3.103 \(\int \frac{1}{(a+b \sinh (c+d x))^3} \, dx\)

Optimal. Leaf size=127 \[ -\frac{\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}}-\frac{3 a b \cosh (c+d x)}{2 d \left (a^2+b^2\right )^2 (a+b \sinh (c+d x))}-\frac{b \cosh (c+d x)}{2 d \left (a^2+b^2\right ) (a+b \sinh (c+d x))^2} \]

[Out]

-(((2*a^2 - b^2)*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d)) - (b*Cosh[c + d*x]
)/(2*(a^2 + b^2)*d*(a + b*Sinh[c + d*x])^2) - (3*a*b*Cosh[c + d*x])/(2*(a^2 + b^2)^2*d*(a + b*Sinh[c + d*x]))

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Rubi [A]  time = 0.122238, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2664, 2754, 12, 2660, 618, 204} \[ -\frac{\left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{5/2}}-\frac{3 a b \cosh (c+d x)}{2 d \left (a^2+b^2\right )^2 (a+b \sinh (c+d x))}-\frac{b \cosh (c+d x)}{2 d \left (a^2+b^2\right ) (a+b \sinh (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x])^(-3),x]

[Out]

-(((2*a^2 - b^2)*ArcTanh[(b - a*Tanh[(c + d*x)/2])/Sqrt[a^2 + b^2]])/((a^2 + b^2)^(5/2)*d)) - (b*Cosh[c + d*x]
)/(2*(a^2 + b^2)*d*(a + b*Sinh[c + d*x])^2) - (3*a*b*Cosh[c + d*x])/(2*(a^2 + b^2)^2*d*(a + b*Sinh[c + d*x]))

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 2754

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

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{1}{(a+b \sinh (c+d x))^3} \, dx &=-\frac{b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac{\int \frac{-2 a+b \sinh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx}{2 \left (a^2+b^2\right )}\\ &=-\frac{b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac{3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}+\frac{\int \frac{2 a^2-b^2}{a+b \sinh (c+d x)} \, dx}{2 \left (a^2+b^2\right )^2}\\ &=-\frac{b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac{3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}+\frac{\left (2 a^2-b^2\right ) \int \frac{1}{a+b \sinh (c+d x)} \, dx}{2 \left (a^2+b^2\right )^2}\\ &=-\frac{b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac{3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}-\frac{\left (i \left (2 a^2-b^2\right )\right ) \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 )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac{b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac{3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}+\frac{\left (2 i \left (2 a^2-b^2\right )\right ) \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 )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac{\left (2 a^2-b^2\right ) \tanh ^{-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 )^{5/2} d}-\frac{b \cosh (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \sinh (c+d x))^2}-\frac{3 a b \cosh (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \sinh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.272079, size = 117, normalized size = 0.92 \[ \frac{\frac{2 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-\frac{b \cosh (c+d x) \left (4 a^2+3 a b \sinh (c+d x)+b^2\right )}{(a+b \sinh (c+d x))^2}}{2 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[c + d*x])^(-3),x]

[Out]

((2*(2*a^2 - b^2)*ArcTan[(b - a*Tanh[(c + d*x)/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] - (b*Cosh[c + d*x]*(4*a
^2 + b^2 + 3*a*b*Sinh[c + d*x]))/(a + b*Sinh[c + d*x])^2)/(2*(a^2 + b^2)^2*d)

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Maple [B]  time = 0.049, size = 280, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ( -2\,{\frac{1}{ \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) ^{2}} \left ( -1/2\,{\frac{{b}^{2} \left ( 5\,{a}^{2}+2\,{b}^{2} \right ) \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}-1/2\,{\frac{b \left ( 4\,{a}^{4}-7\,{a}^{2}{b}^{2}-2\,{b}^{4} \right ) \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ){a}^{2}}}+1/2\,{\frac{{b}^{2} \left ( 11\,{a}^{2}+2\,{b}^{2} \right ) \tanh \left ( 1/2\,dx+c/2 \right ) }{a \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) }}+1/2\,{\frac{b \left ( 4\,{a}^{2}+{b}^{2} \right ) }{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}} \right ) }+{\frac{2\,{a}^{2}-{b}^{2}}{{a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4}}{\it Artanh} \left ({\frac{1}{2} \left ( 2\,a\tanh \left ( 1/2\,dx+c/2 \right ) -2\,b \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sinh(d*x+c))^3,x)

[Out]

1/d*(-2*(-1/2*b^2*(5*a^2+2*b^2)/a/(a^4+2*a^2*b^2+b^4)*tanh(1/2*d*x+1/2*c)^3-1/2*b*(4*a^4-7*a^2*b^2-2*b^4)/(a^4
+2*a^2*b^2+b^4)/a^2*tanh(1/2*d*x+1/2*c)^2+1/2*b^2*(11*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4)/a*tanh(1/2*d*x+1/2*c)+1/2
*b*(4*a^2+b^2)/(a^4+2*a^2*b^2+b^4))/(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)^2+(2*a^2-b^2)/(a^4+2*a
^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))

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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(1/(a+b*sinh(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.65862, size = 3005, normalized size = 23.66 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(6*a^3*b^2 + 6*a*b^4 + 2*(2*a^4*b + a^2*b^3 - b^5)*cosh(d*x + c)^3 + 2*(2*a^4*b + a^2*b^3 - b^5)*sinh(d*x
+ c)^3 + 6*(2*a^5 + a^3*b^2 - a*b^4)*cosh(d*x + c)^2 + 6*(2*a^5 + a^3*b^2 - a*b^4 + (2*a^4*b + a^2*b^3 - b^5)*
cosh(d*x + c))*sinh(d*x + c)^2 - ((2*a^2*b^2 - b^4)*cosh(d*x + c)^4 + (2*a^2*b^2 - b^4)*sinh(d*x + c)^4 + 2*a^
2*b^2 - b^4 + 4*(2*a^3*b - a*b^3)*cosh(d*x + c)^3 + 4*(2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*cosh(d*x + c))*sinh
(d*x + c)^3 + 2*(4*a^4 - 4*a^2*b^2 + b^4)*cosh(d*x + c)^2 + 2*(4*a^4 - 4*a^2*b^2 + b^4 + 3*(2*a^2*b^2 - b^4)*c
osh(d*x + c)^2 + 6*(2*a^3*b - a*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 4*(2*a^3*b - a*b^3)*cosh(d*x + c) - 4*(2
*a^3*b - a*b^3 - (2*a^2*b^2 - b^4)*cosh(d*x + c)^3 - 3*(2*a^3*b - a*b^3)*cosh(d*x + c)^2 - (4*a^4 - 4*a^2*b^2
+ b^4)*cosh(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*co
sh(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*(10*a^4*b + 11*a^2*b^3 + b^5)*cosh(d*x + c) - 2*(10*a^4*b + 11*a^2*b^3 + b^5 - 3*(2*a^4*b
+ a^2*b^3 - b^5)*cosh(d*x + c)^2 - 6*(2*a^5 + a^3*b^2 - a*b^4)*cosh(d*x + c))*sinh(d*x + c))/((a^6*b^2 + 3*a^4
*b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c)^4 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*sinh(d*x + c)^4 + 4*(a^7
*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*x + c)^3 + 2*(2*a^8 + 5*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - b^8)*d*co
sh(d*x + c)^2 + 4*((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c) + (a^7*b + 3*a^5*b^3 + 3*a^3*b^5 +
a*b^7)*d)*sinh(d*x + c)^3 - 4*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*x + c) + 2*(3*(a^6*b^2 + 3*a^4*
b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c)^2 + 6*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*x + c) + (2*a^8
+ 5*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - b^8)*d)*sinh(d*x + c)^2 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d + 4*((
a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d*cosh(d*x + c)^3 + 3*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*cosh(d*
x + c)^2 + (2*a^8 + 5*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - b^8)*d*cosh(d*x + c) - (a^7*b + 3*a^5*b^3 + 3*a^3*b^5 +
a*b^7)*d)*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(1/(a+b*sinh(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.31144, size = 319, normalized size = 2.51 \begin{align*} -\frac{{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac{{\left | -2 \, b e^{\left (d x + c\right )} - 2 \, a - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | -2 \, b e^{\left (d x + c\right )} - 2 \, a + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{2 \,{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} - b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, a^{2} b e^{\left (d x + c\right )} - b^{3} e^{\left (d x + c\right )} + 3 \, a b^{2}}{{\left (a^{4} d + 2 \, a^{2} b^{2} d + b^{4} d\right )}{\left (b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a e^{\left (d x + c\right )} - b\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(2*a^2 - b^2)*log(abs(-2*b*e^(d*x + c) - 2*a - 2*sqrt(a^2 + b^2))/abs(-2*b*e^(d*x + c) - 2*a + 2*sqrt(a^2
+ b^2)))/((a^4*d + 2*a^2*b^2*d + b^4*d)*sqrt(a^2 + b^2)) + (2*a^2*b*e^(3*d*x + 3*c) - b^3*e^(3*d*x + 3*c) + 6
*a^3*e^(2*d*x + 2*c) - 3*a*b^2*e^(2*d*x + 2*c) - 10*a^2*b*e^(d*x + c) - b^3*e^(d*x + c) + 3*a*b^2)/((a^4*d + 2
*a^2*b^2*d + b^4*d)*(b*e^(2*d*x + 2*c) + 2*a*e^(d*x + c) - b)^2)