3.327 \(\int \frac{(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac{a f \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d^2 \left (a^2+b^2\right )^{3/2}}-\frac{f \cosh (c+d x)}{2 d^2 \left (a^2+b^2\right ) (a+b \sinh (c+d x))}-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2} \]

[Out]

-((a*f*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b*(a^2 + b^2)^(3/2)*d^2)) - (e + f*x)/(2*b*d*(a +
b*Sinh[c + d*x])^2) - (f*Cosh[c + d*x])/(2*(a^2 + b^2)*d^2*(a + b*Sinh[c + d*x]))

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Rubi [A]  time = 0.0983786, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5464, 2664, 12, 2660, 618, 204} \[ -\frac{a f \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b d^2 \left (a^2+b^2\right )^{3/2}}-\frac{f \cosh (c+d x)}{2 d^2 \left (a^2+b^2\right ) (a+b \sinh (c+d x))}-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Cosh[c + d*x])/(a + b*Sinh[c + d*x])^3,x]

[Out]

-((a*f*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(b*(a^2 + b^2)^(3/2)*d^2)) - (e + f*x)/(2*b*d*(a +
b*Sinh[c + d*x])^2) - (f*Cosh[c + d*x])/(2*(a^2 + b^2)*d^2*(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 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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))^3} \, dx &=-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2}+\frac{f \int \frac{1}{(a+b \sinh (c+d x))^2} \, dx}{2 b d}\\ &=-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2}-\frac{f \cosh (c+d x)}{2 \left (a^2+b^2\right ) d^2 (a+b \sinh (c+d x))}+\frac{f \int \frac{a}{a+b \sinh (c+d x)} \, dx}{2 b \left (a^2+b^2\right ) d}\\ &=-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2}-\frac{f \cosh (c+d x)}{2 \left (a^2+b^2\right ) d^2 (a+b \sinh (c+d x))}+\frac{(a f) \int \frac{1}{a+b \sinh (c+d x)} \, dx}{2 b \left (a^2+b^2\right ) d}\\ &=-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2}-\frac{f \cosh (c+d x)}{2 \left (a^2+b^2\right ) d^2 (a+b \sinh (c+d x))}-\frac{(i a 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 \left (a^2+b^2\right ) d^2}\\ &=-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2}-\frac{f \cosh (c+d x)}{2 \left (a^2+b^2\right ) d^2 (a+b \sinh (c+d x))}+\frac{(2 i a 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 \left (a^2+b^2\right ) d^2}\\ &=-\frac{a f \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{b \left (a^2+b^2\right )^{3/2} d^2}-\frac{e+f x}{2 b d (a+b \sinh (c+d x))^2}-\frac{f \cosh (c+d x)}{2 \left (a^2+b^2\right ) d^2 (a+b \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.16218, size = 112, normalized size = 1. \[ -\frac{\frac{\frac{2 a f \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\frac{d (e+f x)}{(a+b \sinh (c+d x))^2}}{b}+\frac{f \cosh (c+d x)}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*Cosh[c + d*x])/(a + b*Sinh[c + d*x])^3,x]

[Out]

-((f*Cosh[c + d*x])/((a^2 + b^2)*(a + b*Sinh[c + d*x])) + ((2*a*f*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 -
b^2]])/(-a^2 - b^2)^(3/2) + (d*(e + f*x))/(a + b*Sinh[c + d*x])^2)/b)/(2*d^2)

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Maple [B]  time = 0.319, size = 308, normalized size = 2.8 \begin{align*} -{\frac{2\,{a}^{2}dfx{{\rm e}^{2\,dx+2\,c}}+2\,{b}^{2}dfx{{\rm e}^{2\,dx+2\,c}}+2\,{a}^{2}de{{\rm e}^{2\,dx+2\,c}}-abf{{\rm e}^{3\,dx+3\,c}}+2\,{b}^{2}de{{\rm e}^{2\,dx+2\,c}}-2\,{a}^{2}f{{\rm e}^{2\,dx+2\,c}}+{b}^{2}f{{\rm e}^{2\,dx+2\,c}}+3\,fa{{\rm e}^{dx+c}}b-f{b}^{2}}{{d}^{2}b \left ( b{{\rm e}^{2\,dx+2\,c}}+2\,a{{\rm e}^{dx+c}}-b \right ) ^{2} \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{af}{2\,{d}^{2}b}\ln \left ({{\rm e}^{dx+c}}+{\frac{1}{b} \left ( a \left ({a}^{2}+{b}^{2} \right ) ^{{\frac{3}{2}}}-{a}^{4}-2\,{a}^{2}{b}^{2}-{b}^{4} \right ) \left ({a}^{2}+{b}^{2} \right ) ^{-{\frac{3}{2}}}} \right ) \left ({a}^{2}+{b}^{2} \right ) ^{-{\frac{3}{2}}}}-{\frac{af}{2\,{d}^{2}b}\ln \left ({{\rm e}^{dx+c}}+{\frac{1}{b} \left ( a \left ({a}^{2}+{b}^{2} \right ) ^{{\frac{3}{2}}}+{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ({a}^{2}+{b}^{2} \right ) ^{-{\frac{3}{2}}}} \right ) \left ({a}^{2}+{b}^{2} \right ) ^{-{\frac{3}{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))^3,x)

[Out]

-1/b*(2*a^2*d*f*x*exp(2*d*x+2*c)+2*b^2*d*f*x*exp(2*d*x+2*c)+2*a^2*d*e*exp(2*d*x+2*c)-a*b*f*exp(3*d*x+3*c)+2*b^
2*d*e*exp(2*d*x+2*c)-2*a^2*f*exp(2*d*x+2*c)+b^2*f*exp(2*d*x+2*c)+3*f*a*exp(d*x+c)*b-f*b^2)/d^2/(b*exp(2*d*x+2*
c)+2*a*exp(d*x+c)-b)^2/(a^2+b^2)+1/2/(a^2+b^2)^(3/2)*f*a/d^2/b*ln(exp(d*x+c)+(a*(a^2+b^2)^(3/2)-a^4-2*a^2*b^2-
b^4)/(a^2+b^2)^(3/2)/b)-1/2/(a^2+b^2)^(3/2)*f*a/d^2/b*ln(exp(d*x+c)+(a*(a^2+b^2)^(3/2)+a^4+2*a^2*b^2+b^4)/(a^2
+b^2)^(3/2)/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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.3144, size = 2811, normalized size = 25.1 \begin{align*} \text{result too large to display} \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))^3,x, algorithm="fricas")

[Out]

1/2*(2*(a^3*b + a*b^3)*f*cosh(d*x + c)^3 + 2*(a^3*b + a*b^3)*f*sinh(d*x + c)^3 - 6*(a^3*b + a*b^3)*f*cosh(d*x
+ c) - 2*(2*(a^4 + 2*a^2*b^2 + b^4)*d*f*x + 2*(a^4 + 2*a^2*b^2 + b^4)*d*e - (2*a^4 + a^2*b^2 - b^4)*f)*cosh(d*
x + c)^2 - 2*(2*(a^4 + 2*a^2*b^2 + b^4)*d*f*x + 2*(a^4 + 2*a^2*b^2 + b^4)*d*e - 3*(a^3*b + a*b^3)*f*cosh(d*x +
c) - (2*a^4 + a^2*b^2 - b^4)*f)*sinh(d*x + c)^2 + (a*b^2*f*cosh(d*x + c)^4 + a*b^2*f*sinh(d*x + c)^4 + 4*a^2*
b*f*cosh(d*x + c)^3 - 4*a^2*b*f*cosh(d*x + c) + a*b^2*f + 2*(2*a^3 - a*b^2)*f*cosh(d*x + c)^2 + 4*(a*b^2*f*cos
h(d*x + c) + a^2*b*f)*sinh(d*x + c)^3 + 2*(3*a*b^2*f*cosh(d*x + c)^2 + 6*a^2*b*f*cosh(d*x + c) + (2*a^3 - a*b^
2)*f)*sinh(d*x + c)^2 + 4*(a*b^2*f*cosh(d*x + c)^3 + 3*a^2*b*f*cosh(d*x + c)^2 - a^2*b*f + (2*a^3 - a*b^2)*f*c
osh(d*x + c))*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 + b^4)*f + 2*(3*(a^3*b + a*b^3)*f*cosh(d*x + c)^2 - 3*(a^3*b + a*b^3)*f - 2*(2*(a^4 + 2*a
^2*b^2 + b^4)*d*f*x + 2*(a^4 + 2*a^2*b^2 + b^4)*d*e - (2*a^4 + a^2*b^2 - b^4)*f)*cosh(d*x + c))*sinh(d*x + c))
/((a^4*b^3 + 2*a^2*b^5 + b^7)*d^2*cosh(d*x + c)^4 + (a^4*b^3 + 2*a^2*b^5 + b^7)*d^2*sinh(d*x + c)^4 + 4*(a^5*b
^2 + 2*a^3*b^4 + a*b^6)*d^2*cosh(d*x + c)^3 + 2*(2*a^6*b + 3*a^4*b^3 - b^7)*d^2*cosh(d*x + c)^2 - 4*(a^5*b^2 +
2*a^3*b^4 + a*b^6)*d^2*cosh(d*x + c) + 4*((a^4*b^3 + 2*a^2*b^5 + b^7)*d^2*cosh(d*x + c) + (a^5*b^2 + 2*a^3*b^
4 + a*b^6)*d^2)*sinh(d*x + c)^3 + (a^4*b^3 + 2*a^2*b^5 + b^7)*d^2 + 2*(3*(a^4*b^3 + 2*a^2*b^5 + b^7)*d^2*cosh(
d*x + c)^2 + 6*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*d^2*cosh(d*x + c) + (2*a^6*b + 3*a^4*b^3 - b^7)*d^2)*sinh(d*x + c
)^2 + 4*((a^4*b^3 + 2*a^2*b^5 + b^7)*d^2*cosh(d*x + c)^3 + 3*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*d^2*cosh(d*x + c)^2
+ (2*a^6*b + 3*a^4*b^3 - b^7)*d^2*cosh(d*x + c) - (a^5*b^2 + 2*a^3*b^4 + a*b^6)*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))**3,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 )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

integrate((f*x + e)*cosh(d*x + c)/(b*sinh(d*x + c) + a)^3, x)