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

Optimal. Leaf size=74 \[ -\frac{2 f \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d^2 \sqrt{a^2+b^2}}-\frac{e+f x}{b d (a+b \sinh (c+d x))} \]

[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]))

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Rubi [A]  time = 0.0700089, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5464, 2660, 618, 204} \[ -\frac{2 f \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d^2 \sqrt{a^2+b^2}}-\frac{e+f x}{b d (a+b \sinh (c+d x))} \]

Antiderivative was successfully verified.

[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 5464

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]

Rule 2660

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 618

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 204

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

Rubi steps

\begin{align*} \int \frac{(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx &=-\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) \operatorname{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) \operatorname{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 \tanh ^{-1}\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]  time = 0.396371, size = 78, normalized size = 1.05 \[ \frac{\frac{2 f \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-\frac{d (e+f x)}{a+b \sinh (c+d x)}}{b d^2} \]

Antiderivative was successfully verified.

[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)

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Maple [B]  time = 0., size = 164, normalized size = 2.2 \begin{align*} -2\,{\frac{ \left ( fx+e \right ){{\rm e}^{dx+c}}}{bd \left ( b{{\rm e}^{2\,dx+2\,c}}+2\,a{{\rm e}^{dx+c}}-b \right ) }}+{\frac{f}{{d}^{2}b}\ln \left ({{\rm e}^{dx+c}}+{\frac{1}{b} \left ( a\sqrt{{a}^{2}+{b}^{2}}-{a}^{2}-{b}^{2} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}}-{\frac{f}{{d}^{2}b}\ln \left ({{\rm e}^{dx+c}}+{\frac{1}{b} \left ( a\sqrt{{a}^{2}+{b}^{2}}+{a}^{2}+{b}^{2} \right ){\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((f*x+e)*cosh(d*x+c)/(a+b*sinh(d*x+c))^2,x)

[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)

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.60907, size = 1015, normalized size = 13.72 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)