3.1.31 \(\int (c+d x) \sec (a+b x) \, dx\) [31]

Optimal. Leaf size=75 \[ -\frac {2 i (c+d x) \text {ArcTan}\left (e^{i (a+b x)}\right )}{b}+\frac {i d \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2} \]

[Out]

-2*I*(d*x+c)*arctan(exp(I*(b*x+a)))/b+I*d*polylog(2,-I*exp(I*(b*x+a)))/b^2-I*d*polylog(2,I*exp(I*(b*x+a)))/b^2

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Rubi [A]
time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4266, 2317, 2438} \begin {gather*} -\frac {2 i (c+d x) \text {ArcTan}\left (e^{i (a+b x)}\right )}{b}+\frac {i d \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sec[a + b*x],x]

[Out]

((-2*I)*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b + (I*d*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 - (I*d*PolyLog[2, I*
E^(I*(a + b*x))])/b^2

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 4266

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

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 87, normalized size = 1.16 \begin {gather*} -\frac {2 i d x \text {ArcTan}\left (e^{i a+i b x}\right )}{b}+\frac {c \tanh ^{-1}(\sin (a+b x))}{b}+\frac {i d \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sec[a + b*x],x]

[Out]

((-2*I)*d*x*ArcTan[E^(I*a + I*b*x)])/b + (c*ArcTanh[Sin[a + b*x]])/b + (I*d*PolyLog[2, (-I)*E^(I*(a + b*x))])/
b^2 - (I*d*PolyLog[2, I*E^(I*(a + b*x))])/b^2

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Maple [A]
time = 0.04, size = 128, normalized size = 1.71

method	result	size
derivativedivides	\(\frac {-\frac {d a \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}+c \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )+\frac {d \left (-\left (b x +a \right ) \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )+\left (b x +a \right ) \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )+i \dilog \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )-i \dilog \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )\right )}{b}}{b}\)	\(128\)
default	\(\frac {-\frac {d a \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}+c \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )+\frac {d \left (-\left (b x +a \right ) \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )+\left (b x +a \right ) \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )+i \dilog \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )-i \dilog \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )\right )}{b}}{b}\)	\(128\)
risch	\(-\frac {2 i c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {i d \dilog \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {i d \dilog \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 i d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*sec(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/b*(-1/b*d*a*ln(sec(b*x+a)+tan(b*x+a))+c*ln(sec(b*x+a)+tan(b*x+a))+1/b*d*(-(b*x+a)*ln(1+I*exp(I*(b*x+a)))+(b*
x+a)*ln(1-I*exp(I*(b*x+a)))+I*dilog(1+I*exp(I*(b*x+a)))-I*dilog(1-I*exp(I*(b*x+a)))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a),x, algorithm="maxima")

[Out]

1/2*(4*b*d*integrate((x*cos(2*b*x + 2*a)*cos(b*x + a) + x*sin(2*b*x + 2*a)*sin(b*x + a) + x*cos(b*x + a))/(cos
(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1), x) + c*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2
*sin(b*x + a) + 1) - c*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*sin(b*x + a) + 1))/b

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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. 306 vs. \(2 (60) = 120\).
time = 0.41, size = 306, normalized size = 4.08 \begin {gather*} \frac {-i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - {\left (b c - a d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) + {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) - {\left (b c - a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a),x, algorithm="fricas")

[Out]

1/2*(-I*d*dilog(I*cos(b*x + a) + sin(b*x + a)) - I*d*dilog(I*cos(b*x + a) - sin(b*x + a)) + I*d*dilog(-I*cos(b
*x + a) + sin(b*x + a)) + I*d*dilog(-I*cos(b*x + a) - sin(b*x + a)) + (b*c - a*d)*log(cos(b*x + a) + I*sin(b*x
+ a) + I) - (b*c - a*d)*log(cos(b*x + a) - I*sin(b*x + a) + I) + (b*d*x + a*d)*log(I*cos(b*x + a) + sin(b*x +
a) + 1) - (b*d*x + a*d)*log(I*cos(b*x + a) - sin(b*x + a) + 1) + (b*d*x + a*d)*log(-I*cos(b*x + a) + sin(b*x
+ a) + 1) - (b*d*x + a*d)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) + (b*c - a*d)*log(-cos(b*x + a) + I*sin(b*x
+ a) + I) - (b*c - a*d)*log(-cos(b*x + a) - I*sin(b*x + a) + I))/b^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (c + d x\right ) \sec {\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a),x)

[Out]

Integral((c + d*x)*sec(a + b*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)*sec(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)*sec(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c+d\,x}{\cos \left (a+b\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/cos(a + b*x),x)

[Out]

int((c + d*x)/cos(a + b*x), x)

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