### 3.61 $$\int \frac{c+d x}{(a+b \tan (e+f x))^2} \, dx$$

Optimal. Leaf size=214 $-\frac{i a b d \text{PolyLog}\left (2,-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f^2 \left (a^2+b^2\right )^2}+\frac{b (2 a c f+2 a d f x+b d) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f^2 \left (a^2+b^2\right )^2}-\frac{b (c+d x)}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{(2 a c f+2 a d f x+b d)^2}{4 a d f^2 (a+i b) \left (a^2+b^2\right )}-\frac{(c+d x)^2}{2 d \left (a^2+b^2\right )}$

[Out]

-(c + d*x)^2/(2*(a^2 + b^2)*d) + (b*d + 2*a*c*f + 2*a*d*f*x)^2/(4*a*(a + I*b)*(a^2 + b^2)*d*f^2) + (b*(b*d + 2
*a*c*f + 2*a*d*f*x)*Log[1 + ((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2])/((a^2 + b^2)^2*f^2) - (I*a*b*d*Pol
yLog[2, -(((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2)])/((a^2 + b^2)^2*f^2) - (b*(c + d*x))/((a^2 + b^2)*f*
(a + b*Tan[e + f*x]))

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Rubi [A]  time = 0.275299, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.278, Rules used = {3733, 3732, 2190, 2279, 2391} $\frac{b (2 a c f+2 a d f x+b d) \log \left (1+\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f^2 \left (a^2+b^2\right )^2}-\frac{b (c+d x)}{f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{(2 a c f+2 a d f x+b d)^2}{4 a d f^2 (a+i b) \left (a^2+b^2\right )}-\frac{(c+d x)^2}{2 d \left (a^2+b^2\right )}-\frac{i a b d \text{Li}_2\left (-\frac{\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f^2 \left (a^2+b^2\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(c + d*x)^2/(2*(a^2 + b^2)*d) + (b*d + 2*a*c*f + 2*a*d*f*x)^2/(4*a*(a + I*b)*(a^2 + b^2)*d*f^2) + (b*(b*d + 2
*a*c*f + 2*a*d*f*x)*Log[1 + ((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2])/((a^2 + b^2)^2*f^2) - (I*a*b*d*Pol
yLog[2, -(((a^2 + b^2)*E^((2*I)*(e + f*x)))/(a + I*b)^2)])/((a^2 + b^2)^2*f^2) - (b*(c + d*x))/((a^2 + b^2)*f*
(a + b*Tan[e + f*x]))

Rule 3733

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

Rule 3732

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^S
imp[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 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 2279

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 2391

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]

Rubi steps

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

Mathematica [B]  time = 6.87, size = 745, normalized size = 3.48 $-\frac{d \sec ^2(e+f x) (a \cos (e+f x)+b \sin (e+f x))^2 \left (\frac{a \left (i \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}\left (\frac{a}{b}\right )+e+f x\right )}\right )+i \left (2 \tan ^{-1}\left (\frac{a}{b}\right )-\pi \right ) (e+f x)-2 \left (\tan ^{-1}\left (\frac{a}{b}\right )+e+f x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{a}{b}\right )+e+f x\right )}\right )+2 \tan ^{-1}\left (\frac{a}{b}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac{a}{b}\right )+e+f x\right )\right )-\pi \log \left (1+e^{-2 i (e+f x)}\right )+\pi \log (\cos (e+f x))\right )}{b \sqrt{\frac{a^2}{b^2}+1}}+e^{i \tan ^{-1}\left (\frac{a}{b}\right )} (e+f x)^2\right )}{f^2 (a-i b) (a+i b) \sqrt{\frac{a^2+b^2}{b^2}} (a+b \tan (e+f x))^2}+\frac{2 b c \sec ^2(e+f x) (a \cos (e+f x)+b \sin (e+f x))^2 (a \log (a \cos (e+f x)+b \sin (e+f x))-b (e+f x))}{f (a-i b) (a+i b) \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}+\frac{b^2 d \sec ^2(e+f x) (a \cos (e+f x)+b \sin (e+f x))^2 (a \log (a \cos (e+f x)+b \sin (e+f x))-b (e+f x))}{a f^2 (a-i b) (a+i b) \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac{2 b d e \sec ^2(e+f x) (a \cos (e+f x)+b \sin (e+f x))^2 (a \log (a \cos (e+f x)+b \sin (e+f x))-b (e+f x))}{f^2 (a-i b) (a+i b) \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}+\frac{\sec ^2(e+f x) (a \cos (e+f x)+b \sin (e+f x)) \left (b^2 c f \sin (e+f x)+b^2 (-d) e \sin (e+f x)+b^2 d (e+f x) \sin (e+f x)\right )}{a f^2 (a-i b) (a+i b) (a+b \tan (e+f x))^2}+\frac{(e+f x) \sec ^2(e+f x) (2 c f+d (e+f x)-2 d e) (a \cos (e+f x)+b \sin (e+f x))^2}{2 f^2 (a-i b) (a+i b) (a+b \tan (e+f x))^2}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((e + f*x)*(-2*d*e + 2*c*f + d*(e + f*x))*Sec[e + f*x]^2*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)/(2*(a - I*b)*(a
+ I*b)*f^2*(a + b*Tan[e + f*x])^2) + (b^2*d*(-(b*(e + f*x)) + a*Log[a*Cos[e + f*x] + b*Sin[e + f*x]])*Sec[e +
f*x]^2*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)/(a*(a - I*b)*(a + I*b)*(a^2 + b^2)*f^2*(a + b*Tan[e + f*x])^2) - (
2*b*d*e*(-(b*(e + f*x)) + a*Log[a*Cos[e + f*x] + b*Sin[e + f*x]])*Sec[e + f*x]^2*(a*Cos[e + f*x] + b*Sin[e + f
*x])^2)/((a - I*b)*(a + I*b)*(a^2 + b^2)*f^2*(a + b*Tan[e + f*x])^2) + (2*b*c*(-(b*(e + f*x)) + a*Log[a*Cos[e
+ f*x] + b*Sin[e + f*x]])*Sec[e + f*x]^2*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)/((a - I*b)*(a + I*b)*(a^2 + b^2)
*f*(a + b*Tan[e + f*x])^2) - (d*(E^(I*ArcTan[a/b])*(e + f*x)^2 + (a*(I*(e + f*x)*(-Pi + 2*ArcTan[a/b]) - Pi*Lo
g[1 + E^((-2*I)*(e + f*x))] - 2*(e + f*x + ArcTan[a/b])*Log[1 - E^((2*I)*(e + f*x + ArcTan[a/b]))] + Pi*Log[Co
s[e + f*x]] + 2*ArcTan[a/b]*Log[Sin[e + f*x + ArcTan[a/b]]] + I*PolyLog[2, E^((2*I)*(e + f*x + ArcTan[a/b]))])
)/(Sqrt[1 + a^2/b^2]*b))*Sec[e + f*x]^2*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)/((a - I*b)*(a + I*b)*Sqrt[(a^2 +
b^2)/b^2]*f^2*(a + b*Tan[e + f*x])^2) + (Sec[e + f*x]^2*(a*Cos[e + f*x] + b*Sin[e + f*x])*(-(b^2*d*e*Sin[e + f
*x]) + b^2*c*f*Sin[e + f*x] + b^2*d*(e + f*x)*Sin[e + f*x]))/(a*(a - I*b)*(a + I*b)*f^2*(a + b*Tan[e + f*x])^2
)

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Maple [B]  time = 0.279, size = 999, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 3.37584, size = 1570, normalized size = 7.34 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^3 - I*a^2*b + a*b^2 - I*b^3)*d*f^2*x^2 + (2*a^3 - 2*I*a^2*b + 2*a*b^2 - 2*I*b^3)*c*f^2*x - 4*(-I*a*b^2 + b
^3)*c*f + ((4*I*a^2*b - 4*a*b^2)*c*f - 2*(-I*a*b^2 + b^3)*d + ((4*I*a^2*b + 4*a*b^2)*c*f - 2*(-I*a*b^2 - b^3)*
d)*cos(2*f*x + 2*e) - (4*(a^2*b - I*a*b^2)*c*f + (2*a*b^2 - 2*I*b^3)*d)*sin(2*f*x + 2*e))*arctan2(-b*cos(2*f*x
+ 2*e) + a*sin(2*f*x + 2*e) + b, a*cos(2*f*x + 2*e) + b*sin(2*f*x + 2*e) + a) + ((-4*I*a^2*b - 4*a*b^2)*d*f*x
*cos(2*f*x + 2*e) + 4*(a^2*b - I*a*b^2)*d*f*x*sin(2*f*x + 2*e) + (-4*I*a^2*b + 4*a*b^2)*d*f*x)*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)) + ((a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*d*f^2*x^2 + ((2*a^3 - 6*I*a^2*b - 6*a*
b^2 + 2*I*b^3)*c*f^2 - 4*(I*a*b^2 + b^3)*d*f)*x)*cos(2*f*x + 2*e) + ((-2*I*a^2*b - 2*a*b^2)*d*cos(2*f*x + 2*e)
+ 2*(a^2*b - I*a*b^2)*d*sin(2*f*x + 2*e) + (-2*I*a^2*b + 2*a*b^2)*d)*dilog((I*a + b)*e^(2*I*f*x + 2*I*e)/(-I*
a + b)) + (2*(a^2*b + I*a*b^2)*c*f + (a*b^2 + I*b^3)*d + (2*(a^2*b - I*a*b^2)*c*f + (a*b^2 - I*b^3)*d)*cos(2*f
*x + 2*e) + ((2*I*a^2*b + 2*a*b^2)*c*f + (I*a*b^2 + b^3)*d)*sin(2*f*x + 2*e))*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)) + (
2*(a^2*b - I*a*b^2)*d*f*x*cos(2*f*x + 2*e) + (2*I*a^2*b + 2*a*b^2)*d*f*x*sin(2*f*x + 2*e) + 2*(a^2*b + I*a*b^2
)*d*f*x)*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))/(a^2 + b^2)) + ((I*a^3 + 3*a^2*b - 3*I*a*b^2 - b^3)*d*f^2*x^2 + ((2*I*a
^3 + 6*a^2*b - 6*I*a*b^2 - 2*b^3)*c*f^2 + (4*a*b^2 - 4*I*b^3)*d*f)*x)*sin(2*f*x + 2*e))/((2*a^5 - 2*I*a^4*b +
4*a^3*b^2 - 4*I*a^2*b^3 + 2*a*b^4 - 2*I*b^5)*f^2*cos(2*f*x + 2*e) + (2*I*a^5 + 2*a^4*b + 4*I*a^3*b^2 + 4*a^2*b
^3 + 2*I*a*b^4 + 2*b^5)*f^2*sin(2*f*x + 2*e) + (2*a^5 + 2*I*a^4*b + 4*a^3*b^2 + 4*I*a^2*b^3 + 2*a*b^4 + 2*I*b^
5)*f^2)

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Fricas [B]  time = 2.01808, size = 1952, normalized size = 9.12 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^3 - a*b^2)*d*f^2*x^2 - 2*b^3*c*f - 2*(b^3*d*f - (a^3 - a*b^2)*c*f^2)*x + (-I*a*b^2*d*tan(f*x + e) - I*
a^2*b*d)*dilog(-((2*I*a*b + 2*b^2)*tan(f*x + e)^2 + 2*a^2 - 2*I*a*b + (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(f*x + e)
)/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2) + 1) + (I*a*b^2*d*tan(f*x + e) + I*a^2*b*d)*dilog(-((-2*I*a*b + 2*b
^2)*tan(f*x + e)^2 + 2*a^2 + 2*I*a*b + (-2*I*a^2 + 4*a*b + 2*I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2
+ a^2 + b^2) + 1) + 2*(a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*tan(f*x + e))*log(((2*I*a*b + 2*b^2
)*tan(f*x + e)^2 + 2*a^2 - 2*I*a*b + (2*I*a^2 + 4*a*b - 2*I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a
^2 + b^2)) + 2*(a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*tan(f*x + e))*log(((-2*I*a*b + 2*b^2)*tan(
f*x + e)^2 + 2*a^2 + 2*I*a*b + (-2*I*a^2 + 4*a*b + 2*I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 +
b^2)) - (2*a^2*b*d*e - 2*a^2*b*c*f - a*b^2*d + (2*a*b^2*d*e - 2*a*b^2*c*f - b^3*d)*tan(f*x + e))*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)) - (2*a^2*b*d*e - 2*a^
2*b*c*f - a*b^2*d + (2*a*b^2*d*e - 2*a*b^2*c*f - b^3*d)*tan(f*x + e))*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 - b^3)*d*f^2*x^2 + 2*a*b^2*c*f + 2*(a*
b^2*d*f + (a^2*b - b^3)*c*f^2)*x)*tan(f*x + e))/((a^4*b + 2*a^2*b^3 + b^5)*f^2*tan(f*x + e) + (a^5 + 2*a^3*b^2
+ a*b^4)*f^2)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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