3.39 \(\int (c+d x) \sec ^3(a+b x) \, dx\)

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 = 0.07, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4185, 4181, 2279, 2391} \[ \frac {i d \text {Li}_2\left (-i e^{i (a+b x)}\right )}{2 b^2}-\frac {i d \text {Li}_2\left (i e^{i (a+b x)}\right )}{2 b^2}-\frac {d \sec (a+b x)}{2 b^2}-\frac {i (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \tan (a+b x) \sec (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sec[a + b*x]^3,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]

Rubi steps

\begin {align*} \int (c+d x) \sec ^3(a+b x) \, dx &=-\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 {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 {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 {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 = 3.68, size = 389, normalized size = 3.32 \[ \frac {d \left (i \left (\text {Li}_2\left (-e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )-\text {Li}_2\left (e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )\right )+\left (-a-b x+\frac {\pi }{2}\right ) \left (\log \left (1-e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )-\log \left (1+e^{i \left (-a-b x+\frac {\pi }{2}\right )}\right )\right )-\left (\frac {\pi }{2}-a\right ) \log \left (\tan \left (\frac {1}{2} \left (-a-b x+\frac {\pi }{2}\right )\right )\right )\right )}{2 b^2}-\frac {d \sin \left (\frac {b x}{2}\right )}{2 b^2 \left (\cos \left (\frac {a}{2}\right )-\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )-\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}+\frac {d \sin \left (\frac {b x}{2}\right )}{2 b^2 \left (\sin \left (\frac {a}{2}\right )+\cos \left (\frac {a}{2}\right )\right ) \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right )+\cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}+\frac {c \tanh ^{-1}(\sin (a+b x))}{2 b}+\frac {c \tan (a+b x) \sec (a+b x)}{2 b}+\frac {d x}{4 b \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )-\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )^2}-\frac {d x}{4 b \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right )+\cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sec[a + b*x]^3,x]

[Out]

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

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fricas [B]  time = 0.96, 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)^3,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 \[ \int {\left (d x + c\right )} \sec \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a)^3,x, algorithm="giac")

[Out]

integrate((d*x + c)*sec(b*x + a)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*sec(b*x+a)^3,x)

[Out]

-I/b^2/(exp(2*I*(b*x+a))+1)^2*(d*x*b*exp(3*I*(b*x+a))+c*b*exp(3*I*(b*x+a))-d*x*b*exp(I*(b*x+a))-c*b*exp(I*(b*x
+a))-I*d*exp(3*I*(b*x+a))-I*d*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)^3,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((c + d*x)/cos(a + b*x)^3,x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \sec ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*sec(b*x+a)**3,x)

[Out]

Integral((c + d*x)*sec(a + b*x)**3, x)

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