,b+C\,a\right )}{2}\right )}{d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (-C\,a^3-3\,B\,a^2\,b+3\,C\,a\,b^2+B\,b^3\right )}{d}+\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {B\,a^3}{2}-\frac {3\,C\,a^2\,b}{2}-\frac {3\,B\,a\,b^2}{2}+\frac {C\,b^3}{2}\right )}{d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (\frac {B\,b^3}{3}+C\,a\,b^2\right )}{d}+\frac {C\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4\,d} \]

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]

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

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sympy [A]  time = 0.67, size = 313, normalized size = 1.90 \[ \begin {cases} \frac {B a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 3 B a^{2} b x + \frac {3 B a^{2} b \tan {\left (c + d x \right )}}{d} - \frac {3 B a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + B b^{3} x + \frac {B b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {B b^{3} \tan {\left (c + d x \right )}}{d} - C a^{3} x + \frac {C a^{3} \tan {\left (c + d x \right )}}{d} - \frac {3 C a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 C a^{2} b \tan ^{2}{\left (c + d x \right )}}{2 d} + 3 C a b^{2} x + \frac {C a b^{2} \tan ^{3}{\left (c + d x \right )}}{d} - \frac {3 C a b^{2} \tan {\left (c + d x \right )}}{d} + \frac {C b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C b^{3} \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {C b^{3} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right )^{3} \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**3*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Piecewise((B*a**3*log(tan(c + d*x)**2 + 1)/(2*d) - 3*B*a**2*b*x + 3*B*a**2*b*tan(c + d*x)/d - 3*B*a*b**2*log(t
an(c + d*x)**2 + 1)/(2*d) + 3*B*a*b**2*tan(c + d*x)**2/(2*d) + B*b**3*x + B*b**3*tan(c + d*x)**3/(3*d) - B*b**
3*tan(c + d*x)/d - C*a**3*x + C*a**3*tan(c + d*x)/d - 3*C*a**2*b*log(tan(c + d*x)**2 + 1)/(2*d) + 3*C*a**2*b*t
an(c + d*x)**2/(2*d) + 3*C*a*b**2*x + C*a*b**2*tan(c + d*x)**3/d - 3*C*a*b**2*tan(c + d*x)/d + C*b**3*log(tan(
c + d*x)**2 + 1)/(2*d) + C*b**3*tan(c + d*x)**4/(4*d) - C*b**3*tan(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan
(c))**3*(B*tan(c) + C*tan(c)**2), True))

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