3.300 \(\int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx\)

Optimal. Leaf size=117 \[ -\frac {i d \text {Li}_2\left (-i e^{i (a+b x)}\right )}{2 b^2}+\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]

[Out]

I*(d*x+c)*arctan(exp(I*(b*x+a)))/b-1/2*I*d*polylog(2,-I*exp(I*(b*x+a)))/b^2+1/2*I*d*polylog(2,I*exp(I*(b*x+a))
)/b^2-1/2*d*sec(b*x+a)/b^2+1/2*(d*x+c)*sec(b*x+a)*tan(b*x+a)/b

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Rubi [A]  time = 0.13, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4413, 4181, 2279, 2391, 4185} \[ -\frac {i d \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac {i d \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4181

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]

Rule 4185

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> -Simp[(b^2*(c + d*x)*Cot[e + f*x]*
(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2
), x], x] - Simp[(b^2*d*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[{b, c, d, e, f}, x] && G
tQ[n, 1] && NeQ[n, 2]

Rule 4413

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]*Tan[(a_.) + (b_.)*(x_)]^(p_), x_Symbol] :> -Int[(c + d*
x)^m*Sec[a + b*x]*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sec[a + b*x]^3*Tan[a + b*x]^(p - 2), x] /; FreeQ[
{a, b, c, d, m}, x] && IGtQ[p/2, 0]

Rubi steps

\begin {align*} \int (c+d x) \sec (a+b x) \tan ^2(a+b x) \, dx &=-\int (c+d x) \sec (a+b x) \, dx+\int (c+d x) \sec ^3(a+b x) \, dx\\ &=\frac {2 i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}+\frac {1}{2} \int (c+d x) \sec (a+b x) \, dx+\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 {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}-\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}+\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}-\frac {d \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{2 b}+\frac {d \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{2 b}\\ &=\frac {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}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}+\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}-\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{2 b^2}\\ &=\frac {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 )}{2 b^2}+\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}+\frac {(c+d x) \sec (a+b x) \tan (a+b x)}{2 b}\\ \end {align*}

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Mathematica [B]  time = 6.50, size = 555, normalized size = 4.74 \[ -\frac {d \sin \left (\frac {1}{2} (a+b x)\right )}{2 b^2 \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}+\frac {d \sin \left (\frac {1}{2} (a+b x)\right )}{2 b^2 \left (\sin \left (\frac {1}{2} (a+b x)\right )+\cos \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {c \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac {c \tan (a+b x) \sec (a+b x)}{2 b}+\frac {d x \left (-i \text {Li}_2\left (\frac {1}{2} \left ((1+i)-(1-i) \tan \left (\frac {1}{2} (a+b x)\right )\right )\right )+i \text {Li}_2\left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (\tan \left (\frac {1}{2} (a+b x)\right )+i\right )\right )-i \text {Li}_2\left (\frac {1}{2} \left ((1-i) \tan \left (\frac {1}{2} (a+b x)\right )+(1+i)\right )\right )+i \text {Li}_2\left (\frac {1}{2} \left ((1+i) \tan \left (\frac {1}{2} (a+b x)\right )+(1-i)\right )\right )+a \log \left (1-\tan \left (\frac {1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (\tan \left (\frac {1}{2} (a+b x)\right )-1\right )\right )-i \log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (-\frac {1}{2}+\frac {i}{2}\right ) \left (\tan \left (\frac {1}{2} (a+b x)\right )-1\right )\right )-i \log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (\frac {1}{2}-\frac {i}{2}\right ) \left (\tan \left (\frac {1}{2} (a+b x)\right )+1\right )\right )+i \log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right ) \log \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \left (\tan \left (\frac {1}{2} (a+b x)\right )+1\right )\right )-a \log \left (\tan \left (\frac {1}{2} (a+b x)\right )+1\right )\right )}{2 b \left (-i \log \left (1-i \tan \left (\frac {1}{2} (a+b x)\right )\right )+i \log \left (1+i \tan \left (\frac {1}{2} (a+b x)\right )\right )+a\right )}+\frac {d x}{4 b \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )^2}-\frac {d x}{4 b \left (\sin \left (\frac {1}{2} (a+b x)\right )+\cos \left (\frac {1}{2} (a+b x)\right )\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(c*ArcTanh[Sin[a + b*x]])/b + (d*x*(a*Log[1 - Tan[(a + b*x)/2]] + I*Log[1 + I*Tan[(a + b*x)/2]]*Log[(-1/2
- I/2)*(-1 + Tan[(a + b*x)/2])] - I*Log[1 - I*Tan[(a + b*x)/2]]*Log[(-1/2 + I/2)*(-1 + Tan[(a + b*x)/2])] - I
*Log[1 + I*Tan[(a + b*x)/2]]*Log[(1/2 - I/2)*(1 + Tan[(a + b*x)/2])] + I*Log[1 - I*Tan[(a + b*x)/2]]*Log[(1/2
+ I/2)*(1 + Tan[(a + b*x)/2])] - a*Log[1 + Tan[(a + b*x)/2]] - I*PolyLog[2, ((1 + I) - (1 - I)*Tan[(a + b*x)/2
])/2] + I*PolyLog[2, (-1/2 - I/2)*(I + Tan[(a + b*x)/2])] - I*PolyLog[2, ((1 + I) + (1 - I)*Tan[(a + b*x)/2])/
2] + I*PolyLog[2, ((1 - I) + (1 + I)*Tan[(a + b*x)/2])/2]))/(2*b*(a - I*Log[1 - I*Tan[(a + b*x)/2]] + I*Log[1
+ I*Tan[(a + b*x)/2]])) + (d*x)/(4*b*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])^2) - (d*Sin[(a + b*x)/2])/(2*b^2*(C
os[(a + b*x)/2] - Sin[(a + b*x)/2])) - (d*x)/(4*b*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])^2) + (d*Sin[(a + b*x)/
2])/(2*b^2*(Cos[(a + b*x)/2] + Sin[(a + b*x)/2])) + (c*Sec[a + b*x]*Tan[a + b*x])/(2*b)

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fricas [B]  time = 0.53, size = 435, normalized size = 3.72 \[ \frac {i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d \cos \left (b x + a\right )^{2} {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \cos \left (b x + a\right )^{2} \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c - a d\right )} \cos \left (b x + a\right )^{2} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 2 \, d \cos \left (b x + a\right ) + 2 \, {\left (b d x + b c\right )} \sin \left (b x + a\right )}{4 \, b^{2} \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \sec \left (b x + a\right ) \tan \left (b x + a\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*sec(b*x + a)*tan(b*x + a)^2, x)

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maple [B]  time = 0.12, size = 267, normalized size = 2.28 \[ -\frac {i \left (b d x \,{\mathrm e}^{3 i \left (b x +a \right )}-i d \,{\mathrm e}^{3 i \left (b x +a \right )}+c b \,{\mathrm e}^{3 i \left (b x +a \right )}-b d x \,{\mathrm e}^{i \left (b x +a \right )}-i d \,{\mathrm e}^{i \left (b x +a \right )}-c b \,{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2} \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )^{2}}+\frac {i c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}+\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{2 b}+\frac {d \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{2 b^{2}}-\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{2 b}-\frac {d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{2 b^{2}}-\frac {i d \dilog \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {i d \dilog \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{2 b^{2}}-\frac {i d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*sec(b*x+a)*tan(b*x+a)^2,x)

[Out]

-I/b^2/(1+exp(2*I*(b*x+a)))^2*(b*d*x*exp(3*I*(b*x+a))-I*d*exp(3*I*(b*x+a))+c*b*exp(3*I*(b*x+a))-b*d*x*exp(I*(b
*x+a))-I*d*exp(I*(b*x+a))-c*b*exp(I*(b*x+a)))+I/b*c*arctan(exp(I*(b*x+a)))+1/2/b*d*ln(1+I*exp(I*(b*x+a)))*x+1/
2/b^2*d*ln(1+I*exp(I*(b*x+a)))*a-1/2/b*d*ln(1-I*exp(I*(b*x+a)))*x-1/2/b^2*d*ln(1-I*exp(I*(b*x+a)))*a-1/2*I/b^2
*d*dilog(1+I*exp(I*(b*x+a)))+1/2*I/b^2*d*dilog(1-I*exp(I*(b*x+a)))-I/b^2*d*a*arctan(exp(I*(b*x+a)))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \tan ^{2}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a)*tan(b*x+a)**2,x)

[Out]

Integral((c + d*x)*tan(a + b*x)**2*sec(a + b*x), x)

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