3.1.55 \(\int \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx\) [55]

Optimal. Leaf size=181 \[ \frac {(c+d x)^3}{3 (a+i b) d}+\frac {b (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac {i b d (c+d x) \text {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f^2}+\frac {b d^2 \text {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^3} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3813, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {i b d (c+d x) \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f^2 \left (a^2+b^2\right )}+\frac {b (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f \left (a^2+b^2\right )}+\frac {b d^2 \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^3 \left (a^2+b^2\right )}+\frac {(c+d x)^3}{3 d (a+i b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^3/(3*(a + I*b)*d) + (b*(c + d*x)^2*Log[1 + ((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2])/((a^2 + b
^2)*f) - (I*b*d*(c + d*x)*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2)])/((a^2 + b^2)*f^2) + (b
*d^2*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2)])/(2*(a^2 + b^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 3813

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Si
mp[2*I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && IGtQ[m, 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 \frac {(c+d x)^2}{a+b \tan (e+f x)} \, dx &=\frac {(c+d x)^3}{3 (a+i b) d}+(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)^2}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (e+f x)}} \, dx\\ &=\frac {(c+d x)^3}{3 (a+i b) d}+\frac {b (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac {(2 b d) \int (c+d x) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f}\\ &=\frac {(c+d x)^3}{3 (a+i b) d}+\frac {b (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac {i b d (c+d x) \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f^2}+\frac {\left (i b d^2\right ) \int \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right ) \, dx}{\left (a^2+b^2\right ) f^2}\\ &=\frac {(c+d x)^3}{3 (a+i b) d}+\frac {b (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac {i b d (c+d x) \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f^2}+\frac {\left (b d^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 \left (a^2+b^2\right ) f^3}\\ &=\frac {(c+d x)^3}{3 (a+i b) d}+\frac {b (c+d x)^2 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f}-\frac {i b d (c+d x) \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) f^2}+\frac {b d^2 \text {Li}_3\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) f^3}\\ \end {align*}

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Mathematica [A]
time = 1.70, size = 321, normalized size = 1.77 \begin {gather*} \frac {b \left (2 f^2 \left (-2 i (a-i b) e^{2 i e} f x \left (3 c^2+3 c d x+d^2 x^2\right )+3 \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) (c+d x)^2 \log \left (1+\frac {(a-i b) e^{2 i (e+f x)}}{a+i b}\right )\right )+6 d \left (b-b e^{2 i e}-i a \left (1+e^{2 i e}\right )\right ) f (c+d x) \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i (e+f x)}}{a+i b}\right )+3 d^2 \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) \text {PolyLog}\left (3,-\frac {(a-i b) e^{2 i (e+f x)}}{a+i b}\right )\right )}{6 \left (a^2+b^2\right ) \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) f^3}+\frac {x \left (3 c^2+3 c d x+d^2 x^2\right ) \cos (e)}{3 (a \cos (e)+b \sin (e))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(2*f^2*((-2*I)*(a - I*b)*E^((2*I)*e)*f*x*(3*c^2 + 3*c*d*x + d^2*x^2) + 3*((-I)*b*(-1 + E^((2*I)*e)) + a*(1
+ E^((2*I)*e)))*(c + d*x)^2*Log[1 + ((a - I*b)*E^((2*I)*(e + f*x)))/(a + I*b)]) + 6*d*(b - b*E^((2*I)*e) - I*a
*(1 + E^((2*I)*e)))*f*(c + d*x)*PolyLog[2, -(((a - I*b)*E^((2*I)*(e + f*x)))/(a + I*b))] + 3*d^2*((-I)*b*(-1 +
E^((2*I)*e)) + a*(1 + E^((2*I)*e)))*PolyLog[3, -(((a - I*b)*E^((2*I)*(e + f*x)))/(a + I*b))]))/(6*(a^2 + b^2)
*((-I)*b*(-1 + E^((2*I)*e)) + a*(1 + E^((2*I)*e)))*f^3) + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Cos[e])/(3*(a*Cos[e]
+ b*Sin[e]))

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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. 939 vs. \(2 (166 ) = 332\).
time = 0.39, size = 940, normalized size = 5.19

method	result	size
risch	\(-\frac {d^{2} x^{3}}{3 \left (i b -a \right )}-\frac {d c \,x^{2}}{i b -a}-\frac {c^{2} x}{i b -a}-\frac {c^{3}}{3 d \left (i b -a \right )}+\frac {i b \,d^{2} e^{2} \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{f^{3} \left (i a +b \right ) \left (-i b -a \right )}-\frac {b \,d^{2} \polylog \left (2, \frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}+\frac {i b \,d^{2} e^{2} \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f^{3} \left (i a +b \right ) \left (i b +a \right )}-\frac {2 i b \,d^{2} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3} \left (i a +b \right ) \left (i b +a \right )}-\frac {2 i b c d e \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f^{2} \left (i a +b \right ) \left (i b +a \right )}+\frac {i b \,c^{2} \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f \left (i a +b \right ) \left (i b +a \right )}-\frac {2 i b c d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) e}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 i b \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (i a +b \right ) \left (i b +a \right )}-\frac {2 i b c d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x}{f \left (i a +b \right ) \left (-i b -a \right )}+\frac {4 i b c d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2} \left (i a +b \right ) \left (i b +a \right )}-\frac {2 b \,d^{2} x^{3}}{3 \left (i a +b \right ) \left (-i b -a \right )}+\frac {2 b \,d^{2} e^{2} x}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}+\frac {4 b \,d^{2} e^{3}}{3 f^{3} \left (i a +b \right ) \left (-i b -a \right )}-\frac {i b \,d^{2} \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x^{2}}{f \left (i a +b \right ) \left (-i b -a \right )}-\frac {i b \,d^{2} \polylog \left (3, \frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{2 f^{3} \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 b c d \,x^{2}}{\left (i a +b \right ) \left (-i b -a \right )}-\frac {4 b c d e x}{f \left (i a +b \right ) \left (-i b -a \right )}-\frac {2 b c d \,e^{2}}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}-\frac {b c d \polylog \left (2, \frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{f^{2} \left (i a +b \right ) \left (-i b -a \right )}\)	\(940\)

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

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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. 755 vs. \(2 (163) = 326\).
time = 0.65, size = 755, normalized size = 4.17 \begin {gather*} -\frac {6 \, c d {\left (\frac {2 \, {\left (f x + e\right )} a}{{\left (a^{2} + b^{2}\right )} f} + \frac {2 \, b \log \left (b \tan \left (f x + e\right ) + a\right )}{{\left (a^{2} + b^{2}\right )} f} - \frac {b \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{{\left (a^{2} + b^{2}\right )} f}\right )} e - 3 \, {\left (\frac {2 \, {\left (f x + e\right )} a}{a^{2} + b^{2}} + \frac {2 \, b \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{2} + b^{2}} - \frac {b \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{2} + b^{2}}\right )} c^{2} - \frac {2 \, {\left (f x + e\right )}^{3} {\left (a - i \, b\right )} d^{2} + 6 i \, b d^{2} \arctan \left (-b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) + b \sin \left (2 \, f x + 2 \, e\right ) + a\right ) e^{2} + 3 \, b d^{2} e^{2} \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) + 6 \, {\left (f x + e\right )} {\left (a e^{2} - i \, b e^{2}\right )} d^{2} + 3 \, b d^{2} {\rm Li}_{3}(\frac {{\left (i \, a + b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{-i \, a + b}) + 6 \, {\left ({\left (a - i \, b\right )} c d f - {\left (a e - i \, b e\right )} d^{2}\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 (\frac {2 \, a b \cos \left (2 \, f x + 2 \, e\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\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 (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{-i \, a + b}\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 (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right )}{{\left (a^{2} + b^{2}\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*c*d*(2*(f*x + e)*a/((a^2 + b^2)*f) + 2*b*log(b*tan(f*x + e) + a)/((a^2 + b^2)*f) - b*log(tan(f*x + e)^
2 + 1)/((a^2 + b^2)*f))*e - 3*(2*(f*x + e)*a/(a^2 + b^2) + 2*b*log(b*tan(f*x + e) + a)/(a^2 + b^2) - b*log(tan
(f*x + e)^2 + 1)/(a^2 + b^2))*c^2 - (2*(f*x + e)^3*(a - I*b)*d^2 + 6*I*b*d^2*arctan2(-b*cos(2*f*x + 2*e) + a*s
in(2*f*x + 2*e) + b, a*cos(2*f*x + 2*e) + b*sin(2*f*x + 2*e) + a)*e^2 + 3*b*d^2*e^2*log((a^2 + b^2)*cos(2*f*x
+ 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2)*sin(2*f*x + 2*e)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*f*x + 2*e
)) + 6*(f*x + e)*(a*e^2 - I*b*e^2)*d^2 + 3*b*d^2*polylog(3, (I*a + b)*e^(2*I*f*x + 2*I*e)/(-I*a + b)) + 6*((a
- I*b)*c*d*f - (a*e - I*b*e)*d^2)*(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((2*a*b*cos(2*f*x + 2*e) - (a^2 - b^2)*sin(2*f*x + 2*e))/(a^2 + b^2), (2*a*b*sin(2*f*x + 2*e) + a^2 + b
^2 + (a^2 - b^2)*cos(2*f*x + 2*e))/(a^2 + b^2)) - 6*(I*(f*x + e)*b*d^2 + I*b*c*d*f - I*b*d^2*e)*dilog((I*a + b
)*e^(2*I*f*x + 2*I*e)/(-I*a + b)) + 3*((f*x + e)^2*b*d^2 + 2*(b*c*d*f - b*d^2*e)*(f*x + e))*log(((a^2 + b^2)*c
os(2*f*x + 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2)*sin(2*f*x + 2*e)^2 + a^2 + b^2 + 2*(a^2 - b^2)*cos(2*
f*x + 2*e))/(a^2 + b^2)))/((a^2 + b^2)*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. 858 vs. \(2 (163) = 326\).
time = 0.39, size = 858, normalized size = 4.74 \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 {{\left (a^{2} + 2 i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - 2 i \, a b + b^{2} - 2 \, {\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 3 \, b d^{2} {\rm polylog}\left (3, \frac {{\left (a^{2} - 2 i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + 2 i \, a b + b^{2} - 2 \, {\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) - 6 \, {\left (-i \, b d^{2} f x - i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) - 6 \, {\left (i \, b d^{2} f x + i \, b c d f\right )} {\rm Li}_2\left (\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) + 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + 2 \, b c d f e - b d^{2} e^{2}\right )} \log \left (-\frac {2 \, {\left ({\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - i \, a b + {\left (i \, a^{2} - 2 \, a b - i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 6 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x + 2 \, b c d f e - b d^{2} e^{2}\right )} \log \left (-\frac {2 \, {\left ({\left (-i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b + {\left (-i \, a^{2} - 2 \, a b + i \, b^{2}\right )} \tan \left (f x + e\right )\right )}}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 6 \, {\left (b c^{2} f^{2} - 2 \, b c d f e + b d^{2} e^{2}\right )} \log \left (\frac {{\left (i \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right ) + 6 \, {\left (b c^{2} f^{2} - 2 \, b c d f e + b d^{2} e^{2}\right )} \log \left (\frac {{\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right )}{12 \, {\left (a^{2} + b^{2}\right )} 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, ((a^2 + 2*I*a*b - b^2)*tan(f*x
+ e)^2 - a^2 - 2*I*a*b + b^2 - 2*(-I*a^2 + 2*a*b + I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^
2)) + 3*b*d^2*polylog(3, ((a^2 - 2*I*a*b - b^2)*tan(f*x + e)^2 - a^2 + 2*I*a*b + b^2 - 2*(I*a^2 + 2*a*b - I*b^
2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2)) - 6*(-I*b*d^2*f*x - I*b*c*d*f)*dilog(2*((I*a*b - b^
2)*tan(f*x + e)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^
2) + 1) - 6*(I*b*d^2*f*x + I*b*c*d*f)*dilog(2*((-I*a*b - b^2)*tan(f*x + e)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b +
I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2) + 1) + 6*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x + 2*b*c
*d*f*e - b*d^2*e^2)*log(-2*((I*a*b - b^2)*tan(f*x + e)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(f*x + e))
/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2)) + 6*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x + 2*b*c*d*f*e - b*d^2*e^2)*log(-
2*((-I*a*b - b^2)*tan(f*x + e)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x +
e)^2 + a^2 + b^2)) + 6*(b*c^2*f^2 - 2*b*c*d*f*e + b*d^2*e^2)*log(((I*a*b + b^2)*tan(f*x + e)^2 - a^2 + I*a*b
+ (I*a^2 + I*b^2)*tan(f*x + e))/(tan(f*x + e)^2 + 1)) + 6*(b*c^2*f^2 - 2*b*c*d*f*e + b*d^2*e^2)*log(((I*a*b -
b^2)*tan(f*x + e)^2 + a^2 + I*a*b + (I*a^2 + I*b^2)*tan(f*x + e))/(tan(f*x + e)^2 + 1)))/((a^2 + b^2)*f^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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