\(\int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx\) [321]

Optimal result

Integrand size = 24, antiderivative size = 74 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=-\frac {2 f \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {e+f x}{b d (a+b \sinh (c+d x))} \]

[Out]

(-f*x-e)/b/d/(a+b*sinh(d*x+c))-2*f*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/b/d^2/(a^2+b^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5572, 2739, 632, 210} \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=-\frac {2 f \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {e+f x}{b d (a+b \sinh (c+d x))} \]

[In]

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

[Out]

(-2*f*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b*Sqrt[a^2 + b^2]*d^2) - (e + f*x)/(b*d*(a + b*Sinh
[c + d*x]))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[2*(e/d), Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 5572

Int[Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)])^(n_.), x_Symbo
l] :> Simp[(e + f*x)^m*((a + b*Sinh[c + d*x])^(n + 1)/(b*d*(n + 1))), x] - Dist[f*(m/(b*d*(n + 1))), Int[(e +
f*x)^(m - 1)*(a + b*Sinh[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n,
-1]

Rubi steps \begin{align*} \text {integral}& = -\frac {e+f x}{b d (a+b \sinh (c+d x))}+\frac {f \int \frac {1}{a+b \sinh (c+d x)} \, dx}{b d} \\ & = -\frac {e+f x}{b d (a+b \sinh (c+d x))}-\frac {(2 i f) \text {Subst}\left (\int \frac {1}{a-2 i b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{b d^2} \\ & = -\frac {e+f x}{b d (a+b \sinh (c+d x))}+\frac {(4 i f) \text {Subst}\left (\int \frac {1}{-4 \left (a^2+b^2\right )-x^2} \, dx,x,-2 i b+2 a \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{b d^2} \\ & = -\frac {2 f \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b \sqrt {a^2+b^2} d^2}-\frac {e+f x}{b d (a+b \sinh (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\frac {\frac {2 f \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\frac {d (e+f x)}{a+b \sinh (c+d x)}}{b d^2} \]

[In]

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

[Out]

((2*f*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - (d*(e + f*x))/(a + b*Sinh[c + d*x
]))/(b*d^2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(163\) vs. \(2(71)=142\).

Time = 3.10 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.22

method	result	size
risch	\(-\frac {2 \left (f x +e \right ) {\mathrm e}^{d x +c}}{b d \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}+\frac {f \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}-a^{2}-b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d^{2} b}-\frac {f \ln \left ({\mathrm e}^{d x +c}+\frac {a \sqrt {a^{2}+b^{2}}+a^{2}+b^{2}}{\sqrt {a^{2}+b^{2}}\, b}\right )}{\sqrt {a^{2}+b^{2}}\, d^{2} b}\)	\(164\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (71) = 142\).

Time = 0.24 (sec) , antiderivative size = 411, normalized size of antiderivative = 5.55 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\frac {{\left (b f \cosh \left (d x + c\right )^{2} + b f \sinh \left (d x + c\right )^{2} + 2 \, a f \cosh \left (d x + c\right ) - b f + 2 \, {\left (b f \cosh \left (d x + c\right ) + a f\right )} \sinh \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) - 2 \, {\left ({\left (a^{2} + b^{2}\right )} d f x + {\left (a^{2} + b^{2}\right )} d e\right )} \cosh \left (d x + c\right ) - 2 \, {\left ({\left (a^{2} + b^{2}\right )} d f x + {\left (a^{2} + b^{2}\right )} d e\right )} \sinh \left (d x + c\right )}{{\left (a^{2} b^{2} + b^{4}\right )} d^{2} \cosh \left (d x + c\right )^{2} + {\left (a^{2} b^{2} + b^{4}\right )} d^{2} \sinh \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} d^{2} \cosh \left (d x + c\right ) - {\left (a^{2} b^{2} + b^{4}\right )} d^{2} + 2 \, {\left ({\left (a^{2} b^{2} + b^{4}\right )} d^{2} \cosh \left (d x + c\right ) + {\left (a^{3} b + a b^{3}\right )} d^{2}\right )} \sinh \left (d x + c\right )} \]

[In]

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

[Out]

((b*f*cosh(d*x + c)^2 + b*f*sinh(d*x + c)^2 + 2*a*f*cosh(d*x + c) - b*f + 2*(b*f*cosh(d*x + c) + a*f)*sinh(d*x
+ c))*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*
(b^2*cosh(d*x + c) + a*b)*sinh(d*x + c) - 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d*x + c) + a))/(b*cosh(d
*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) - 2*((a^2 + b^
2)*d*f*x + (a^2 + b^2)*d*e)*cosh(d*x + c) - 2*((a^2 + b^2)*d*f*x + (a^2 + b^2)*d*e)*sinh(d*x + c))/((a^2*b^2 +
b^4)*d^2*cosh(d*x + c)^2 + (a^2*b^2 + b^4)*d^2*sinh(d*x + c)^2 + 2*(a^3*b + a*b^3)*d^2*cosh(d*x + c) - (a^2*b
^2 + b^4)*d^2 + 2*((a^2*b^2 + b^4)*d^2*cosh(d*x + c) + (a^3*b + a*b^3)*d^2)*sinh(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (71) = 142\).

Time = 0.36 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.12 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=-f {\left (\frac {2 \, x e^{\left (d x + c\right )}}{b^{2} d e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b d e^{\left (d x + c\right )} - b^{2} d} - \frac {\log \left (\frac {b e^{\left (d x + c\right )} + a - \sqrt {a^{2} + b^{2}}}{b e^{\left (d x + c\right )} + a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b d^{2}}\right )} - \frac {2 \, e e^{\left (-d x - c\right )}}{{\left (2 \, a b e^{\left (-d x - c\right )} - b^{2} e^{\left (-2 \, d x - 2 \, c\right )} + b^{2}\right )} d} \]

[In]

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

[Out]

-f*(2*x*e^(d*x + c)/(b^2*d*e^(2*d*x + 2*c) + 2*a*b*d*e^(d*x + c) - b^2*d) - log((b*e^(d*x + c) + a - sqrt(a^2
+ b^2))/(b*e^(d*x + c) + a + sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*b*d^2)) - 2*e*e^(-d*x - c)/((2*a*b*e^(-d*x - c
) - b^2*e^(-2*d*x - 2*c) + b^2)*d)

Giac [F]

\[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\int { \frac {{\left (f x + e\right )} \cosh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.69 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx=\frac {f\,\ln \left (\frac {2\,f\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^2\,d\,\sqrt {a^2+b^2}}-\frac {2\,f\,{\mathrm {e}}^{c+d\,x}}{b^2\,d}\right )}{b\,d^2\,\sqrt {a^2+b^2}}-\frac {f\,\ln \left (-\frac {2\,f\,{\mathrm {e}}^{c+d\,x}}{b^2\,d}-\frac {2\,f\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{b^2\,d\,\sqrt {a^2+b^2}}\right )}{b\,d^2\,\sqrt {a^2+b^2}}-\frac {2\,{\mathrm {e}}^{c+d\,x}\,\left (a^2\,e+b^2\,e+a^2\,f\,x+b^2\,f\,x\right )}{d\,\left (a^2\,b+b^3\right )\,\left (2\,a\,{\mathrm {e}}^{c+d\,x}-b+b\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )} \]

[In]

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

[Out]

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