3.41 \(\int (c+d x) (a+b \tan (e+f x)) \, dx\)

Optimal. Leaf size=84 \[ \frac {a (c+d x)^2}{2 d}-\frac {b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^2}{2 d}+\frac {i b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3722, 3719, 2190, 2279, 2391} \[ \frac {a (c+d x)^2}{2 d}-\frac {b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b (c+d x)^2}{2 d}+\frac {i b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + b*Tan[e + f*x]),x]

[Out]

(a*(c + d*x)^2)/(2*d) + ((I/2)*b*(c + d*x)^2)/d - (b*(c + d*x)*Log[1 + E^((2*I)*(e + f*x))])/f + ((I/2)*b*d*Po
lyLog[2, -E^((2*I)*(e + f*x))])/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 3719

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

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int (c+d x) (a+b \tan (e+f x)) \, dx &=\int (a (c+d x)+b (c+d x) \tan (e+f x)) \, dx\\ &=\frac {a (c+d x)^2}{2 d}+b \int (c+d x) \tan (e+f x) \, dx\\ &=\frac {a (c+d x)^2}{2 d}+\frac {i b (c+d x)^2}{2 d}-(2 i b) \int \frac {e^{2 i (e+f x)} (c+d x)}{1+e^{2 i (e+f x)}} \, dx\\ &=\frac {a (c+d x)^2}{2 d}+\frac {i b (c+d x)^2}{2 d}-\frac {b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {(b d) \int \log \left (1+e^{2 i (e+f x)}\right ) \, dx}{f}\\ &=\frac {a (c+d x)^2}{2 d}+\frac {i b (c+d x)^2}{2 d}-\frac {b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}-\frac {(i b d) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (e+f x)}\right )}{2 f^2}\\ &=\frac {a (c+d x)^2}{2 d}+\frac {i b (c+d x)^2}{2 d}-\frac {b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 87, normalized size = 1.04 \[ a c x+\frac {1}{2} a d x^2-\frac {b c \log (\cos (e+f x))}{f}+\frac {i b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{2 f^2}-\frac {b d x \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {1}{2} i b d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.55, size = 160, normalized size = 1.90 \[ \frac {2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x - i \, b d {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) + i \, b d {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - 2 \, {\left (b d f x + b c f\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \, {\left (b d f x + b c f\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, f^{2}} \]

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

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

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 [A]  time = 0.39, size = 143, normalized size = 1.70 \[ \frac {i b d \,x^{2}}{2}-i b c x +\frac {a d \,x^{2}}{2}+c a x -\frac {b c \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {2 b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {2 i b d e x}{f}+\frac {i b d \,e^{2}}{f^{2}}-\frac {b d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) x}{f}+\frac {i b d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{2 f^{2}}-\frac {2 b d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.75, size = 130, normalized size = 1.55 \[ \frac {{\left (a + i \, b\right )} d f^{2} x^{2} + 2 \, {\left (a + i \, b\right )} c f^{2} x + i \, b d {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) - {\left (2 i \, b d f x + 2 i \, b c f\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - {\left (b d f x + b c f\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}{2 \, f^{2}} \]

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 + I*b*d*dilog(-e^(2*I*f*x + 2*I*e)) - (2*I*b*d*f*x + 2*I*b*c*f)
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - (b*d*f*x + b*c*f)*log(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)
^2 + 2*cos(2*f*x + 2*e) + 1))/f^2

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mupad [B]  time = 3.22, size = 161, normalized size = 1.92 \[ \frac {a\,x\,\left (2\,c+d\,x\right )}{2}-\frac {b\,d\,\left (\pi \,\ln \left (\cos \left (f\,x\right )\right )-\pi \,\ln \left ({\mathrm {e}}^{f\,x\,2{}\mathrm {i}}+1\right )+\mathrm {polylog}\left (2,-{\mathrm {e}}^{-e\,2{}\mathrm {i}}\,{\mathrm {e}}^{-f\,x\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}-\pi \,\ln \left ({\mathrm {e}}^{-e\,2{}\mathrm {i}}\,{\mathrm {e}}^{-f\,x\,2{}\mathrm {i}}+1\right )+2\,e\,\ln \left ({\mathrm {e}}^{-e\,2{}\mathrm {i}}\,{\mathrm {e}}^{-f\,x\,2{}\mathrm {i}}+1\right )-\ln \left (\cos \left (e+f\,x\right )\right )\,\left (2\,e-\pi \right )+f^2\,x^2\,1{}\mathrm {i}+2\,f\,x\,\ln \left ({\mathrm {e}}^{-e\,2{}\mathrm {i}}\,{\mathrm {e}}^{-f\,x\,2{}\mathrm {i}}+1\right )+e\,f\,x\,2{}\mathrm {i}\right )}{2\,f^2}+\frac {b\,c\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))*(c + d*x),x)

[Out]

(a*x*(2*c + d*x))/2 - (b*d*(polylog(2, -exp(-e*2i)*exp(-f*x*2i))*1i - pi*log(exp(f*x*2i) + 1) - log(cos(e + f*
x))*(2*e - pi) - pi*log(exp(-e*2i)*exp(-f*x*2i) + 1) + 2*e*log(exp(-e*2i)*exp(-f*x*2i) + 1) + pi*log(cos(f*x))
+ f^2*x^2*1i + 2*f*x*log(exp(-e*2i)*exp(-f*x*2i) + 1) + e*f*x*2i))/(2*f^2) + (b*c*log(tan(e + f*x)^2 + 1))/(2
*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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