3.175 \(\int \frac {c+d x}{(a+b \cosh (e+f x))^2} \, dx\)

Optimal. Leaf size=274 \[ \frac {a (c+d x) \log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac {a (c+d x) \log \left (\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac {b (c+d x) \sinh (e+f x)}{f \left (a^2-b^2\right ) (a+b \cosh (e+f x))}+\frac {a d \text {Li}_2\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac {a d \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {d \log (a+b \cosh (e+f x))}{f^2 \left (a^2-b^2\right )} \]

[Out]

d*ln(a+b*cosh(f*x+e))/(a^2-b^2)/f^2+a*(d*x+c)*ln(1+b*exp(f*x+e)/(a-(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f-a*(d*x+
c)*ln(1+b*exp(f*x+e)/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f+a*d*polylog(2,-b*exp(f*x+e)/(a-(a^2-b^2)^(1/2)))/(
a^2-b^2)^(3/2)/f^2-a*d*polylog(2,-b*exp(f*x+e)/(a+(a^2-b^2)^(1/2)))/(a^2-b^2)^(3/2)/f^2-b*(d*x+c)*sinh(f*x+e)/
(a^2-b^2)/f/(a+b*cosh(f*x+e))

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Rubi [A]  time = 0.46, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3324, 3320, 2264, 2190, 2279, 2391, 2668, 31} \[ \frac {a d \text {PolyLog}\left (2,-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}-\frac {a d \text {PolyLog}\left (2,-\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \left (a^2-b^2\right )^{3/2}}+\frac {a (c+d x) \log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac {a (c+d x) \log \left (\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}+1\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac {b (c+d x) \sinh (e+f x)}{f \left (a^2-b^2\right ) (a+b \cosh (e+f x))}+\frac {d \log (a+b \cosh (e+f x))}{f^2 \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*Cosh[e + f*x])^2,x]

[Out]

(a*(c + d*x)*Log[1 + (b*E^(e + f*x))/(a - Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) - (a*(c + d*x)*Log[1 + (b*E
^(e + f*x))/(a + Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) + (d*Log[a + b*Cosh[e + f*x]])/((a^2 - b^2)*f^2) + (
a*d*PolyLog[2, -((b*E^(e + f*x))/(a - Sqrt[a^2 - b^2]))])/((a^2 - b^2)^(3/2)*f^2) - (a*d*PolyLog[2, -((b*E^(e
+ f*x))/(a + Sqrt[a^2 - b^2]))])/((a^2 - b^2)^(3/2)*f^2) - (b*(c + d*x)*Sinh[e + f*x])/((a^2 - b^2)*f*(a + b*C
osh[e + f*x]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 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 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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3320

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

Rule 3324

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

Rubi steps

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

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Mathematica [A]  time = 5.04, size = 509, normalized size = 1.86 \[ \frac {\frac {\left (a^2-b^2\right ) \left (2 a c f \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {a+b e^{e+f x}}{\sqrt {a^2-b^2}}\right )-a d \sqrt {b^2-a^2} \text {Li}_2\left (\frac {b e^{e+f x}}{\sqrt {a^2-b^2}-a}\right )+a d \sqrt {b^2-a^2} \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )+d \sqrt {-\left (a^2-b^2\right )^2} (e+f x)-a d \sqrt {b^2-a^2} (e+f x) \log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}+1\right )+a d \sqrt {b^2-a^2} (e+f x) \log \left (\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}+1\right )-d \sqrt {-\left (a^2-b^2\right )^2} \log \left (2 a e^{e+f x}+b e^{2 (e+f x)}+b\right )-2 a d \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a+b e^{e+f x}}{\sqrt {b^2-a^2}}\right )-2 a d \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {a+b e^{e+f x}}{\sqrt {a^2-b^2}}\right )-2 a d e \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {a+b e^{e+f x}}{\sqrt {a^2-b^2}}\right )\right )}{\left (-\left (a^2-b^2\right )^2\right )^{3/2}}-\frac {b f (c+d x) \sinh (e+f x)}{(a-b) (a+b) (a+b \cosh (e+f x))}}{f^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)/(a + b*Cosh[e + f*x])^2,x]

[Out]

(((a^2 - b^2)*(Sqrt[-(a^2 - b^2)^2]*d*(e + f*x) - 2*a*Sqrt[a^2 - b^2]*d*ArcTan[(a + b*E^(e + f*x))/Sqrt[-a^2 +
b^2]] - 2*a*Sqrt[-a^2 + b^2]*d*ArcTanh[(a + b*E^(e + f*x))/Sqrt[a^2 - b^2]] - 2*a*Sqrt[-a^2 + b^2]*d*e*ArcTan
h[(a + b*E^(e + f*x))/Sqrt[a^2 - b^2]] + 2*a*Sqrt[-a^2 + b^2]*c*f*ArcTanh[(a + b*E^(e + f*x))/Sqrt[a^2 - b^2]]
- a*Sqrt[-a^2 + b^2]*d*(e + f*x)*Log[1 + (b*E^(e + f*x))/(a - Sqrt[a^2 - b^2])] + a*Sqrt[-a^2 + b^2]*d*(e + f
*x)*Log[1 + (b*E^(e + f*x))/(a + Sqrt[a^2 - b^2])] - Sqrt[-(a^2 - b^2)^2]*d*Log[b + 2*a*E^(e + f*x) + b*E^(2*(
e + f*x))] - a*Sqrt[-a^2 + b^2]*d*PolyLog[2, (b*E^(e + f*x))/(-a + Sqrt[a^2 - b^2])] + a*Sqrt[-a^2 + b^2]*d*Po
lyLog[2, -((b*E^(e + f*x))/(a + Sqrt[a^2 - b^2]))]))/(-(a^2 - b^2)^2)^(3/2) - (b*f*(c + d*x)*Sinh[e + f*x])/((
a - b)*(a + b)*(a + b*Cosh[e + f*x])))/f^2

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fricas [B]  time = 0.86, size = 1765, normalized size = 6.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cosh(f*x+e))^2,x, algorithm="fricas")

[Out]

-(2*(a^2*b - b^3)*d*e - 2*(a^2*b - b^3)*c*f + 2*((a^2*b - b^3)*d*f*x + (a^2*b - b^3)*d*e)*cosh(f*x + e)^2 + 2*
((a^2*b - b^3)*d*f*x + (a^2*b - b^3)*d*e)*sinh(f*x + e)^2 - (a*b^2*d*cosh(f*x + e)^2 + a*b^2*d*sinh(f*x + e)^2
+ 2*a^2*b*d*cosh(f*x + e) + a*b^2*d + 2*(a*b^2*d*cosh(f*x + e) + a^2*b*d)*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2
)*dilog(-(a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b
+ 1) + (a*b^2*d*cosh(f*x + e)^2 + a*b^2*d*sinh(f*x + e)^2 + 2*a^2*b*d*cosh(f*x + e) + a*b^2*d + 2*(a*b^2*d*co
sh(f*x + e) + a^2*b*d)*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cos
h(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b + 1) - (a*b^2*d*f*x + a*b^2*d*e + (a*b^2*d*f*x + a*
b^2*d*e)*cosh(f*x + e)^2 + (a*b^2*d*f*x + a*b^2*d*e)*sinh(f*x + e)^2 + 2*(a^2*b*d*f*x + a^2*b*d*e)*cosh(f*x +
e) + 2*(a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*cosh(f*x + e))*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2
)*log((a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b) +
(a*b^2*d*f*x + a*b^2*d*e + (a*b^2*d*f*x + a*b^2*d*e)*cosh(f*x + e)^2 + (a*b^2*d*f*x + a*b^2*d*e)*sinh(f*x + e
)^2 + 2*(a^2*b*d*f*x + a^2*b*d*e)*cosh(f*x + e) + 2*(a^2*b*d*f*x + a^2*b*d*e + (a*b^2*d*f*x + a*b^2*d*e)*cosh(
f*x + e))*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2)*log((a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f*x + e) + b*s
inh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b) + 2*((a^3 - a*b^2)*d*f*x + 2*(a^3 - a*b^2)*d*e - (a^3 - a*b^2)*c*f
)*cosh(f*x + e) - ((a^2*b - b^3)*d*cosh(f*x + e)^2 + (a^2*b - b^3)*d*sinh(f*x + e)^2 + 2*(a^3 - a*b^2)*d*cosh(
f*x + e) + (a^2*b - b^3)*d + 2*((a^2*b - b^3)*d*cosh(f*x + e) + (a^3 - a*b^2)*d)*sinh(f*x + e) + (a*b^2*d*e -
a*b^2*c*f + (a*b^2*d*e - a*b^2*c*f)*cosh(f*x + e)^2 + (a*b^2*d*e - a*b^2*c*f)*sinh(f*x + e)^2 + 2*(a^2*b*d*e -
a^2*b*c*f)*cosh(f*x + e) + 2*(a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e - a*b^2*c*f)*cosh(f*x + e))*sinh(f*x + e))*s
qrt((a^2 - b^2)/b^2))*log(2*b*cosh(f*x + e) + 2*b*sinh(f*x + e) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) - ((a^2*b -
b^3)*d*cosh(f*x + e)^2 + (a^2*b - b^3)*d*sinh(f*x + e)^2 + 2*(a^3 - a*b^2)*d*cosh(f*x + e) + (a^2*b - b^3)*d
+ 2*((a^2*b - b^3)*d*cosh(f*x + e) + (a^3 - a*b^2)*d)*sinh(f*x + e) - (a*b^2*d*e - a*b^2*c*f + (a*b^2*d*e - a*
b^2*c*f)*cosh(f*x + e)^2 + (a*b^2*d*e - a*b^2*c*f)*sinh(f*x + e)^2 + 2*(a^2*b*d*e - a^2*b*c*f)*cosh(f*x + e) +
2*(a^2*b*d*e - a^2*b*c*f + (a*b^2*d*e - a*b^2*c*f)*cosh(f*x + e))*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2))*log(2
*b*cosh(f*x + e) + 2*b*sinh(f*x + e) - 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) + 2*((a^3 - a*b^2)*d*f*x + 2*(a^3 - a*
b^2)*d*e - (a^3 - a*b^2)*c*f + 2*((a^2*b - b^3)*d*f*x + (a^2*b - b^3)*d*e)*cosh(f*x + e))*sinh(f*x + e))/((a^4
*b - 2*a^2*b^3 + b^5)*f^2*cosh(f*x + e)^2 + (a^4*b - 2*a^2*b^3 + b^5)*f^2*sinh(f*x + e)^2 + 2*(a^5 - 2*a^3*b^2
+ a*b^4)*f^2*cosh(f*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*f^2 + 2*((a^4*b - 2*a^2*b^3 + b^5)*f^2*cosh(f*x + e) +
(a^5 - 2*a^3*b^2 + a*b^4)*f^2)*sinh(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cosh(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*x + c)/(b*cosh(f*x + e) + a)^2, x)

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maple [B]  time = 0.30, size = 585, normalized size = 2.14 \[ \frac {2 \left (d x +c \right ) \left (a \,{\mathrm e}^{f x +e}+b \right )}{f \left (a^{2}-b^{2}\right ) \left (b \,{\mathrm e}^{2 f x +2 e}+2 a \,{\mathrm e}^{f x +e}+b \right )}+\frac {2 a c \arctan \left (\frac {2 b \,{\mathrm e}^{f x +e}+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {d a \ln \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{f \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}+\frac {d a \ln \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) e}{f^{2} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}-\frac {d a \ln \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{f \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}-\frac {d a \ln \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) e}{f^{2} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}+\frac {d a \dilog \left (\frac {-b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{f^{2} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}-\frac {d a \dilog \left (\frac {b \,{\mathrm e}^{f x +e}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{f^{2} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}+\frac {d \ln \left (b \,{\mathrm e}^{2 f x +2 e}+2 a \,{\mathrm e}^{f x +e}+b \right )}{f^{2} \left (a^{2}-b^{2}\right )}-\frac {2 d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2} \left (a^{2}-b^{2}\right )}-\frac {2 a d e \arctan \left (\frac {2 b \,{\mathrm e}^{f x +e}+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{f^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(a+b*cosh(f*x+e))^2,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cosh(f*x+e))^2,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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {c+d\,x}{{\left (a+b\,\mathrm {cosh}\left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*cosh(e + f*x))^2,x)

[Out]

int((c + d*x)/(a + b*cosh(e + f*x))^2, x)

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sympy [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)/(a+b*cosh(f*x+e))**2,x)

[Out]

Timed out

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