3.56 \(\int \frac {c+d x}{a+b \tan (e+f x)} \, dx\)

Optimal. Leaf size=125 \[ \frac {b (c+d x) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f \left (a^2+b^2\right )}-\frac {i b d \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac {(c+d x)^2}{2 d (a+i b)} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3732, 2190, 2279, 2391} \[ \frac {b (c+d x) \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{f \left (a^2+b^2\right )}-\frac {i b d \text {Li}_2\left (-\frac {\left (a^2+b^2\right ) e^{2 i (e+f x)}}{(a+i b)^2}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac {(c+d x)^2}{2 d (a+i b)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*Tan[e + f*x]),x]

[Out]

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

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]

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]

Rubi steps

\begin {align*} \int \frac {c+d x}{a+b \tan (e+f x)} \, dx &=\frac {(c+d x)^2}{2 (a+i b) d}+(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (e+f x)}} \, dx\\ &=\frac {(c+d x)^2}{2 (a+i b) d}+\frac {b (c+d 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 ) f}-\frac {(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 ) f}\\ &=\frac {(c+d x)^2}{2 (a+i b) d}+\frac {b (c+d 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 ) f}+\frac {(i 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 )}{2 \left (a^2+b^2\right ) f^2}\\ &=\frac {(c+d x)^2}{2 (a+i b) d}+\frac {b (c+d 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 ) f}-\frac {i b d \text {Li}_2\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^2}\\ \end {align*}

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Mathematica [A]  time = 1.92, size = 177, normalized size = 1.42 \[ \frac {b (c+d x) \log \left (1+\frac {(a+i b) e^{-2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac {i b d \text {Li}_2\left (\frac {(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}+\frac {b (c+d x)^2}{d (a-i b) \left (-i a \left (1+e^{2 i e}\right )+b \left (-e^{2 i e}\right )+b\right )}+\frac {x \cos (e) (2 c+d x)}{2 (a \cos (e)+b \sin (e))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 533, normalized size = 4.26 \[ \frac {2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x + i \, b d {\rm Li}_2\left (\frac {2 \, {\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 2 i \, a b - {\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) - i \, b d {\rm Li}_2\left (\frac {2 \, {\left (-i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} + 2 i \, a b - {\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}} + 1\right ) + 2 \, {\left (b d f x + b d e\right )} \log \left (-\frac {2 \, {\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} - 2 i \, a b - {\left (-2 i \, a^{2} + 4 \, a b + 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) + 2 \, {\left (b d f x + b d e\right )} \log \left (-\frac {2 \, {\left (-i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - 2 \, a^{2} + 2 i \, a b - {\left (2 i \, a^{2} + 4 \, a b - 2 i \, b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{2} + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + b^{2}}\right ) - 2 \, {\left (b d e - b c f\right )} \log \left (\frac {{\left (i \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, {\left (b d e - b c f\right )} \log \left (\frac {{\left (i \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + i \, a b + {\left (i \, a^{2} + i \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, {\left (a^{2} + b^{2}\right )} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.96, size = 462, normalized size = 3.70 \[ -\frac {d \,x^{2}}{2 \left (i b -a \right )}-\frac {c x}{i b -a}-\frac {2 b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (-i b +a \right ) \left (i b +a \right )}+\frac {b c \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f \left (-i b +a \right ) \left (i b +a \right )}-\frac {b d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x}{f \left (-i b +a \right ) \left (-i b -a \right )}-\frac {b d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) e}{f^{2} \left (-i b +a \right ) \left (-i b -a \right )}+\frac {i b d \,x^{2}}{\left (-i b +a \right ) \left (-i b -a \right )}+\frac {2 i b d e x}{f \left (-i b +a \right ) \left (-i b -a \right )}+\frac {i b d \,e^{2}}{f^{2} \left (-i b +a \right ) \left (-i b -a \right )}+\frac {i b d \polylog \left (2, \frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{2 f^{2} \left (-i b +a \right ) \left (-i b -a \right )}+\frac {2 b d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2} \left (-i b +a \right ) \left (i b +a \right )}-\frac {b d e \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f^{2} \left (-i b +a \right ) \left (i b +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.76, size = 399, normalized size = 3.19 \[ \frac {{\left (a - i \, b\right )} d f^{2} x^{2} + 2 \, {\left (a - i \, b\right )} c f^{2} x - 2 i \, b d f x \arctan \left (\frac {2 \, a b \cos \left (2 \, f x + 2 \, e\right ) - {\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac {2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} + {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) + b d f x \log \left (\frac {{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) + 2 i \, b c f \arctan \left (-b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) + b \sin \left (2 \, f x + 2 \, e\right ) + a\right ) + b c f \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) + {\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} + 2 \, {\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) - i \, b d {\rm Li}_2\left (\frac {{\left (i \, a + b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{-i \, a + b}\right )}{2 \, {\left (a^{2} + b^{2}\right )} f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a - I*b)*d*f^2*x^2 + 2*(a - I*b)*c*f^2*x - 2*I*b*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)) +
b*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)) + 2*I*b*c*f*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) + b*c*f*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)) - I*b*d*dilog((I*a +
b)*e^(2*I*f*x + 2*I*e)/(-I*a + b)))/((a^2 + b^2)*f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*tan(e + f*x)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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