3.1.56 \(\int \frac {c+d x}{a+b \tan (e+f x)} \, dx\) [56]

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

[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 = 0.11, 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 = {3813, 2221, 2317, 2438} \begin {gather*} \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)} \end {gather*}

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 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 2317

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 2438

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 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]

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) \text {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 = 1.02, size = 113, normalized size = 0.90 \begin {gather*} \frac {f \left ((a-i b) f x (2 c+d x)+2 b (c+d x) \log \left (1+\frac {(a-i b) e^{2 i (e+f x)}}{a+i b}\right )\right )-i b d \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i (e+f x)}}{a+i b}\right )}{2 \left (a^2+b^2\right ) f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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. 461 vs. \(2 (113 ) = 226\).
time = 0.39, size = 462, normalized size = 3.70

method	result	size
risch	\(-\frac {d \,x^{2}}{2 \left (i b -a \right )}-\frac {c x}{i b -a}+\frac {b c \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f \left (-i b +a \right ) \left (i b +a \right )}-\frac {2 b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (-i b +a \right ) \left (i b +a \right )}-\frac {b d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) x}{f \left (-i b +a \right ) \left (-i b -a \right )}-\frac {b d \ln \left (1-\frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right ) e}{f^{2} \left (-i b +a \right ) \left (-i b -a \right )}+\frac {i b d \,x^{2}}{\left (-i b +a \right ) \left (-i b -a \right )}+\frac {2 i b d e x}{f \left (-i b +a \right ) \left (-i b -a \right )}+\frac {i b d \,e^{2}}{f^{2} \left (-i b +a \right ) \left (-i b -a \right )}+\frac {i b d \polylog \left (2, \frac {\left (-i b +a \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{-i b -a}\right )}{2 f^{2} \left (-i b +a \right ) \left (-i b -a \right )}-\frac {b d e \ln \left (i {\mathrm e}^{2 i \left (f x +e \right )} b -a \,{\mathrm e}^{2 i \left (f x +e \right )}-i b -a \right )}{f^{2} \left (-i b +a \right ) \left (i b +a \right )}+\frac {2 b d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2} \left (-i b +a \right ) \left (i b +a \right )}\)	\(462\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 422 vs. \(2 (110) = 220\).
time = 0.62, size = 422, normalized size = 3.38 \begin {gather*} \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 e^{\left (2 i \, e\right )} + b e^{\left (2 i \, e\right )}\right )} e^{\left (2 i \, f x\right )}}{-i \, a + b}\right )}{2 \, {\left (a^{2} + b^{2}\right )} f^{2}} \end {gather*}

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*e
^(2*I*e) + b*e^(2*I*e))*e^(2*I*f*x)/(-I*a + b)))/((a^2 + b^2)*f^2)

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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. 549 vs. \(2 (110) = 220\).
time = 0.40, size = 549, normalized size = 4.39 \begin {gather*} \frac {2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x + i \, b d {\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 ) - i \, b d {\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 ) + 2 \, {\left (b d f x + b d e\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 ) + 2 \, {\left (b d f x + b d e\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 ) + 2 \, {\left (b c f - b d e\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 c f - b d e\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}} \end {gather*}

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 - 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) - I*b*d*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) + 2
*(b*d*f*x + b*d*e)*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)) + 2*(b*d*f*x + b*d*e)*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)) + 2*(b*c*f - b*d*e)*l
og(((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*c*
f - b*d*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^2)*f^2)

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

[Out]

Integral((c + d*x)/(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)/(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c+d\,x}{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)/(a + b*tan(e + f*x)),x)

[Out]

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

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