3.50 \(\int (a+b \tan (e+f x))^3 (c+d \tan (e+f x)) (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=353 \[ \frac{b \tan (e+f x) \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac{\log (\cos (e+f x)) \left (3 a^2 b (A c-B d-c C)+a^3 (d (A-C)+B c)-3 a b^2 (d (A-C)+B c)-b^3 (A c-B d-c C)\right )}{f}+x \left (-3 a^2 b (d (A-C)+B c)+a^3 (A c-B d-c C)-3 a b^2 (A c-B d-c C)+b^3 (d (A-C)+B c)\right )+\frac{(d (A-C)+B c) (a+b \tan (e+f x))^3}{3 f}+\frac{(a+b \tan (e+f x))^2 (a A d+a B c-a C d+A b c-b B d-b c C)}{2 f}-\frac{(a C d-5 b (B d+c C)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f} \]

[Out]

(a^3*(A*c - c*C - B*d) - 3*a*b^2*(A*c - c*C - B*d) - 3*a^2*b*(B*c + (A - C)*d) + b^3*(B*c + (A - C)*d))*x - ((
3*a^2*b*(A*c - c*C - B*d) - b^3*(A*c - c*C - B*d) + a^3*(B*c + (A - C)*d) - 3*a*b^2*(B*c + (A - C)*d))*Log[Cos
[e + f*x]])/f + (b*(2*a*b*(A*c - c*C - B*d) + a^2*(B*c + (A - C)*d) - b^2*(B*c + (A - C)*d))*Tan[e + f*x])/f +
((A*b*c + a*B*c - b*c*C + a*A*d - b*B*d - a*C*d)*(a + b*Tan[e + f*x])^2)/(2*f) + ((B*c + (A - C)*d)*(a + b*Ta
n[e + f*x])^3)/(3*f) - ((a*C*d - 5*b*(c*C + B*d))*(a + b*Tan[e + f*x])^4)/(20*b^2*f) + (C*d*Tan[e + f*x]*(a +
b*Tan[e + f*x])^4)/(5*b*f)

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Rubi [A]  time = 0.785304, antiderivative size = 353, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3637, 3630, 3528, 3525, 3475} \[ \frac{b \tan (e+f x) \left (a^2 (d (A-C)+B c)+2 a b (A c-B d-c C)-b^2 (d (A-C)+B c)\right )}{f}-\frac{\log (\cos (e+f x)) \left (3 a^2 b (A c-B d-c C)+a^3 (d (A-C)+B c)-3 a b^2 (d (A-C)+B c)-b^3 (A c-B d-c C)\right )}{f}+x \left (-3 a^2 b (d (A-C)+B c)+a^3 (A c-B d-c C)-3 a b^2 (A c-B d-c C)+b^3 (d (A-C)+B c)\right )+\frac{(d (A-C)+B c) (a+b \tan (e+f x))^3}{3 f}+\frac{(a+b \tan (e+f x))^2 (a A d+a B c-a C d+A b c-b B d-b c C)}{2 f}-\frac{(a C d-5 b (B d+c C)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(A*c - c*C - B*d) - 3*a*b^2*(A*c - c*C - B*d) - 3*a^2*b*(B*c + (A - C)*d) + b^3*(B*c + (A - C)*d))*x - ((
3*a^2*b*(A*c - c*C - B*d) - b^3*(A*c - c*C - B*d) + a^3*(B*c + (A - C)*d) - 3*a*b^2*(B*c + (A - C)*d))*Log[Cos
[e + f*x]])/f + (b*(2*a*b*(A*c - c*C - B*d) + a^2*(B*c + (A - C)*d) - b^2*(B*c + (A - C)*d))*Tan[e + f*x])/f +
((A*b*c + a*B*c - b*c*C + a*A*d - b*B*d - a*C*d)*(a + b*Tan[e + f*x])^2)/(2*f) + ((B*c + (A - C)*d)*(a + b*Ta
n[e + f*x])^3)/(3*f) - ((a*C*d - 5*b*(c*C + B*d))*(a + b*Tan[e + f*x])^4)/(20*b^2*f) + (C*d*Tan[e + f*x]*(a +
b*Tan[e + f*x])^4)/(5*b*f)

Rule 3637

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*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

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]

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 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 3475

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

Rubi steps

\begin{align*} \int (a+b \tan (e+f x))^3 (c+d \tan (e+f x)) \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac{\int (a+b \tan (e+f x))^3 \left (-5 A b c+a C d-5 b (B c+(A-C) d) \tan (e+f x)+(a C d-5 b (c C+B d)) \tan ^2(e+f x)\right ) \, dx}{5 b}\\ &=-\frac{(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac{\int (a+b \tan (e+f x))^3 (-5 b (A c-c C-B d)-5 b (B c+(A-C) d) \tan (e+f x)) \, dx}{5 b}\\ &=\frac{(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac{(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac{\int (a+b \tan (e+f x))^2 (5 b (b B c+b (A-C) d-a (A c-c C-B d))-5 b (A b c+a B c-b c C+a A d-b B d-a C d) \tan (e+f x)) \, dx}{5 b}\\ &=\frac{(A b c+a B c-b c C+a A d-b B d-a C d) (a+b \tan (e+f x))^2}{2 f}+\frac{(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac{(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac{\int (a+b \tan (e+f x)) \left (-5 b \left (a^2 (A c-c C-B d)-b^2 (A c-c C-B d)-2 a b (B c+(A-C) d)\right )-5 b \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \tan (e+f x)\right ) \, dx}{5 b}\\ &=\left (a^3 (A c-c C-B d)-3 a b^2 (A c-c C-B d)-3 a^2 b (B c+(A-C) d)+b^3 (B c+(A-C) d)\right ) x+\frac{b \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \tan (e+f x)}{f}+\frac{(A b c+a B c-b c C+a A d-b B d-a C d) (a+b \tan (e+f x))^2}{2 f}+\frac{(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac{(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\left (-3 a^2 b (A c-c C-B d)+b^3 (A c-c C-B d)-a^3 (B c+(A-C) d)+3 a b^2 (B c+(A-C) d)\right ) \int \tan (e+f x) \, dx\\ &=\left (a^3 (A c-c C-B d)-3 a b^2 (A c-c C-B d)-3 a^2 b (B c+(A-C) d)+b^3 (B c+(A-C) d)\right ) x-\frac{\left (3 a^2 b (A c-c C-B d)-b^3 (A c-c C-B d)+a^3 (B c+(A-C) d)-3 a b^2 (B c+(A-C) d)\right ) \log (\cos (e+f x))}{f}+\frac{b \left (2 a b (A c-c C-B d)+a^2 (B c+(A-C) d)-b^2 (B c+(A-C) d)\right ) \tan (e+f x)}{f}+\frac{(A b c+a B c-b c C+a A d-b B d-a C d) (a+b \tan (e+f x))^2}{2 f}+\frac{(B c+(A-C) d) (a+b \tan (e+f x))^3}{3 f}-\frac{(a C d-5 b (c C+B d)) (a+b \tan (e+f x))^4}{20 b^2 f}+\frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}\\ \end{align*}

Mathematica [C]  time = 6.36469, size = 300, normalized size = 0.85 \[ \frac{C d \tan (e+f x) (a+b \tan (e+f x))^4}{5 b f}-\frac{\frac{(a C d-5 b (B d+c C)) (a+b \tan (e+f x))^4}{4 b f}-\frac{5 \left (3 (-a A d-a B c+a C d+A b c-b B d-b c C) \left (6 a b^2 \tan (e+f x)+(-b+i a)^3 \log (-\tan (e+f x)+i)-(b+i a)^3 \log (\tan (e+f x)+i)+b^3 \tan ^2(e+f x)\right )-(d (A-C)+B c) \left (-6 b^2 \left (6 a^2-b^2\right ) \tan (e+f x)-12 a b^3 \tan ^2(e+f x)-3 i (a-i b)^4 \log (\tan (e+f x)+i)+3 i (a+i b)^4 \log (-\tan (e+f x)+i)-2 b^4 \tan ^3(e+f x)\right )\right )}{6 f}}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*d*Tan[e + f*x]*(a + b*Tan[e + f*x])^4)/(5*b*f) - (((a*C*d - 5*b*(c*C + B*d))*(a + b*Tan[e + f*x])^4)/(4*b*f
) - (5*(3*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C*d)*((I*a - b)^3*Log[I - Tan[e + f*x]] - (I*a + b)^3*Log
[I + Tan[e + f*x]] + 6*a*b^2*Tan[e + f*x] + b^3*Tan[e + f*x]^2) - (B*c + (A - C)*d)*((3*I)*(a + I*b)^4*Log[I -
Tan[e + f*x]] - (3*I)*(a - I*b)^4*Log[I + Tan[e + f*x]] - 6*b^2*(6*a^2 - b^2)*Tan[e + f*x] - 12*a*b^3*Tan[e +
f*x]^2 - 2*b^4*Tan[e + f*x]^3)))/(6*f))/(5*b)

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Maple [B]  time = 0.018, size = 994, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*B*tan(f*x+e)^4*b^3*d+1/4/f*C*tan(f*x+e)^4*b^3*c+1/3/f*A*tan(f*x+e)^3*b^3*d-1/f*C*arctan(tan(f*x+e))*b^3*
d+1/2/f*ln(1+tan(f*x+e)^2)*A*a^3*d-1/2/f*ln(1+tan(f*x+e)^2)*A*b^3*c+1/2/f*ln(1+tan(f*x+e)^2)*B*a^3*c+1/2/f*ln(
1+tan(f*x+e)^2)*B*b^3*d+1/2/f*ln(1+tan(f*x+e)^2)*C*b^3*c+1/f*C*a^3*c*tan(f*x+e)+3/f*C*arctan(tan(f*x+e))*a*b^2
*c+3/f*A*a^2*b*d*tan(f*x+e)+3/2/f*ln(1+tan(f*x+e)^2)*C*a*b^2*d+1/f*C*tan(f*x+e)^3*a*b^2*c+1/f*B*tan(f*x+e)^3*a
*b^2*d+1/f*A*arctan(tan(f*x+e))*a^3*c+1/f*A*arctan(tan(f*x+e))*b^3*d-1/f*B*arctan(tan(f*x+e))*a^3*d+1/f*B*arct
an(tan(f*x+e))*b^3*c-1/f*C*arctan(tan(f*x+e))*a^3*c+1/f*B*a^3*d*tan(f*x+e)-1/f*B*b^3*c*tan(f*x+e)+1/3/f*B*tan(
f*x+e)^3*b^3*c-1/3/f*C*tan(f*x+e)^3*b^3*d+1/f*C*b^3*d*tan(f*x+e)-1/f*A*b^3*d*tan(f*x+e)+1/2/f*C*tan(f*x+e)^2*a
^3*d-1/2/f*C*tan(f*x+e)^2*b^3*c+1/2/f*A*tan(f*x+e)^2*b^3*c-1/2/f*B*tan(f*x+e)^2*b^3*d+1/5/f*C*b^3*d*tan(f*x+e)
^5-1/2/f*ln(1+tan(f*x+e)^2)*a^3*C*d-3/f*A*arctan(tan(f*x+e))*a^2*b*d-3/f*C*a^2*b*d*tan(f*x+e)+3/2/f*B*tan(f*x+
e)^2*a*b^2*c+3/2/f*A*tan(f*x+e)^2*a*b^2*d+3/f*A*a*b^2*c*tan(f*x+e)+3/f*B*a^2*b*c*tan(f*x+e)-3/f*B*a*b^2*d*tan(
f*x+e)-3/f*A*arctan(tan(f*x+e))*a*b^2*c-3/f*B*arctan(tan(f*x+e))*a^2*b*c+3/f*B*arctan(tan(f*x+e))*a*b^2*d+1/f*
C*tan(f*x+e)^3*a^2*b*d-3/2/f*ln(1+tan(f*x+e)^2)*B*a*b^2*c-3/2/f*ln(1+tan(f*x+e)^2)*C*a^2*b*c+3/4/f*C*tan(f*x+e
)^4*a*b^2*d+3/f*C*arctan(tan(f*x+e))*a^2*b*d-3/f*C*a*b^2*c*tan(f*x+e)+3/2/f*B*tan(f*x+e)^2*a^2*b*d-3/2/f*C*tan
(f*x+e)^2*a*b^2*d+3/2/f*C*tan(f*x+e)^2*a^2*b*c+3/2/f*ln(1+tan(f*x+e)^2)*A*a^2*b*c-3/2/f*ln(1+tan(f*x+e)^2)*A*a
*b^2*d-3/2/f*ln(1+tan(f*x+e)^2)*B*a^2*b*d

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Maxima [A]  time = 1.50679, size = 562, normalized size = 1.59 \begin{align*} \frac{12 \, C b^{3} d \tan \left (f x + e\right )^{5} + 15 \,{\left (C b^{3} c +{\left (3 \, C a b^{2} + B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{4} + 20 \,{\left ({\left (3 \, C a b^{2} + B b^{3}\right )} c +{\left (3 \, C a^{2} b + 3 \, B a b^{2} +{\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )^{3} + 30 \,{\left ({\left (3 \, C a^{2} b + 3 \, B a b^{2} +{\left (A - C\right )} b^{3}\right )} c +{\left (C a^{3} + 3 \, B a^{2} b + 3 \,{\left (A - C\right )} a b^{2} - B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{2} + 60 \,{\left ({\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \,{\left (A - C\right )} a b^{2} + B b^{3}\right )} c -{\left (B a^{3} + 3 \,{\left (A - C\right )} a^{2} b - 3 \, B a b^{2} -{\left (A - C\right )} b^{3}\right )} d\right )}{\left (f x + e\right )} + 30 \,{\left ({\left (B a^{3} + 3 \,{\left (A - C\right )} a^{2} b - 3 \, B a b^{2} -{\left (A - C\right )} b^{3}\right )} c +{\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \,{\left (A - C\right )} a b^{2} + B b^{3}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 60 \,{\left ({\left (C a^{3} + 3 \, B a^{2} b + 3 \,{\left (A - C\right )} a b^{2} - B b^{3}\right )} c +{\left (B a^{3} + 3 \,{\left (A - C\right )} a^{2} b - 3 \, B a b^{2} -{\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*C*b^3*d*tan(f*x + e)^5 + 15*(C*b^3*c + (3*C*a*b^2 + B*b^3)*d)*tan(f*x + e)^4 + 20*((3*C*a*b^2 + B*b^3
)*c + (3*C*a^2*b + 3*B*a*b^2 + (A - C)*b^3)*d)*tan(f*x + e)^3 + 30*((3*C*a^2*b + 3*B*a*b^2 + (A - C)*b^3)*c +
(C*a^3 + 3*B*a^2*b + 3*(A - C)*a*b^2 - B*b^3)*d)*tan(f*x + e)^2 + 60*(((A - C)*a^3 - 3*B*a^2*b - 3*(A - C)*a*b
^2 + B*b^3)*c - (B*a^3 + 3*(A - C)*a^2*b - 3*B*a*b^2 - (A - C)*b^3)*d)*(f*x + e) + 30*((B*a^3 + 3*(A - C)*a^2*
b - 3*B*a*b^2 - (A - C)*b^3)*c + ((A - C)*a^3 - 3*B*a^2*b - 3*(A - C)*a*b^2 + B*b^3)*d)*log(tan(f*x + e)^2 + 1
) + 60*((C*a^3 + 3*B*a^2*b + 3*(A - C)*a*b^2 - B*b^3)*c + (B*a^3 + 3*(A - C)*a^2*b - 3*B*a*b^2 - (A - C)*b^3)*
d)*tan(f*x + e))/f

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Fricas [A]  time = 1.22339, size = 915, normalized size = 2.59 \begin{align*} \frac{12 \, C b^{3} d \tan \left (f x + e\right )^{5} + 15 \,{\left (C b^{3} c +{\left (3 \, C a b^{2} + B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{4} + 20 \,{\left ({\left (3 \, C a b^{2} + B b^{3}\right )} c +{\left (3 \, C a^{2} b + 3 \, B a b^{2} +{\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )^{3} + 60 \,{\left ({\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \,{\left (A - C\right )} a b^{2} + B b^{3}\right )} c -{\left (B a^{3} + 3 \,{\left (A - C\right )} a^{2} b - 3 \, B a b^{2} -{\left (A - C\right )} b^{3}\right )} d\right )} f x + 30 \,{\left ({\left (3 \, C a^{2} b + 3 \, B a b^{2} +{\left (A - C\right )} b^{3}\right )} c +{\left (C a^{3} + 3 \, B a^{2} b + 3 \,{\left (A - C\right )} a b^{2} - B b^{3}\right )} d\right )} \tan \left (f x + e\right )^{2} - 30 \,{\left ({\left (B a^{3} + 3 \,{\left (A - C\right )} a^{2} b - 3 \, B a b^{2} -{\left (A - C\right )} b^{3}\right )} c +{\left ({\left (A - C\right )} a^{3} - 3 \, B a^{2} b - 3 \,{\left (A - C\right )} a b^{2} + B b^{3}\right )} d\right )} \log \left (\frac{1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 60 \,{\left ({\left (C a^{3} + 3 \, B a^{2} b + 3 \,{\left (A - C\right )} a b^{2} - B b^{3}\right )} c +{\left (B a^{3} + 3 \,{\left (A - C\right )} a^{2} b - 3 \, B a b^{2} -{\left (A - C\right )} b^{3}\right )} d\right )} \tan \left (f x + e\right )}{60 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*C*b^3*d*tan(f*x + e)^5 + 15*(C*b^3*c + (3*C*a*b^2 + B*b^3)*d)*tan(f*x + e)^4 + 20*((3*C*a*b^2 + B*b^3
)*c + (3*C*a^2*b + 3*B*a*b^2 + (A - C)*b^3)*d)*tan(f*x + e)^3 + 60*(((A - C)*a^3 - 3*B*a^2*b - 3*(A - C)*a*b^2
+ B*b^3)*c - (B*a^3 + 3*(A - C)*a^2*b - 3*B*a*b^2 - (A - C)*b^3)*d)*f*x + 30*((3*C*a^2*b + 3*B*a*b^2 + (A - C
)*b^3)*c + (C*a^3 + 3*B*a^2*b + 3*(A - C)*a*b^2 - B*b^3)*d)*tan(f*x + e)^2 - 30*((B*a^3 + 3*(A - C)*a^2*b - 3*
B*a*b^2 - (A - C)*b^3)*c + ((A - C)*a^3 - 3*B*a^2*b - 3*(A - C)*a*b^2 + B*b^3)*d)*log(1/(tan(f*x + e)^2 + 1))
+ 60*((C*a^3 + 3*B*a^2*b + 3*(A - C)*a*b^2 - B*b^3)*c + (B*a^3 + 3*(A - C)*a^2*b - 3*B*a*b^2 - (A - C)*b^3)*d)
*tan(f*x + e))/f

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Sympy [A]  time = 5.48655, size = 1001, normalized size = 2.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((A*a**3*c*x + A*a**3*d*log(tan(e + f*x)**2 + 1)/(2*f) + 3*A*a**2*b*c*log(tan(e + f*x)**2 + 1)/(2*f)
- 3*A*a**2*b*d*x + 3*A*a**2*b*d*tan(e + f*x)/f - 3*A*a*b**2*c*x + 3*A*a*b**2*c*tan(e + f*x)/f - 3*A*a*b**2*d*l
og(tan(e + f*x)**2 + 1)/(2*f) + 3*A*a*b**2*d*tan(e + f*x)**2/(2*f) - A*b**3*c*log(tan(e + f*x)**2 + 1)/(2*f) +
A*b**3*c*tan(e + f*x)**2/(2*f) + A*b**3*d*x + A*b**3*d*tan(e + f*x)**3/(3*f) - A*b**3*d*tan(e + f*x)/f + B*a*
*3*c*log(tan(e + f*x)**2 + 1)/(2*f) - B*a**3*d*x + B*a**3*d*tan(e + f*x)/f - 3*B*a**2*b*c*x + 3*B*a**2*b*c*tan
(e + f*x)/f - 3*B*a**2*b*d*log(tan(e + f*x)**2 + 1)/(2*f) + 3*B*a**2*b*d*tan(e + f*x)**2/(2*f) - 3*B*a*b**2*c*
log(tan(e + f*x)**2 + 1)/(2*f) + 3*B*a*b**2*c*tan(e + f*x)**2/(2*f) + 3*B*a*b**2*d*x + B*a*b**2*d*tan(e + f*x)
**3/f - 3*B*a*b**2*d*tan(e + f*x)/f + B*b**3*c*x + B*b**3*c*tan(e + f*x)**3/(3*f) - B*b**3*c*tan(e + f*x)/f +
B*b**3*d*log(tan(e + f*x)**2 + 1)/(2*f) + B*b**3*d*tan(e + f*x)**4/(4*f) - B*b**3*d*tan(e + f*x)**2/(2*f) - C*
a**3*c*x + C*a**3*c*tan(e + f*x)/f - C*a**3*d*log(tan(e + f*x)**2 + 1)/(2*f) + C*a**3*d*tan(e + f*x)**2/(2*f)
- 3*C*a**2*b*c*log(tan(e + f*x)**2 + 1)/(2*f) + 3*C*a**2*b*c*tan(e + f*x)**2/(2*f) + 3*C*a**2*b*d*x + C*a**2*b
*d*tan(e + f*x)**3/f - 3*C*a**2*b*d*tan(e + f*x)/f + 3*C*a*b**2*c*x + C*a*b**2*c*tan(e + f*x)**3/f - 3*C*a*b**
2*c*tan(e + f*x)/f + 3*C*a*b**2*d*log(tan(e + f*x)**2 + 1)/(2*f) + 3*C*a*b**2*d*tan(e + f*x)**4/(4*f) - 3*C*a*
b**2*d*tan(e + f*x)**2/(2*f) + C*b**3*c*log(tan(e + f*x)**2 + 1)/(2*f) + C*b**3*c*tan(e + f*x)**4/(4*f) - C*b*
*3*c*tan(e + f*x)**2/(2*f) - C*b**3*d*x + C*b**3*d*tan(e + f*x)**5/(5*f) - C*b**3*d*tan(e + f*x)**3/(3*f) + C*
b**3*d*tan(e + f*x)/f, Ne(f, 0)), (x*(a + b*tan(e))**3*(c + d*tan(e))*(A + B*tan(e) + C*tan(e)**2), True))

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out