\(\int \frac {c+d x}{a+b \sec (e+f x)} \, dx\) [36]

Optimal result

Integrand size = 18, antiderivative size = 257 \[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\frac {(c+d x)^2}{2 a d}+\frac {i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}-\frac {i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f}+\frac {b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2}-\frac {b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} f^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4276, 3402, 2296, 2221, 2317, 2438} \[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\frac {i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f \sqrt {b^2-a^2}}-\frac {i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{\sqrt {b^2-a^2}+b}\right )}{a f \sqrt {b^2-a^2}}+\frac {b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a f^2 \sqrt {b^2-a^2}}-\frac {b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a f^2 \sqrt {b^2-a^2}}+\frac {(c+d x)^2}{2 a d} \]

[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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 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 3402

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 4276

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]

Rubi steps \begin{align*} \text {integral}& = \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) \text {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) \text {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 \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,-\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] (verified)

Time = 0.72 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.83 \[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\frac {f \left (\sqrt {-a^2+b^2} f x (2 c+d x)+2 i b (c+d x) \log \left (1-\frac {a e^{i (e+f x)}}{-b+\sqrt {-a^2+b^2}}\right )-2 i b (c+d x) \log \left (1+\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )\right )+2 b d \operatorname {PolyLog}\left (2,\frac {a e^{i (e+f x)}}{-b+\sqrt {-a^2+b^2}}\right )-2 b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {-a^2+b^2}}\right )}{2 a \sqrt {-a^2+b^2} f^2} \]

[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)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (229 ) = 458\).

Time = 0.56 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.01

method	result	size
risch	\(\frac {d \,x^{2}}{2 a}+\frac {c x}{a}+\frac {2 i b c \arctan \left (\frac {2 a \,{\mathrm e}^{i \left (f x +e \right )}+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{f a \sqrt {a^{2}-b^{2}}}+\frac {i b d \ln \left (\frac {-a \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}-b}{-b +\sqrt {-a^{2}+b^{2}}}\right ) x}{f a \sqrt {-a^{2}+b^{2}}}-\frac {i b d \ln \left (\frac {a \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}+b}{b +\sqrt {-a^{2}+b^{2}}}\right ) x}{f a \sqrt {-a^{2}+b^{2}}}+\frac {i b d \ln \left (\frac {-a \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}-b}{-b +\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {i b d \ln \left (\frac {a \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}+b}{b +\sqrt {-a^{2}+b^{2}}}\right ) e}{f^{2} a \sqrt {-a^{2}+b^{2}}}+\frac {b d \operatorname {dilog}\left (\frac {-a \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}-b}{-b +\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {b d \operatorname {dilog}\left (\frac {a \,{\mathrm e}^{i \left (f x +e \right )}+\sqrt {-a^{2}+b^{2}}+b}{b +\sqrt {-a^{2}+b^{2}}}\right )}{f^{2} a \sqrt {-a^{2}+b^{2}}}-\frac {2 i b d e \arctan \left (\frac {2 a \,{\mathrm e}^{i \left (f x +e \right )}+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{f^{2} a \sqrt {a^{2}-b^{2}}}\)	\(516\)

[In]

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

[Out]

1/2*d/a*x^2+c/a*x+2*I/f/a*b*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/f/a*b*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/f/a*b*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/f^2/a*b*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/f^2/a*b*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+1/f^2/a*b*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)))-1/f^2/a*b*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/f^2/a*b*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))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1041 vs. \(2 (225) = 450\).

Time = 0.45 (sec) , antiderivative size = 1041, normalized size of antiderivative = 4.05 \[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\text {Too large to display} \]

[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(-(b*cos(f*x + e) + I*b
*sin(f*x + e) + (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + a*b*d*sqrt(-(a^2 - b^
2)/a^2)*dilog(-(b*cos(f*x + e) + I*b*sin(f*x + e) - (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2)
+ a)/a + 1) - a*b*d*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) - I*b*sin(f*x + e) + (a*cos(f*x + e) - I*a*
sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a + 1) + a*b*d*sqrt(-(a^2 - b^2)/a^2)*dilog(-(b*cos(f*x + e) - I*b*s
in(f*x + e) - (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^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*sq
rt(-(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*si
n(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((b*cos(f
*x + e) + I*b*sin(f*x + e) + (a*cos(f*x + e) + I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a) - (-I*a*b*d*f*
x - I*a*b*d*e)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(f*x + e) + I*b*sin(f*x + e) - (a*cos(f*x + e) + I*a*sin(f*x +
e))*sqrt(-(a^2 - b^2)/a^2) + a)/a) - (-I*a*b*d*f*x - I*a*b*d*e)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(f*x + e) -
I*b*sin(f*x + e) + (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-(a^2 - b^2)/a^2) + a)/a) - (I*a*b*d*f*x + I*a*b*d
*e)*sqrt(-(a^2 - b^2)/a^2)*log((b*cos(f*x + e) - I*b*sin(f*x + e) - (a*cos(f*x + e) - I*a*sin(f*x + e))*sqrt(-
(a^2 - b^2)/a^2) + a)/a))/((a^3 - a*b^2)*f^2)

Sympy [F]

\[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\int \frac {c + d x}{a + b \sec {\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more de

Giac [F]

\[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\int { \frac {d x + c}{b \sec \left (f x + e\right ) + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x}{a+b \sec (e+f x)} \, dx=\int \frac {c+d\,x}{a+\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]

[In]

int((c + d*x)/(a + b/cos(e + f*x)),x)

[Out]

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