\(\int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx\) [330]

Optimal result

Integrand size = 24, antiderivative size = 112 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=-\frac {a f \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b \left (a^2+b^2\right )^{3/2} d^2}-\frac {e+f x}{2 b d (a+b \sinh (c+d x))^2}-\frac {f \cosh (c+d x)}{2 \left (a^2+b^2\right ) d^2 (a+b \sinh (c+d x))} \]

[Out]

-a*f*arctanh((b-a*tanh(1/2*d*x+1/2*c))/(a^2+b^2)^(1/2))/b/(a^2+b^2)^(3/2)/d^2+1/2*(-f*x-e)/b/d/(a+b*sinh(d*x+c
))^2-1/2*f*cosh(d*x+c)/(a^2+b^2)/d^2/(a+b*sinh(d*x+c))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5572, 2743, 12, 2739, 632, 210} \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=-\frac {a f \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{b d^2 \left (a^2+b^2\right )^{3/2}}-\frac {f \cosh (c+d x)}{2 d^2 \left (a^2+b^2\right ) (a+b \sinh (c+d x))}-\frac {e+f x}{2 b d (a+b \sinh (c+d x))^2} \]

[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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2743

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] && Integ
erQ[2*n]

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}{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) \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 \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) \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 \left (a^2+b^2\right ) d^2} \\ & = -\frac {a f \text {arctanh}\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] (verified)

Time = 0.63 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=-\frac {\frac {f \cosh (c+d x)}{\left (a^2+b^2\right ) (a+b \sinh (c+d x))}+\frac {\frac {2 a f \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}+\frac {d (e+f x)}{(a+b \sinh (c+d x))^2}}{b}}{2 d^2} \]

[In]

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

[Out]

-1/2*((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)/d^2

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(307\) vs. \(2(106)=212\).

Time = 14.55 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.75

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{2 d x +2 c} a^{2} d f x +2 b^{2} d f x \,{\mathrm e}^{2 d x +2 c}+2 \,{\mathrm e}^{2 d x +2 c} a^{2} d e -a b f \,{\mathrm e}^{3 d x +3 c}+2 b^{2} d e \,{\mathrm e}^{2 d x +2 c}-2 a^{2} f \,{\mathrm e}^{2 d x +2 c}+b^{2} f \,{\mathrm e}^{2 d x +2 c}+3 a f \,{\mathrm e}^{d x +c} b -f \,b^{2}}{b \,d^{2} \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )^{2}}+\frac {f a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{2 \left (a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2} b}-\frac {f a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{2 \left (a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2} b}\)	\(308\)

[In]

int((f*x+e)*cosh(d*x+c)/(a+b*sinh(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1230 vs. \(2 (105) = 210\).

Time = 0.26 (sec) , antiderivative size = 1230, normalized size of antiderivative = 10.98 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=\text {Too large to display} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (105) = 210\).

Time = 0.43 (sec) , antiderivative size = 413, normalized size of antiderivative = 3.69 \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=\frac {1}{2} \, f {\left (\frac {2 \, {\left (a b e^{\left (3 \, d x + 3 \, c\right )} - 3 \, a b e^{\left (d x + c\right )} + b^{2} + {\left (2 \, a^{2} e^{\left (2 \, c\right )} - b^{2} e^{\left (2 \, c\right )} - 2 \, {\left (a^{2} d e^{\left (2 \, c\right )} + b^{2} d e^{\left (2 \, c\right )}\right )} x\right )} e^{\left (2 \, d x\right )}\right )}}{a^{2} b^{3} d^{2} + b^{5} d^{2} + {\left (a^{2} b^{3} d^{2} e^{\left (4 \, c\right )} + b^{5} d^{2} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 4 \, {\left (a^{3} b^{2} d^{2} e^{\left (3 \, c\right )} + a b^{4} d^{2} e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + 2 \, {\left (2 \, a^{4} b d^{2} e^{\left (2 \, c\right )} + a^{2} b^{3} d^{2} e^{\left (2 \, c\right )} - b^{5} d^{2} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - 4 \, {\left (a^{3} b^{2} d^{2} e^{c} + a b^{4} d^{2} e^{c}\right )} e^{\left (d x\right )}} + \frac {a \log \left (\frac {b e^{\left (d x + 2 \, c\right )} + a e^{c} - \sqrt {a^{2} + b^{2}} e^{c}}{b e^{\left (d x + 2 \, c\right )} + a e^{c} + \sqrt {a^{2} + b^{2}} e^{c}}\right )}{{\left (a^{2} b + b^{3}\right )} \sqrt {a^{2} + b^{2}} d^{2}}\right )} - \frac {2 \, e e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (4 \, a b^{2} e^{\left (-d x - c\right )} - 4 \, a b^{2} e^{\left (-3 \, d x - 3 \, c\right )} + b^{3} e^{\left (-4 \, d x - 4 \, c\right )} + b^{3} + 2 \, {\left (2 \, a^{2} b - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} \]

[In]

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

[Out]

1/2*f*(2*(a*b*e^(3*d*x + 3*c) - 3*a*b*e^(d*x + c) + b^2 + (2*a^2*e^(2*c) - b^2*e^(2*c) - 2*(a^2*d*e^(2*c) + b^
2*d*e^(2*c))*x)*e^(2*d*x))/(a^2*b^3*d^2 + b^5*d^2 + (a^2*b^3*d^2*e^(4*c) + b^5*d^2*e^(4*c))*e^(4*d*x) + 4*(a^3
*b^2*d^2*e^(3*c) + a*b^4*d^2*e^(3*c))*e^(3*d*x) + 2*(2*a^4*b*d^2*e^(2*c) + a^2*b^3*d^2*e^(2*c) - b^5*d^2*e^(2*
c))*e^(2*d*x) - 4*(a^3*b^2*d^2*e^c + a*b^4*d^2*e^c)*e^(d*x)) + a*log((b*e^(d*x + 2*c) + a*e^c - sqrt(a^2 + b^2
)*e^c)/(b*e^(d*x + 2*c) + a*e^c + sqrt(a^2 + b^2)*e^c))/((a^2*b + b^3)*sqrt(a^2 + b^2)*d^2)) - 2*e*e^(-2*d*x -
2*c)/((4*a*b^2*e^(-d*x - c) - 4*a*b^2*e^(-3*d*x - 3*c) + b^3*e^(-4*d*x - 4*c) + b^3 + 2*(2*a^2*b - b^3)*e^(-2
*d*x - 2*c))*d)

Giac [F]

\[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )} \cosh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x) \cosh (c+d x)}{(a+b \sinh (c+d x))^3} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (e+f\,x\right )}{{\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^3} \,d x \]

[In]

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

[Out]

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