3.40 \(\int (c+d x)^2 (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=115 \[ \frac {a (c+d x)^3}{3 d}+\frac {i b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^3}{3 d}-\frac {b d^2 \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3} \]

[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 = 0.21, 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 = {3722, 3719, 2190, 2531, 2282, 6589} \[ \frac {a (c+d x)^3}{3 d}+\frac {i b d (c+d x) \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {b (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^3}{3 d}-\frac {b d^2 \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3} \]

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

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 2531

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 3719

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 3722

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 6589

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 ) \operatorname {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 [A]  time = 0.13, size = 191, normalized size = 1.66 \[ a c^2 x+a c d x^2+\frac {1}{3} a d^2 x^3-\frac {b c^2 \log (\cos (e+f x))}{f}+\frac {i b c d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {2 b c d x \log \left (1+e^{2 i (e+f x)}\right )}{f}+i b c d x^2-\frac {b d^2 \text {Li}_3\left (-e^{2 i (e+f x)}\right )}{2 f^3}+\frac {i b d^2 x \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}-\frac {b d^2 x^2 \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {1}{3} i b d^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.56, size = 315, normalized size = 2.74 \[ \frac {4 \, a d^{2} f^{3} x^{3} + 12 \, a c d f^{3} x^{2} + 12 \, a c^{2} f^{3} x - 3 \, b d^{2} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} + 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) - 3 \, b d^{2} {\rm polylog}\left (3, \frac {\tan \left (f x + e\right )^{2} - 2 i \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) + {\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) + {\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + b c^{2} f^{2}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, f^{3}} \]

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 - 6*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 +
6*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} {\left (b \tan \left (f x + e\right ) + a\right )}\,{d x} \]

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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maple [B]  time = 0.43, size = 295, normalized size = 2.57 \[ \frac {i b c d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}+\frac {i b \,d^{2} x^{3}}{3}+\frac {a \,d^{2} x^{3}}{3}+i b c d \,x^{2}+a c d \,x^{2}+a \,c^{2} x -\frac {b \,c^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 i b \,d^{2} e^{2} x}{f^{2}}+\frac {2 i b c d \,e^{2}}{f^{2}}-\frac {4 i b \,d^{2} e^{3}}{3 f^{3}}-\frac {b \,d^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x^{2}}{f}+\frac {i b \,d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right ) x}{f^{2}}-\frac {b \,d^{2} \polylog \left (3, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{3}}-\frac {4 b c d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-i b \,c^{2} x +\frac {4 i b c d e x}{f}-\frac {2 b c d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.23, size = 369, normalized size = 3.21 \[ \frac {6 \, {\left (f x + e\right )} a c^{2} + \frac {2 \, {\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac {6 \, {\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac {6 \, {\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac {6 \, {\left (f x + e\right )}^{2} a c d}{f} - \frac {12 \, {\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sec \left (f x + e\right )\right ) + \frac {6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right )\right )}{f^{2}} - \frac {12 \, b c d e \log \left (\sec \left (f x + e\right )\right )}{f} - \frac {-2 i \, {\left (f x + e\right )}^{3} b d^{2} + 3 \, b d^{2} {\rm Li}_{3}(-e^{\left (2 i \, f x + 2 i \, e\right )}) + {\left (6 i \, b d^{2} e - 6 i \, b c d f\right )} {\left (f x + e\right )}^{2} + {\left (6 i \, {\left (f x + e\right )}^{2} b d^{2} + {\left (-12 i \, b d^{2} e + 12 i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + {\left (-6 i \, {\left (f x + e\right )} b d^{2} + 6 i \, b d^{2} e - 6 i \, b c d f\right )} {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) + 3 \, {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}{f^{2}}}{6 \, f} \]

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*d^2*e/f^2 + 6*(f*x + e)*a*d^2*e^2/f^2 + 6*(
f*x + e)^2*a*c*d/f - 12*(f*x + e)*a*c*d*e/f + 6*b*c^2*log(sec(f*x + e)) + 6*b*d^2*e^2*log(sec(f*x + e))/f^2 -
12*b*c*d*e*log(sec(f*x + e))/f - (-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*d
^2*e - 6*I*b*c*d*f)*(f*x + e)^2 + (6*I*(f*x + e)^2*b*d^2 + (-12*I*b*d^2*e + 12*I*b*c*d*f)*(f*x + e))*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) + (-6*I*(f*x + e)*b*d^2 + 6*I*b*d^2*e - 6*I*b*c*d*f)*dilog(-e^(2*I*f*x
+ 2*I*e)) + 3*((f*x + e)^2*b*d^2 - 2*(b*d^2*e - b*c*d*f)*(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^2 \,d x \]

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

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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