3.1.40 \(\int (c+d x)^2 (a+b \tan (e+f x)) \, dx\) [40]

Optimal. Leaf size=115 \[ \frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}-\frac {b d^2 \text {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^3} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3803, 3800, 2221, 2611, 2320, 6724} \begin {gather*} \frac {a (c+d x)^3}{3 d}+\frac {i b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^3}{3 d}-\frac {b d^2 \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \tan (e+f x)\right ) \, dx\\ &=\frac {a (c+d x)^3}{3 d}+b \int (c+d x)^2 \tan (e+f x) \, dx\\ &=\frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)^2}{1+e^{2 i (e+f x)}} \, dx\\ &=\frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {(2 b d) \int (c+d x) \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f}\\ &=\frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {\left (i b d^2\right ) \int \text {Li}_2\left (-e^{2 i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^3}\\ &=\frac {a (c+d x)^3}{3 d}+\frac {i b (c+d x)^3}{3 d}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {b d^2 \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}\\ \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. \(324\) vs. \(2(115)=230\).
time = 6.32, size = 324, normalized size = 2.82 \begin {gather*} \frac {1}{12} \sec (e) \left (\frac {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 )}{f^3}+4 x \left (3 c^2+3 c d x+d^2 x^2\right ) (a \cos (e)+b \sin (e))-\frac {12 b c^2 (\cos (e) \log (\cos (e+f x))+f x \sin (e))}{f}+12 b c d \csc (e) \left (-\frac {e^{-i \text {ArcTan}(\cot (e))} x^2}{\sqrt {\csc ^2(e)}}+\frac {\cos (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 ) \sin (e)}{f^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e]*((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
))]))/(E^(I*e)*f^3) + 4*x*(3*c^2 + 3*c*d*x + d^2*x^2)*(a*Cos[e] + b*Sin[e]) - (12*b*c^2*(Cos[e]*Log[Cos[e + f*
x]] + f*x*Sin[e]))/f + 12*b*c*d*Csc[e]*(-(x^2/(E^(I*ArcTan[Cot[e]])*Sqrt[Csc[e]^2])) + (Cos[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 - Arc
Tan[Cot[e]]))])*Sin[e])/f^2)))/12

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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. 313 vs. \(2 (101 ) = 202\).
time = 0.21, size = 314, normalized size = 2.73

method	result	size
risch	\(-\frac {i b \,c^{3}}{3 d}-i b \,c^{2} x +\frac {d^{2} a \,x^{3}}{3}+\frac {2 i b c d \,e^{2}}{f^{2}}+d a c \,x^{2}+\frac {i b \,d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f^{2}}+a \,c^{2} x +\frac {a \,c^{3}}{3 d}+\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}-\frac {b \,c^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {4 i b \,d^{2} e^{3}}{3 f^{3}}-\frac {2 b c d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x}{f}+\frac {4 i b c d e x}{f}-\frac {b \,d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{2}}{f}-\frac {2 i b \,d^{2} e^{2} x}{f^{2}}-\frac {b \,d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{3}}-\frac {4 b c d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+i d b c \,x^{2}+\frac {i b c d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}+\frac {i d^{2} b \,x^{3}}{3}\)	\(314\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I/d*b*c^3-I*b*c^2*x+1/3*d^2*a*x^3+2*I/f^2*b*c*d*e^2+d*a*c*x^2+I/f^2*b*d^2*polylog(2,-exp(2*I*(f*x+e)))*x+
a*c^2*x+1/3/d*a*c^3+2/f*b*c^2*ln(exp(I*(f*x+e)))-1/f*b*c^2*ln(exp(2*I*(f*x+e))+1)+2/f^3*b*d^2*e^2*ln(exp(I*(f*
x+e)))-4/3*I/f^3*b*d^2*e^3-2/f*b*c*d*ln(exp(2*I*(f*x+e))+1)*x+4*I/f*b*c*d*e*x-1/f*b*d^2*ln(exp(2*I*(f*x+e))+1)
*x^2-2*I/f^2*b*d^2*e^2*x-1/2*b*d^2*polylog(3,-exp(2*I*(f*x+e)))/f^3-4/f^2*b*c*d*e*ln(exp(I*(f*x+e)))+I*d*b*c*x
^2+I/f^2*b*c*d*polylog(2,-exp(2*I*(f*x+e)))+1/3*I*d^2*b*x^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. 401 vs. \(2 (101) = 202\).
time = 0.55, size = 401, normalized size = 3.49 \begin {gather*} \frac {6 \, {\left (f x + e\right )} a c^{2} + \frac {2 \, {\left (f x + e\right )}^{3} a d^{2}}{f^{2}} + \frac {6 \, {\left (f x + e\right )}^{2} a c d}{f} - \frac {6 \, {\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} - \frac {12 \, {\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sec \left (f x + e\right )\right ) - \frac {12 \, b c d e \log \left (\sec \left (f x + e\right )\right )}{f} + \frac {6 \, {\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac {6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} - \frac {-2 i \, {\left (f x + e\right )}^{3} b d^{2} + 3 \, b d^{2} {\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )}) - 6 \, {\left (i \, b c d f - i \, b d^{2} e\right )} {\left (f x + e\right )}^{2} - 6 \, {\left (-i \, {\left (f x + e\right )}^{2} b d^{2} + 2 \, {\left (-i \, b c d f + i \, b d^{2} e\right )} {\left (f x + e\right )}\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 6 \, {\left (i \, {\left (f x + e\right )} b d^{2} + i \, b c d f - i \, b d^{2} e\right )} {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 3 \, {\left ({\left (f x + e\right )}^{2} b d^{2} + 2 \, {\left (b c d f - b d^{2} e\right )} {\left (f x + 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 )}{f^{2}}}{6 \, 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)),x, algorithm="maxima")

[Out]

1/6*(6*(f*x + e)*a*c^2 + 2*(f*x + e)^3*a*d^2/f^2 + 6*(f*x + e)^2*a*c*d/f - 6*(f*x + e)^2*a*d^2*e/f^2 - 12*(f*x
+ e)*a*c*d*e/f + 6*b*c^2*log(sec(f*x + e)) - 12*b*c*d*e*log(sec(f*x + e))/f + 6*(f*x + e)*a*d^2*e^2/f^2 + 6*b
*d^2*e^2*log(sec(f*x + e))/f^2 - (-2*I*(f*x + e)^3*b*d^2 + 3*b*d^2*polylog(3, -e^(2*I*f*x + 2*I*e)) - 6*(I*b*c
*d*f - I*b*d^2*e)*(f*x + e)^2 - 6*(-I*(f*x + e)^2*b*d^2 + 2*(-I*b*c*d*f + I*b*d^2*e)*(f*x + e))*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e) + 1) - 6*(I*(f*x + e)*b*d^2 + I*b*c*d*f - I*b*d^2*e)*dilog(-e^(2*I*f*x + 2*I*e))
+ 3*((f*x + e)^2*b*d^2 + 2*(b*c*d*f - b*d^2*e)*(f*x + e))*log(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(
2*f*x + 2*e) + 1))/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. 331 vs. \(2 (101) = 202\).
time = 0.40, size = 331, normalized size = 2.88 \begin {gather*} \frac {4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x - 3 \, 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 \, 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 ) - 6 \, {\left (i \, b d^{2} f x + i \, b c d f\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 (-i \, b d^{2} f x - i \, b c d f\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 (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\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 d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, 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)),x, algorithm="fricas")

[Out]

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

[Out]

Integral((a + b*tan(e + f*x))*(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)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\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))*(c + d*x)^2,x)

[Out]

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

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