3.36 \(\int \frac{c+d x}{a+b \sec (e+f x)} \, dx\)

Optimal. Leaf size=257 \[ \frac{b d \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f^2 \sqrt{b^2-a^2}}-\frac{b d \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f^2 \sqrt{b^2-a^2}}+\frac{i b (c+d x) \log \left (1+\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f \sqrt{b^2-a^2}}-\frac{i b (c+d x) \log \left (1+\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f \sqrt{b^2-a^2}}+\frac{(c+d x)^2}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.490205, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4191, 3321, 2264, 2190, 2279, 2391} \[ \frac{b d \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f^2 \sqrt{b^2-a^2}}-\frac{b d \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f^2 \sqrt{b^2-a^2}}+\frac{i b (c+d x) \log \left (1+\frac{a e^{i (e+f x)}}{b-\sqrt{b^2-a^2}}\right )}{a f \sqrt{b^2-a^2}}-\frac{i b (c+d x) \log \left (1+\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )}{a f \sqrt{b^2-a^2}}+\frac{(c+d x)^2}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*Sec[e + f*x]),x]

[Out]

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

Rule 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 3321

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

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

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]

Rubi steps

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

Mathematica [A]  time = 0.456709, size = 214, normalized size = 0.83 \[ \frac{2 b d \text{PolyLog}\left (2,\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}-b}\right )-2 b d \text{PolyLog}\left (2,-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )+f \left (2 i b (c+d x) \log \left (1-\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}-b}\right )-2 i b (c+d x) \log \left (1+\frac{a e^{i (e+f x)}}{\sqrt{b^2-a^2}+b}\right )+f x \sqrt{b^2-a^2} (2 c+d x)\right )}{2 a f^2 \sqrt{b^2-a^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.132, size = 516, normalized size = 2. \begin{align*}{\frac{d{x}^{2}}{2\,a}}+{\frac{cx}{a}}+{\frac{2\,ibc}{af}\arctan \left ({\frac{2\,a{{\rm e}^{i \left ( fx+e \right ) }}+2\,b}{2}{\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}}+{\frac{ibdx}{af}\ln \left ({ \left ( -a{{\rm e}^{i \left ( fx+e \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}-b \right ) \left ( -b+\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}+{\frac{ibde}{a{f}^{2}}\ln \left ({ \left ( -a{{\rm e}^{i \left ( fx+e \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}-b \right ) \left ( -b+\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}-{\frac{ibdx}{af}\ln \left ({ \left ( a{{\rm e}^{i \left ( fx+e \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}+b \right ) \left ( b+\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}-{\frac{ibde}{a{f}^{2}}\ln \left ({ \left ( a{{\rm e}^{i \left ( fx+e \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}+b \right ) \left ( b+\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}+{\frac{bd}{a{f}^{2}}{\it dilog} \left ({ \left ( -a{{\rm e}^{i \left ( fx+e \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}-b \right ) \left ( -b+\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}-{\frac{bd}{a{f}^{2}}{\it dilog} \left ({ \left ( a{{\rm e}^{i \left ( fx+e \right ) }}+\sqrt{-{a}^{2}+{b}^{2}}+b \right ) \left ( b+\sqrt{-{a}^{2}+{b}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{a}^{2}+{b}^{2}}}}}-{\frac{2\,ibde}{a{f}^{2}}\arctan \left ({\frac{2\,a{{\rm e}^{i \left ( fx+e \right ) }}+2\,b}{2}{\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.55067, size = 2569, normalized size = 10. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2 - b^2)*d*f^2*x^2 + 2*(a^2 - b^2)*c*f^2*x - a*b*d*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(f*x + e)
+ 2*I*b*sin(f*x + e) + 2*(a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) + a*b*d*sqr
t(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) + I*a*sin(f*x + e))*
sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) - a*b*d*sqrt(-(a^2 - b^2)/a^2)*dilog(-1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f
*x + e) + 2*(a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a + 1) + a*b*d*sqrt(-(a^2 - b^2)
/a^2)*dilog(-1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b
^2)/a^2) + 2*a)/a + 1) - (I*a*b*d*e - I*a*b*c*f)*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(f*x + e) + 2*I*a*sin(f*x +
e) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - (-I*a*b*d*e + I*a*b*c*f)*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(f*x + e)
- 2*I*a*sin(f*x + e) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - (I*a*b*d*e - I*a*b*c*f)*sqrt(-(a^2 - b^2)/a^2)*log
(-2*a*cos(f*x + e) + 2*I*a*sin(f*x + e) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - (-I*a*b*d*e + I*a*b*c*f)*sqrt(-(
a^2 - b^2)/a^2)*log(-2*a*cos(f*x + e) - 2*I*a*sin(f*x + e) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) - (I*a*b*d*f*x
+ I*a*b*d*e)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(f*x + e) + 2*I*b*sin(f*x + e) + 2*(a*cos(f*x + e) + I*a*s
in(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - (-I*a*b*d*f*x - I*a*b*d*e)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b
*cos(f*x + e) + 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) -
(-I*a*b*d*f*x - I*a*b*d*e)*sqrt(-(a^2 - b^2)/a^2)*log(1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) + 2*(a*cos(f*
x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + 2*a)/a) - (I*a*b*d*f*x + I*a*b*d*e)*sqrt(-(a^2 - b^2)/a^2)
*log(1/2*(2*b*cos(f*x + e) - 2*I*b*sin(f*x + e) - 2*(a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2)
+ 2*a)/a))/((a^3 - a*b^2)*f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{b \sec \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*sec(f*x+e)),x, algorithm="giac")

[Out]

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