3.1.45 \(\int (c+d x)^2 (a+b \tan (e+f x))^2 \, dx\) [45]

Optimal. Leaf size=229 \[ -\frac {i b^2 (c+d x)^2}{f}+\frac {a^2 (c+d x)^3}{3 d}+\frac {2 i a b (c+d x)^3}{3 d}-\frac {b^2 (c+d x)^3}{3 d}+\frac {2 b^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}-\frac {2 a b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {i b^2 d^2 \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}+\frac {2 i a b d (c+d x) \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {a b d^2 \text {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{f^3}+\frac {b^2 (c+d x)^2 \tan (e+f x)}{f} \]

[Out]

-I*b^2*(d*x+c)^2/f+1/3*a^2*(d*x+c)^3/d+2/3*I*a*b*(d*x+c)^3/d-1/3*b^2*(d*x+c)^3/d+2*b^2*d*(d*x+c)*ln(1+exp(2*I*
(f*x+e)))/f^2-2*a*b*(d*x+c)^2*ln(1+exp(2*I*(f*x+e)))/f-I*b^2*d^2*polylog(2,-exp(2*I*(f*x+e)))/f^3+2*I*a*b*d*(d
*x+c)*polylog(2,-exp(2*I*(f*x+e)))/f^2-a*b*d^2*polylog(3,-exp(2*I*(f*x+e)))/f^3+b^2*(d*x+c)^2*tan(f*x+e)/f

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Rubi [A]
time = 0.26, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3803, 3800, 2221, 2611, 2320, 6724, 3801, 2317, 2438, 32} \begin {gather*} \frac {a^2 (c+d x)^3}{3 d}+\frac {2 i a b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {2 a b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {2 i a b (c+d x)^3}{3 d}-\frac {a b d^2 \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{f^3}+\frac {2 b^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x)^2 \tan (e+f x)}{f}-\frac {i b^2 (c+d x)^2}{f}-\frac {b^2 (c+d x)^3}{3 d}-\frac {i b^2 d^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*b^2*(c + d*x)^2)/f + (a^2*(c + d*x)^3)/(3*d) + (((2*I)/3)*a*b*(c + d*x)^3)/d - (b^2*(c + d*x)^3)/(3*d) +
(2*b^2*d*(c + d*x)*Log[1 + E^((2*I)*(e + f*x))])/f^2 - (2*a*b*(c + d*x)^2*Log[1 + E^((2*I)*(e + f*x))])/f - (
I*b^2*d^2*PolyLog[2, -E^((2*I)*(e + f*x))])/f^3 + ((2*I)*a*b*d*(c + d*x)*PolyLog[2, -E^((2*I)*(e + f*x))])/f^2
- (a*b*d^2*PolyLog[3, -E^((2*I)*(e + f*x))])/f^3 + (b^2*(c + d*x)^2*Tan[e + f*x])/f

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[2*I, Int[(c + d*x)^m*(E^(2*I*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 3801

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

Rule 3803

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int (c+d x)^2 (a+b \tan (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)^2+2 a b (c+d x)^2 \tan (e+f x)+b^2 (c+d x)^2 \tan ^2(e+f x)\right ) \, dx\\ &=\frac {a^2 (c+d x)^3}{3 d}+(2 a b) \int (c+d x)^2 \tan (e+f x) \, dx+b^2 \int (c+d x)^2 \tan ^2(e+f x) \, dx\\ &=\frac {a^2 (c+d x)^3}{3 d}+\frac {2 i a b (c+d x)^3}{3 d}+\frac {b^2 (c+d x)^2 \tan (e+f x)}{f}-(4 i a b) \int \frac {e^{2 i (e+f x)} (c+d x)^2}{1+e^{2 i (e+f x)}} \, dx-b^2 \int (c+d x)^2 \, dx-\frac {\left (2 b^2 d\right ) \int (c+d x) \tan (e+f x) \, dx}{f}\\ &=-\frac {i b^2 (c+d x)^2}{f}+\frac {a^2 (c+d x)^3}{3 d}+\frac {2 i a b (c+d x)^3}{3 d}-\frac {b^2 (c+d x)^3}{3 d}-\frac {2 a b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^2 (c+d x)^2 \tan (e+f x)}{f}+\frac {(4 a b d) \int (c+d x) \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f}+\frac {\left (4 i b^2 d\right ) \int \frac {e^{2 i (e+f x)} (c+d x)}{1+e^{2 i (e+f x)}} \, dx}{f}\\ &=-\frac {i b^2 (c+d x)^2}{f}+\frac {a^2 (c+d x)^3}{3 d}+\frac {2 i a b (c+d x)^3}{3 d}-\frac {b^2 (c+d x)^3}{3 d}+\frac {2 b^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}-\frac {2 a b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {2 i a b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x)^2 \tan (e+f x)}{f}-\frac {\left (2 i a b d^2\right ) \int \text {Li}_2\left (-e^{2 i (e+f x)}\right ) \, dx}{f^2}-\frac {\left (2 b^2 d^2\right ) \int \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f^2}\\ &=-\frac {i b^2 (c+d x)^2}{f}+\frac {a^2 (c+d x)^3}{3 d}+\frac {2 i a b (c+d x)^3}{3 d}-\frac {b^2 (c+d x)^3}{3 d}+\frac {2 b^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}-\frac {2 a b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {2 i a b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x)^2 \tan (e+f x)}{f}-\frac {\left (a b d^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{f^3}+\frac {\left (i b^2 d^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{f^3}\\ &=-\frac {i b^2 (c+d x)^2}{f}+\frac {a^2 (c+d x)^3}{3 d}+\frac {2 i a b (c+d x)^3}{3 d}-\frac {b^2 (c+d x)^3}{3 d}+\frac {2 b^2 d (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f^2}-\frac {2 a b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {i b^2 d^2 \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^3}+\frac {2 i a b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {a b d^2 \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{f^3}+\frac {b^2 (c+d x)^2 \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(656\) vs. \(2(229)=458\).
time = 6.95, size = 656, normalized size = 2.86 \begin {gather*} \frac {a b d^2 e^{-i e} \left (2 i f^2 x^2 \left (2 e^{2 i e} f x+3 i \left (1+e^{2 i e}\right ) \log \left (1+e^{2 i (e+f x)}\right )\right )+6 i \left (1+e^{2 i e}\right ) f x \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )-3 \left (1+e^{2 i e}\right ) \text {PolyLog}\left (3,-e^{2 i (e+f x)}\right )\right ) \sec (e)}{6 f^3}+\frac {1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right ) \sec (e) \left (a^2 \cos (e)-b^2 \cos (e)+2 a b \sin (e)\right )+\frac {2 b^2 c d \sec (e) (\cos (e) \log (\cos (e) \cos (f x)-\sin (e) \sin (f x))+f x \sin (e))}{f^2 \left (\cos ^2(e)+\sin ^2(e)\right )}-\frac {2 a b c^2 \sec (e) (\cos (e) \log (\cos (e) \cos (f x)-\sin (e) \sin (f x))+f x \sin (e))}{f \left (\cos ^2(e)+\sin ^2(e)\right )}+\frac {b^2 d^2 \csc (e) \left (e^{-i \text {ArcTan}(\cot (e))} f^2 x^2-\frac {\cot (e) \left (i f x (-\pi -2 \text {ArcTan}(\cot (e)))-\pi \log \left (1+e^{-2 i f x}\right )-2 (f x-\text {ArcTan}(\cot (e))) \log \left (1-e^{2 i (f x-\text {ArcTan}(\cot (e)))}\right )+\pi \log (\cos (f x))-2 \text {ArcTan}(\cot (e)) \log (\sin (f x-\text {ArcTan}(\cot (e))))+i \text {PolyLog}\left (2,e^{2 i (f x-\text {ArcTan}(\cot (e)))}\right )\right )}{\sqrt {1+\cot ^2(e)}}\right ) \sec (e)}{f^3 \sqrt {\csc ^2(e) \left (\cos ^2(e)+\sin ^2(e)\right )}}-\frac {2 a b c d \csc (e) \left (e^{-i \text {ArcTan}(\cot (e))} f^2 x^2-\frac {\cot (e) \left (i f x (-\pi -2 \text {ArcTan}(\cot (e)))-\pi \log \left (1+e^{-2 i f x}\right )-2 (f x-\text {ArcTan}(\cot (e))) \log \left (1-e^{2 i (f x-\text {ArcTan}(\cot (e)))}\right )+\pi \log (\cos (f x))-2 \text {ArcTan}(\cot (e)) \log (\sin (f x-\text {ArcTan}(\cot (e))))+i \text {PolyLog}\left (2,e^{2 i (f x-\text {ArcTan}(\cot (e)))}\right )\right )}{\sqrt {1+\cot ^2(e)}}\right ) \sec (e)}{f^2 \sqrt {\csc ^2(e) \left (\cos ^2(e)+\sin ^2(e)\right )}}+\frac {\sec (e) \sec (e+f x) \left (b^2 c^2 \sin (f x)+2 b^2 c d x \sin (f x)+b^2 d^2 x^2 \sin (f x)\right )}{f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*b*d^2*((2*I)*f^2*x^2*(2*E^((2*I)*e)*f*x + (3*I)*(1 + E^((2*I)*e))*Log[1 + E^((2*I)*(e + f*x))]) + (6*I)*(1
+ E^((2*I)*e))*f*x*PolyLog[2, -E^((2*I)*(e + f*x))] - 3*(1 + E^((2*I)*e))*PolyLog[3, -E^((2*I)*(e + f*x))])*Se
c[e])/(6*E^(I*e)*f^3) + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Sec[e]*(a^2*Cos[e] - b^2*Cos[e] + 2*a*b*Sin[e]))/3 + (2
*b^2*c*d*Sec[e]*(Cos[e]*Log[Cos[e]*Cos[f*x] - Sin[e]*Sin[f*x]] + f*x*Sin[e]))/(f^2*(Cos[e]^2 + Sin[e]^2)) - (2
*a*b*c^2*Sec[e]*(Cos[e]*Log[Cos[e]*Cos[f*x] - Sin[e]*Sin[f*x]] + f*x*Sin[e]))/(f*(Cos[e]^2 + Sin[e]^2)) + (b^2
*d^2*Csc[e]*((f^2*x^2)/E^(I*ArcTan[Cot[e]]) - (Cot[e]*(I*f*x*(-Pi - 2*ArcTan[Cot[e]]) - Pi*Log[1 + E^((-2*I)*f
*x)] - 2*(f*x - ArcTan[Cot[e]])*Log[1 - E^((2*I)*(f*x - ArcTan[Cot[e]]))] + Pi*Log[Cos[f*x]] - 2*ArcTan[Cot[e]
]*Log[Sin[f*x - ArcTan[Cot[e]]]] + I*PolyLog[2, E^((2*I)*(f*x - ArcTan[Cot[e]]))]))/Sqrt[1 + Cot[e]^2])*Sec[e]
)/(f^3*Sqrt[Csc[e]^2*(Cos[e]^2 + Sin[e]^2)]) - (2*a*b*c*d*Csc[e]*((f^2*x^2)/E^(I*ArcTan[Cot[e]]) - (Cot[e]*(I*
f*x*(-Pi - 2*ArcTan[Cot[e]]) - Pi*Log[1 + E^((-2*I)*f*x)] - 2*(f*x - ArcTan[Cot[e]])*Log[1 - E^((2*I)*(f*x - A
rcTan[Cot[e]]))] + Pi*Log[Cos[f*x]] - 2*ArcTan[Cot[e]]*Log[Sin[f*x - ArcTan[Cot[e]]]] + I*PolyLog[2, E^((2*I)*
(f*x - ArcTan[Cot[e]]))]))/Sqrt[1 + Cot[e]^2])*Sec[e])/(f^2*Sqrt[Csc[e]^2*(Cos[e]^2 + Sin[e]^2)]) + (Sec[e]*Se
c[e + f*x]*(b^2*c^2*Sin[f*x] + 2*b^2*c*d*x*Sin[f*x] + b^2*d^2*x^2*Sin[f*x]))/f

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 574 vs. \(2 (209 ) = 418\).
time = 0.27, size = 575, normalized size = 2.51

method	result	size
risch	\(-\frac {a b \,d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{3}}-\frac {i b^{2} d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 i b^{2} d^{2} x^{2}}{f}-\frac {2 i b^{2} d^{2} e^{2}}{f^{3}}+\frac {2 i d^{2} a b \,x^{3}}{3}-2 i a b \,c^{2} x -\frac {2 i a b \,c^{3}}{3 d}+\frac {2 i b^{2} \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {2 b^{2} d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x}{f^{2}}+\frac {4 b^{2} d^{2} e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+d \,a^{2} c \,x^{2}+a^{2} c^{2} x -d \,b^{2} c \,x^{2}+\frac {8 i b a c d e x}{f}-\frac {4 b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a c d x}{f}-\frac {4 i b a \,d^{2} e^{2} x}{f^{2}}+\frac {4 i b a c d \,e^{2}}{f^{2}}+\frac {2 i b a \,d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {2 i b a c d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b a \,c^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {4 b a \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {2 b^{2} c d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f^{2}}-\frac {4 b^{2} c d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {d^{2} a^{2} x^{3}}{3}+\frac {a^{2} c^{3}}{3 d}-\frac {d^{2} b^{2} x^{3}}{3}-b^{2} c^{2} x -\frac {b^{2} c^{3}}{3 d}-\frac {8 b a c d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b a \,d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{2}}{f}+\frac {4 b a \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {8 i b a \,d^{2} e^{3}}{3 f^{3}}-\frac {4 i b^{2} d^{2} e x}{f^{2}}+2 i d a b c \,x^{2}\)	\(575\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

d*a^2*c*x^2+a^2*c^2*x-d*b^2*c*x^2-2/f*b*a*d^2*ln(exp(2*I*(f*x+e))+1)*x^2+4/f^3*b*a*d^2*e^2*ln(exp(I*(f*x+e)))-
8/3*I/f^3*b*a*d^2*e^3-4*I/f^2*b^2*d^2*e*x-2*I/f*b^2*d^2*x^2-2*I/f^3*b^2*d^2*e^2+2/3*I*d^2*a*b*x^3-8/f^2*b*a*c*
d*e*ln(exp(I*(f*x+e)))-4/f*b*ln(exp(2*I*(f*x+e))+1)*a*c*d*x+8*I/f*b*a*c*d*e*x-2*I*a*b*c^2*x-2/3*I/d*a*b*c^3-4*
I/f^2*b*a*d^2*e^2*x+4*I/f^2*b*a*c*d*e^2+2*I/f^2*b*a*d^2*polylog(2,-exp(2*I*(f*x+e)))*x+2*I/f^2*b*a*c*d*polylog
(2,-exp(2*I*(f*x+e)))+2*I*b^2*(d^2*x^2+2*c*d*x+c^2)/f/(exp(2*I*(f*x+e))+1)+2/f^2*b^2*d^2*ln(exp(2*I*(f*x+e))+1
)*x+4/f^3*b^2*d^2*e*ln(exp(I*(f*x+e)))-2/f*b*a*c^2*ln(exp(2*I*(f*x+e))+1)+4/f*b*a*c^2*ln(exp(I*(f*x+e)))+2/f^2
*b^2*c*d*ln(exp(2*I*(f*x+e))+1)-4/f^2*b^2*c*d*ln(exp(I*(f*x+e)))+1/3*d^2*a^2*x^3+1/3/d*a^2*c^3-1/3*d^2*b^2*x^3
-b^2*c^2*x-1/3/d*b^2*c^3+2*I*d*a*b*c*x^2-a*b*d^2*polylog(3,-exp(2*I*(f*x+e)))/f^3-I*b^2*d^2*polylog(2,-exp(2*I
*(f*x+e)))/f^3

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1295 vs. \(2 (209) = 418\).
time = 0.72, size = 1295, normalized size = 5.66 \begin {gather*} \frac {3 \, {\left (f x + e\right )} a^{2} c^{2} + \frac {{\left (f x + e\right )}^{3} a^{2} d^{2}}{f^{2}} + \frac {3 \, {\left (f x + e\right )}^{2} a^{2} c d}{f} - \frac {3 \, {\left (f x + e\right )}^{2} a^{2} d^{2} e}{f^{2}} - \frac {6 \, {\left (f x + e\right )} a^{2} c d e}{f} + 6 \, a b c^{2} \log \left (\sec \left (f x + e\right )\right ) - \frac {12 \, a b c d e \log \left (\sec \left (f x + e\right )\right )}{f} + \frac {3 \, {\left (f x + e\right )} a^{2} d^{2} e^{2}}{f^{2}} + \frac {6 \, a b d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} + \frac {3 \, {\left ({\left (2 \, a b + i \, b^{2}\right )} {\left (f x + e\right )}^{3} d^{2} + 6 \, b^{2} c^{2} f^{2} - 12 \, b^{2} c d f e + 6 \, b^{2} d^{2} e^{2} + 3 \, {\left ({\left (2 \, a b + i \, b^{2}\right )} c d f - {\left (2 \, a b e + i \, b^{2} e\right )} d^{2}\right )} {\left (f x + e\right )}^{2} + 3 \, {\left (i \, b^{2} c^{2} f^{2} - 2 i \, b^{2} c d f e + i \, b^{2} d^{2} e^{2}\right )} {\left (f x + e\right )} - 6 \, {\left ({\left (f x + e\right )}^{2} a b d^{2} - b^{2} c d f + b^{2} d^{2} e + {\left (2 \, a b c d f - {\left (2 \, a b e + b^{2}\right )} d^{2}\right )} {\left (f x + e\right )} + {\left ({\left (f x + e\right )}^{2} a b d^{2} - b^{2} c d f + b^{2} d^{2} e + {\left (2 \, a b c d f - {\left (2 \, a b e + b^{2}\right )} d^{2}\right )} {\left (f x + e\right )}\right )} \cos \left (2 \, f x + 2 \, e\right ) - {\left (-i \, {\left (f x + e\right )}^{2} a b d^{2} + i \, b^{2} c d f - i \, b^{2} d^{2} e + {\left (-2 i \, a b c d f + {\left (2 i \, a b e + i \, b^{2}\right )} d^{2}\right )} {\left (f x + e\right )}\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + {\left ({\left (2 \, a b + i \, b^{2}\right )} {\left (f x + e\right )}^{3} d^{2} + 3 \, {\left ({\left (2 \, a b + i \, b^{2}\right )} c d f + {\left (b^{2} {\left (-i \, e - 2\right )} - 2 \, a b e\right )} d^{2}\right )} {\left (f x + e\right )}^{2} + 3 \, {\left (i \, b^{2} c^{2} f^{2} + 2 \, b^{2} c d f {\left (-i \, e - 2\right )} + b^{2} d^{2} {\left (i \, e^{2} + 4 \, e\right )}\right )} {\left (f x + e\right )}\right )} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, {\left (2 \, {\left (f x + e\right )} a b d^{2} + 2 \, a b c d f - {\left (2 \, a b e + b^{2}\right )} d^{2} + {\left (2 \, {\left (f x + e\right )} a b d^{2} + 2 \, a b c d f - {\left (2 \, a b e + b^{2}\right )} d^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (2 i \, {\left (f x + e\right )} a b d^{2} + 2 i \, a b c d f + {\left (-2 i \, a b e - i \, b^{2}\right )} d^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 3 \, {\left (i \, {\left (f x + e\right )}^{2} a b d^{2} - i \, b^{2} c d f + i \, b^{2} d^{2} e + {\left (2 i \, a b c d f + {\left (-2 i \, a b e - i \, b^{2}\right )} d^{2}\right )} {\left (f x + e\right )} + {\left (i \, {\left (f x + e\right )}^{2} a b d^{2} - i \, b^{2} c d f + i \, b^{2} d^{2} e + {\left (2 i \, a b c d f + {\left (-2 i \, a b e - i \, b^{2}\right )} d^{2}\right )} {\left (f x + e\right )}\right )} \cos \left (2 \, f x + 2 \, e\right ) - {\left ({\left (f x + e\right )}^{2} a b d^{2} - b^{2} c d f + b^{2} d^{2} e + {\left (2 \, a b c d f - {\left (2 \, a b e + b^{2}\right )} d^{2}\right )} {\left (f x + e\right )}\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + 3 \, {\left (i \, a b d^{2} \cos \left (2 \, f x + 2 \, e\right ) - a b d^{2} \sin \left (2 \, f x + 2 \, e\right ) + i \, a b d^{2}\right )} {\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )}) - {\left ({\left (-2 i \, a b + b^{2}\right )} {\left (f x + e\right )}^{3} d^{2} - 3 \, {\left ({\left (2 i \, a b - b^{2}\right )} c d f + {\left (b^{2} {\left (e - 2 i\right )} - 2 i \, a b e\right )} d^{2}\right )} {\left (f x + e\right )}^{2} + 3 \, {\left (b^{2} c^{2} f^{2} - 2 \, b^{2} c d f {\left (e - 2 i\right )} + b^{2} d^{2} {\left (e^{2} - 4 i \, e\right )}\right )} {\left (f x + e\right )}\right )} \sin \left (2 \, f x + 2 \, e\right )\right )}}{-3 i \, f^{2} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, f^{2} \sin \left (2 \, f x + 2 \, e\right ) - 3 i \, f^{2}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(f*x + e)*a^2*c^2 + (f*x + e)^3*a^2*d^2/f^2 + 3*(f*x + e)^2*a^2*c*d/f - 3*(f*x + e)^2*a^2*d^2*e/f^2 - 6
*(f*x + e)*a^2*c*d*e/f + 6*a*b*c^2*log(sec(f*x + e)) - 12*a*b*c*d*e*log(sec(f*x + e))/f + 3*(f*x + e)*a^2*d^2*
e^2/f^2 + 6*a*b*d^2*e^2*log(sec(f*x + e))/f^2 + 3*((2*a*b + I*b^2)*(f*x + e)^3*d^2 + 6*b^2*c^2*f^2 - 12*b^2*c*
d*f*e + 6*b^2*d^2*e^2 + 3*((2*a*b + I*b^2)*c*d*f - (2*a*b*e + I*b^2*e)*d^2)*(f*x + e)^2 + 3*(I*b^2*c^2*f^2 - 2
*I*b^2*c*d*f*e + I*b^2*d^2*e^2)*(f*x + e) - 6*((f*x + e)^2*a*b*d^2 - b^2*c*d*f + b^2*d^2*e + (2*a*b*c*d*f - (2
*a*b*e + b^2)*d^2)*(f*x + e) + ((f*x + e)^2*a*b*d^2 - b^2*c*d*f + b^2*d^2*e + (2*a*b*c*d*f - (2*a*b*e + b^2)*d
^2)*(f*x + e))*cos(2*f*x + 2*e) - (-I*(f*x + e)^2*a*b*d^2 + I*b^2*c*d*f - I*b^2*d^2*e + (-2*I*a*b*c*d*f + (2*I
*a*b*e + I*b^2)*d^2)*(f*x + e))*sin(2*f*x + 2*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) + ((2*a*b +
I*b^2)*(f*x + e)^3*d^2 + 3*((2*a*b + I*b^2)*c*d*f + (b^2*(-I*e - 2) - 2*a*b*e)*d^2)*(f*x + e)^2 + 3*(I*b^2*c^2
*f^2 + 2*b^2*c*d*f*(-I*e - 2) + b^2*d^2*(I*e^2 + 4*e))*(f*x + e))*cos(2*f*x + 2*e) + 3*(2*(f*x + e)*a*b*d^2 +
2*a*b*c*d*f - (2*a*b*e + b^2)*d^2 + (2*(f*x + e)*a*b*d^2 + 2*a*b*c*d*f - (2*a*b*e + b^2)*d^2)*cos(2*f*x + 2*e)
+ (2*I*(f*x + e)*a*b*d^2 + 2*I*a*b*c*d*f + (-2*I*a*b*e - I*b^2)*d^2)*sin(2*f*x + 2*e))*dilog(-e^(2*I*f*x + 2*
I*e)) + 3*(I*(f*x + e)^2*a*b*d^2 - I*b^2*c*d*f + I*b^2*d^2*e + (2*I*a*b*c*d*f + (-2*I*a*b*e - I*b^2)*d^2)*(f*x
+ e) + (I*(f*x + e)^2*a*b*d^2 - I*b^2*c*d*f + I*b^2*d^2*e + (2*I*a*b*c*d*f + (-2*I*a*b*e - I*b^2)*d^2)*(f*x +
e))*cos(2*f*x + 2*e) - ((f*x + e)^2*a*b*d^2 - b^2*c*d*f + b^2*d^2*e + (2*a*b*c*d*f - (2*a*b*e + b^2)*d^2)*(f*
x + e))*sin(2*f*x + 2*e))*log(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1) + 3*(I*a*b*d^2
*cos(2*f*x + 2*e) - a*b*d^2*sin(2*f*x + 2*e) + I*a*b*d^2)*polylog(3, -e^(2*I*f*x + 2*I*e)) - ((-2*I*a*b + b^2)
*(f*x + e)^3*d^2 - 3*((2*I*a*b - b^2)*c*d*f + (b^2*(e - 2*I) - 2*I*a*b*e)*d^2)*(f*x + e)^2 + 3*(b^2*c^2*f^2 -
2*b^2*c*d*f*(e - 2*I) + b^2*d^2*(e^2 - 4*I*e))*(f*x + e))*sin(2*f*x + 2*e))/(-3*I*f^2*cos(2*f*x + 2*e) + 3*f^2
*sin(2*f*x + 2*e) - 3*I*f^2))/f

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 465 vs. \(2 (209) = 418\).
time = 0.39, size = 465, normalized size = 2.03 \begin {gather*} \frac {2 \, {\left (a^{2} - b^{2}\right )} d^{2} f^{3} x^{3} + 6 \, {\left (a^{2} - b^{2}\right )} c d f^{3} x^{2} + 6 \, {\left (a^{2} - b^{2}\right )} c^{2} f^{3} x - 3 \, a b d^{2} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, a b d^{2} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, {\left (2 i \, a b d^{2} f x + 2 i \, a b c d f - i \, b^{2} d^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 3 \, {\left (-2 i \, a b d^{2} f x - 2 i \, a b c d f + i \, b^{2} d^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \, {\left (a b d^{2} f^{2} x^{2} + a b c^{2} f^{2} - b^{2} c d f + {\left (2 \, a b c d f^{2} - b^{2} d^{2} f\right )} x\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (a b d^{2} f^{2} x^{2} + a b c^{2} f^{2} - b^{2} c d f + {\left (2 \, a b c d f^{2} - b^{2} d^{2} f\right )} x\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 6 \, {\left (b^{2} d^{2} f^{2} x^{2} + 2 \, b^{2} c d f^{2} x + b^{2} c^{2} f^{2}\right )} \tan \left (f x + e\right )}{6 \, f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*(a^2 - b^2)*d^2*f^3*x^3 + 6*(a^2 - b^2)*c*d*f^3*x^2 + 6*(a^2 - b^2)*c^2*f^3*x - 3*a*b*d^2*polylog(3, (t
an(f*x + e)^2 + 2*I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 3*a*b*d^2*polylog(3, (tan(f*x + e)^2 - 2*I*tan(f
*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 3*(2*I*a*b*d^2*f*x + 2*I*a*b*c*d*f - I*b^2*d^2)*dilog(2*(I*tan(f*x + e) -
1)/(tan(f*x + e)^2 + 1) + 1) - 3*(-2*I*a*b*d^2*f*x - 2*I*a*b*c*d*f + I*b^2*d^2)*dilog(2*(-I*tan(f*x + e) - 1)
/(tan(f*x + e)^2 + 1) + 1) - 6*(a*b*d^2*f^2*x^2 + a*b*c^2*f^2 - b^2*c*d*f + (2*a*b*c*d*f^2 - b^2*d^2*f)*x)*log
(-2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - 6*(a*b*d^2*f^2*x^2 + a*b*c^2*f^2 - b^2*c*d*f + (2*a*b*c*d*f^2
- b^2*d^2*f)*x)*log(-2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) + 6*(b^2*d^2*f^2*x^2 + 2*b^2*c*d*f^2*x + b
^2*c^2*f^2)*tan(f*x + e))/f^3

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \left (c + d x\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*(a+b*tan(f*x+e))**2,x)

[Out]

Integral((a + b*tan(e + f*x))**2*(c + d*x)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^2*(b*tan(f*x + e) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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