3.43 \(\int \frac{\sinh ^3(c+d x)}{(a+b \sinh ^2(c+d x))^2} \, dx\)
Optimal. Leaf size=90 \[ \frac{(a-2 b) \tan ^{-1}\left (\frac{\sqrt{b} \cosh (c+d x)}{\sqrt{a-b}}\right )}{2 b^{3/2} d (a-b)^{3/2}}-\frac{a \cosh (c+d x)}{2 b d (a-b) \left (a+b \cosh ^2(c+d x)-b\right )} \]
[Out]
((a - 2*b)*ArcTan[(Sqrt[b]*Cosh[c + d*x])/Sqrt[a - b]])/(2*(a - b)^(3/2)*b^(3/2)*d) - (a*Cosh[c + d*x])/(2*(a
- b)*b*d*(a - b + b*Cosh[c + d*x]^2))
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Rubi [A] time = 0.112523, antiderivative size = 90, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used =
{3186, 385, 205} \[ \frac{(a-2 b) \tan ^{-1}\left (\frac{\sqrt{b} \cosh (c+d x)}{\sqrt{a-b}}\right )}{2 b^{3/2} d (a-b)^{3/2}}-\frac{a \cosh (c+d x)}{2 b d (a-b) \left (a+b \cosh ^2(c+d x)-b\right )} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[c + d*x]^3/(a + b*Sinh[c + d*x]^2)^2,x]
[Out]
((a - 2*b)*ArcTan[(Sqrt[b]*Cosh[c + d*x])/Sqrt[a - b]])/(2*(a - b)^(3/2)*b^(3/2)*d) - (a*Cosh[c + d*x])/(2*(a
- b)*b*d*(a - b + b*Cosh[c + d*x]^2))
Rule 3186
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos
[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rule 385
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{\sinh ^3(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (a-b+b x^2\right )^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{a \cosh (c+d x)}{2 (a-b) b d \left (a-b+b \cosh ^2(c+d x)\right )}+\frac{(a-2 b) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\cosh (c+d x)\right )}{2 (a-b) b d}\\ &=\frac{(a-2 b) \tan ^{-1}\left (\frac{\sqrt{b} \cosh (c+d x)}{\sqrt{a-b}}\right )}{2 (a-b)^{3/2} b^{3/2} d}-\frac{a \cosh (c+d x)}{2 (a-b) b d \left (a-b+b \cosh ^2(c+d x)\right )}\\ \end{align*}
Mathematica [C] time = 0.596782, size = 141, normalized size = 1.57 \[ \frac{-\frac{2 a \sqrt{b} \cosh (c+d x)}{(a-b) (2 a+b \cosh (2 (c+d x))-b)}+\frac{(a-2 b) \left (\tan ^{-1}\left (\frac{\sqrt{b}-i \sqrt{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a-b}}\right )+\tan ^{-1}\left (\frac{\sqrt{b}+i \sqrt{a} \tanh \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a-b}}\right )\right )}{(a-b)^{3/2}}}{2 b^{3/2} d} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[c + d*x]^3/(a + b*Sinh[c + d*x]^2)^2,x]
[Out]
(((a - 2*b)*(ArcTan[(Sqrt[b] - I*Sqrt[a]*Tanh[(c + d*x)/2])/Sqrt[a - b]] + ArcTan[(Sqrt[b] + I*Sqrt[a]*Tanh[(c
+ d*x)/2])/Sqrt[a - b]]))/(a - b)^(3/2) - (2*a*Sqrt[b]*Cosh[c + d*x])/((a - b)*(2*a - b + b*Cosh[2*(c + d*x)]
)))/(2*b^(3/2)*d)
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Maple [B] time = 0.032, size = 341, normalized size = 3.8 \begin{align*} 16\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a}{d \left ( 16\,ab-16\,{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }}-32\,{\frac{ \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b}{d \left ( 16\,ab-16\,{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }}-16\,{\frac{a}{d \left ( 16\,ab-16\,{b}^{2} \right ) \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+4\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a \right ) }}+8\,{\frac{a}{d \left ( 16\,ab-16\,{b}^{2} \right ) \sqrt{ab-{b}^{2}}}\arctan \left ( 1/4\,{\frac{2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,a+4\,b}{\sqrt{ab-{b}^{2}}}} \right ) }-16\,{\frac{b}{d \left ( 16\,ab-16\,{b}^{2} \right ) \sqrt{ab-{b}^{2}}}\arctan \left ( 1/4\,{\frac{2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\,a+4\,b}{\sqrt{ab-{b}^{2}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(d*x+c)^3/(a+b*sinh(d*x+c)^2)^2,x)
[Out]
16/d/(16*a*b-16*b^2)/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2*b+a)*a*tanh(1/
2*d*x+1/2*c)^2-32/d/(16*a*b-16*b^2)/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1/2*d*x+1/2*c)^2
*b+a)*tanh(1/2*d*x+1/2*c)^2*b-16/d/(16*a*b-16*b^2)/(tanh(1/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*tanh(1
/2*d*x+1/2*c)^2*b+a)*a+8/d/(16*a*b-16*b^2)/(a*b-b^2)^(1/2)*arctan(1/4*(2*tanh(1/2*d*x+1/2*c)^2*a-2*a+4*b)/(a*b
-b^2)^(1/2))*a-16/d/(16*a*b-16*b^2)/(a*b-b^2)^(1/2)*arctan(1/4*(2*tanh(1/2*d*x+1/2*c)^2*a-2*a+4*b)/(a*b-b^2)^(
1/2))*b
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a e^{\left (3 \, d x + 3 \, c\right )} + a e^{\left (d x + c\right )}}{a b^{2} d - b^{3} d +{\left (a b^{2} d e^{\left (4 \, c\right )} - b^{3} d e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 2 \,{\left (2 \, a^{2} b d e^{\left (2 \, c\right )} - 3 \, a b^{2} d e^{\left (2 \, c\right )} + b^{3} d e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}} + \frac{1}{8} \, \int \frac{8 \,{\left ({\left (a e^{\left (3 \, c\right )} - 2 \, b e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} -{\left (a e^{c} - 2 \, b e^{c}\right )} e^{\left (d x\right )}\right )}}{a b^{2} - b^{3} +{\left (a b^{2} e^{\left (4 \, c\right )} - b^{3} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 2 \,{\left (2 \, a^{2} b e^{\left (2 \, c\right )} - 3 \, a b^{2} e^{\left (2 \, c\right )} + b^{3} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^3/(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
-(a*e^(3*d*x + 3*c) + a*e^(d*x + c))/(a*b^2*d - b^3*d + (a*b^2*d*e^(4*c) - b^3*d*e^(4*c))*e^(4*d*x) + 2*(2*a^2
*b*d*e^(2*c) - 3*a*b^2*d*e^(2*c) + b^3*d*e^(2*c))*e^(2*d*x)) + 1/8*integrate(8*((a*e^(3*c) - 2*b*e^(3*c))*e^(3
*d*x) - (a*e^c - 2*b*e^c)*e^(d*x))/(a*b^2 - b^3 + (a*b^2*e^(4*c) - b^3*e^(4*c))*e^(4*d*x) + 2*(2*a^2*b*e^(2*c)
- 3*a*b^2*e^(2*c) + b^3*e^(2*c))*e^(2*d*x)), x)
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Fricas [B] time = 2.43195, size = 4536, normalized size = 50.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^3/(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
[-1/4*(4*(a^2*b - a*b^2)*cosh(d*x + c)^3 + 12*(a^2*b - a*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + 4*(a^2*b - a*b^2
)*sinh(d*x + c)^3 + ((a*b - 2*b^2)*cosh(d*x + c)^4 + 4*(a*b - 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a*b - 2*
b^2)*sinh(d*x + c)^4 + 2*(2*a^2 - 5*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(a*b - 2*b^2)*cosh(d*x + c)^2 + 2*a^2
- 5*a*b + 2*b^2)*sinh(d*x + c)^2 + a*b - 2*b^2 + 4*((a*b - 2*b^2)*cosh(d*x + c)^3 + (2*a^2 - 5*a*b + 2*b^2)*co
sh(d*x + c))*sinh(d*x + c))*sqrt(-a*b + b^2)*log((b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*si
nh(d*x + c)^4 - 2*(2*a - 3*b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 - 2*a + 3*b)*sinh(d*x + c)^2 + 4*(b*cos
h(d*x + c)^3 - (2*a - 3*b)*cosh(d*x + c))*sinh(d*x + c) - 4*(cosh(d*x + c)^3 + 3*cosh(d*x + c)*sinh(d*x + c)^2
+ sinh(d*x + c)^3 + (3*cosh(d*x + c)^2 + 1)*sinh(d*x + c) + cosh(d*x + c))*sqrt(-a*b + b^2) + b)/(b*cosh(d*x
+ c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(2*a - b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x
+ c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 + (2*a - b)*cosh(d*x + c))*sinh(d*x + c) + b)) + 4*(
a^2*b - a*b^2)*cosh(d*x + c) + 4*(a^2*b - a*b^2 + 3*(a^2*b - a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c))/((a^2*b^3
- 2*a*b^4 + b^5)*d*cosh(d*x + c)^4 + 4*(a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2*b^3 -
2*a*b^4 + b^5)*d*sinh(d*x + c)^4 + 2*(2*a^3*b^2 - 5*a^2*b^3 + 4*a*b^4 - b^5)*d*cosh(d*x + c)^2 + 2*(3*(a^2*b^3
- 2*a*b^4 + b^5)*d*cosh(d*x + c)^2 + (2*a^3*b^2 - 5*a^2*b^3 + 4*a*b^4 - b^5)*d)*sinh(d*x + c)^2 + (a^2*b^3 -
2*a*b^4 + b^5)*d + 4*((a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)^3 + (2*a^3*b^2 - 5*a^2*b^3 + 4*a*b^4 - b^5)*d*
cosh(d*x + c))*sinh(d*x + c)), -1/2*(2*(a^2*b - a*b^2)*cosh(d*x + c)^3 + 6*(a^2*b - a*b^2)*cosh(d*x + c)*sinh(
d*x + c)^2 + 2*(a^2*b - a*b^2)*sinh(d*x + c)^3 - ((a*b - 2*b^2)*cosh(d*x + c)^4 + 4*(a*b - 2*b^2)*cosh(d*x + c
)*sinh(d*x + c)^3 + (a*b - 2*b^2)*sinh(d*x + c)^4 + 2*(2*a^2 - 5*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(a*b - 2*
b^2)*cosh(d*x + c)^2 + 2*a^2 - 5*a*b + 2*b^2)*sinh(d*x + c)^2 + a*b - 2*b^2 + 4*((a*b - 2*b^2)*cosh(d*x + c)^3
+ (2*a^2 - 5*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a*b - b^2)*arctan(-1/2*(b*cosh(d*x + c)^3 + 3*b*
cosh(d*x + c)*sinh(d*x + c)^2 + b*sinh(d*x + c)^3 + (4*a - 3*b)*cosh(d*x + c) + (3*b*cosh(d*x + c)^2 + 4*a - 3
*b)*sinh(d*x + c))/sqrt(a*b - b^2)) + ((a*b - 2*b^2)*cosh(d*x + c)^4 + 4*(a*b - 2*b^2)*cosh(d*x + c)*sinh(d*x
+ c)^3 + (a*b - 2*b^2)*sinh(d*x + c)^4 + 2*(2*a^2 - 5*a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(3*(a*b - 2*b^2)*cosh(d
*x + c)^2 + 2*a^2 - 5*a*b + 2*b^2)*sinh(d*x + c)^2 + a*b - 2*b^2 + 4*((a*b - 2*b^2)*cosh(d*x + c)^3 + (2*a^2 -
5*a*b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(a*b - b^2)*arctan(-1/2*sqrt(a*b - b^2)*(cosh(d*x + c) + sin
h(d*x + c))/(a - b)) + 2*(a^2*b - a*b^2)*cosh(d*x + c) + 2*(a^2*b - a*b^2 + 3*(a^2*b - a*b^2)*cosh(d*x + c)^2)
*sinh(d*x + c))/((a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)^4 + 4*(a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)*sin
h(d*x + c)^3 + (a^2*b^3 - 2*a*b^4 + b^5)*d*sinh(d*x + c)^4 + 2*(2*a^3*b^2 - 5*a^2*b^3 + 4*a*b^4 - b^5)*d*cosh(
d*x + c)^2 + 2*(3*(a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)^2 + (2*a^3*b^2 - 5*a^2*b^3 + 4*a*b^4 - b^5)*d)*sin
h(d*x + c)^2 + (a^2*b^3 - 2*a*b^4 + b^5)*d + 4*((a^2*b^3 - 2*a*b^4 + b^5)*d*cosh(d*x + c)^3 + (2*a^3*b^2 - 5*a
^2*b^3 + 4*a*b^4 - b^5)*d*cosh(d*x + c))*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(sinh(d*x+c)**3/(a+b*sinh(d*x+c)**2)**2,x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(d*x+c)^3/(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")
[Out]
Exception raised: TypeError