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

Optimal. Leaf size=157 \[ \frac {a (c+d x)^3}{3 d}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac {2 b d^2 \text {Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {Li}_3\left (i e^{i (e+f x)}\right )}{f^3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4190, 4181, 2531, 2282, 6589} \[ \frac {2 i b d (c+d x) \text {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {2 b d^2 \text {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4181

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

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[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 \sec (e+f x)) \, dx &=\int \left (a (c+d x)^2+b (c+d x)^2 \sec (e+f x)\right ) \, dx\\ &=\frac {a (c+d x)^3}{3 d}+b \int (c+d x)^2 \sec (e+f x) \, dx\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac {(2 b d) \int (c+d x) \log \left (1-i e^{i (e+f x)}\right ) \, dx}{f}+\frac {(2 b d) \int (c+d x) \log \left (1+i e^{i (e+f x)}\right ) \, dx}{f}\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {\left (2 i b d^2\right ) \int \text {Li}_2\left (-i e^{i (e+f x)}\right ) \, dx}{f^2}+\frac {\left (2 i b d^2\right ) \int \text {Li}_2\left (i e^{i (e+f x)}\right ) \, dx}{f^2}\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {\left (2 b d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3}+\frac {\left (2 b d^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i (e+f x)}\right )}{f^3}\\ &=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {2 b d^2 \text {Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {Li}_3\left (i e^{i (e+f x)}\right )}{f^3}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 203, normalized size = 1.29 \[ a c^2 x+a c d x^2+\frac {1}{3} a d^2 x^3+\frac {b c^2 \tanh ^{-1}(\sin (e+f x))}{f}+\frac {2 i b d (c+d x) \text {Li}_2\left (-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \text {Li}_2\left (i e^{i (e+f x)}\right )}{f^2}-\frac {4 i b c d x \tan ^{-1}\left (e^{i (e+f x)}\right )}{f}-\frac {2 b d^2 \text {Li}_3\left (-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \text {Li}_3\left (i e^{i (e+f x)}\right )}{f^3}-\frac {2 i b d^2 x^2 \tan ^{-1}\left (e^{i (e+f x)}\right )}{f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*c^2*x + a*c*d*x^2 + (a*d^2*x^3)/3 - ((4*I)*b*c*d*x*ArcTan[E^(I*(e + f*x))])/f - ((2*I)*b*d^2*x^2*ArcTan[E^(I
*(e + f*x))])/f + (b*c^2*ArcTanh[Sin[e + f*x]])/f + ((2*I)*b*d*(c + d*x)*PolyLog[2, (-I)*E^(I*(e + f*x))])/f^2
- ((2*I)*b*d*(c + d*x)*PolyLog[2, I*E^(I*(e + f*x))])/f^2 - (2*b*d^2*PolyLog[3, (-I)*E^(I*(e + f*x))])/f^3 +
(2*b*d^2*PolyLog[3, I*E^(I*(e + f*x))])/f^3

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fricas [C]  time = 0.80, size = 671, normalized size = 4.27 \[ \frac {2 \, a d^{2} f^{3} x^{3} + 6 \, a c d f^{3} x^{2} + 6 \, a c^{2} f^{3} x - 6 \, b d^{2} {\rm polylog}\left (3, i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 6 \, b d^{2} {\rm polylog}\left (3, i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) - 6 \, b d^{2} {\rm polylog}\left (3, -i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + 6 \, b d^{2} {\rm polylog}\left (3, -i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + {\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + {\left (-6 i \, b d^{2} f x - 6 i \, b c d f\right )} {\rm Li}_2\left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + {\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right )\right ) + {\left (6 i \, b d^{2} f x + 6 i \, b c d f\right )} {\rm Li}_2\left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right )\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-i \, \cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-i \, \cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + i\right ) - 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + i\right )}{6 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*d^2*f^3*x^3 + 6*a*c*d*f^3*x^2 + 6*a*c^2*f^3*x - 6*b*d^2*polylog(3, I*cos(f*x + e) + sin(f*x + e)) + 6
*b*d^2*polylog(3, I*cos(f*x + e) - sin(f*x + e)) - 6*b*d^2*polylog(3, -I*cos(f*x + e) + sin(f*x + e)) + 6*b*d^
2*polylog(3, -I*cos(f*x + e) - sin(f*x + e)) + (-6*I*b*d^2*f*x - 6*I*b*c*d*f)*dilog(I*cos(f*x + e) + sin(f*x +
e)) + (-6*I*b*d^2*f*x - 6*I*b*c*d*f)*dilog(I*cos(f*x + e) - sin(f*x + e)) + (6*I*b*d^2*f*x + 6*I*b*c*d*f)*dil
og(-I*cos(f*x + e) + sin(f*x + e)) + (6*I*b*d^2*f*x + 6*I*b*c*d*f)*dilog(-I*cos(f*x + e) - sin(f*x + e)) + 3*(
b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(cos(f*x + e) + I*sin(f*x + e) + I) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c
^2*f^2)*log(cos(f*x + e) - I*sin(f*x + e) + I) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*l
og(I*cos(f*x + e) + sin(f*x + e) + 1) - 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(I*cos(
f*x + e) - sin(f*x + e) + 1) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(-I*cos(f*x + e)
+ sin(f*x + e) + 1) - 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(-I*cos(f*x + e) - sin(f
*x + e) + 1) + 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(-cos(f*x + e) + I*sin(f*x + e) + I) - 3*(b*d^2*e^2
- 2*b*c*d*e*f + b*c^2*f^2)*log(-cos(f*x + e) - I*sin(f*x + e) + I))/f^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.94, size = 431, normalized size = 2.75 \[ \frac {a \,d^{2} x^{3}}{3}+a c d \,x^{2}+a \,c^{2} x -\frac {b \,d^{2} e^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {2 b c d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 i b \,d^{2} \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {2 b c d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {2 i b \,c^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}-\frac {2 i b \,d^{2} e^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 b c d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {4 i b c d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b c d \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) e}{f^{2}}+\frac {2 b \,d^{2} \polylog \left (3, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {b \,d^{2} e^{2} \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right )}{f^{3}}-\frac {2 b \,d^{2} \polylog \left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {b \,d^{2} \ln \left (i {\mathrm e}^{i \left (f x +e \right )}+1\right ) x^{2}}{f}+\frac {2 i b \,d^{2} \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}+\frac {b \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f}+\frac {2 i b c d \polylog \left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 i b c d \polylog \left (2, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.23, size = 510, normalized size = 3.25 \[ \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 ) + \tan \left (f x + e\right )\right ) + \frac {6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f^{2}} - \frac {12 \, b c d e \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f} + \frac {3 \, {\left (4 \, b d^{2} {\rm Li}_{3}(i \, e^{\left (i \, f x + i \, e\right )}) - 4 \, b d^{2} {\rm Li}_{3}(-i \, e^{\left (i \, f x + i \, e\right )}) + {\left (-2 i \, {\left (f x + e\right )}^{2} b d^{2} + {\left (4 i \, b d^{2} e - 4 i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + {\left (-2 i \, {\left (f x + e\right )}^{2} b d^{2} + {\left (4 i \, b d^{2} e - 4 i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + {\left (-4 i \, {\left (f x + e\right )} b d^{2} + 4 i \, b d^{2} e - 4 i \, b c d f\right )} {\rm Li}_2\left (i \, e^{\left (i \, f x + i \, e\right )}\right ) + {\left (4 i \, {\left (f x + e\right )} b d^{2} - 4 i \, b d^{2} e + 4 i \, b c d f\right )} {\rm Li}_2\left (-i \, e^{\left (i \, f x + i \, e\right )}\right ) + {\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 (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - {\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 (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )}}{f^{2}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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