3.98 \(\int \frac{\sinh ^3(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)
Optimal. Leaf size=83 \[ \frac{\cosh (e+f x) \sqrt{a+b \cosh ^2(e+f x)-b}}{2 b f}-\frac{(a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \cosh (e+f x)}{\sqrt{a+b \cosh ^2(e+f x)-b}}\right )}{2 b^{3/2} f} \]
[Out]
-((a + b)*ArcTanh[(Sqrt[b]*Cosh[e + f*x])/Sqrt[a - b + b*Cosh[e + f*x]^2]])/(2*b^(3/2)*f) + (Cosh[e + f*x]*Sqr
t[a - b + b*Cosh[e + f*x]^2])/(2*b*f)
________________________________________________________________________________________
Rubi [A] time = 0.0992391, antiderivative size = 83, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{3186, 388, 217, 206} \[ \frac{\cosh (e+f x) \sqrt{a+b \cosh ^2(e+f x)-b}}{2 b f}-\frac{(a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \cosh (e+f x)}{\sqrt{a+b \cosh ^2(e+f x)-b}}\right )}{2 b^{3/2} f} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[e + f*x]^3/Sqrt[a + b*Sinh[e + f*x]^2],x]
[Out]
-((a + b)*ArcTanh[(Sqrt[b]*Cosh[e + f*x])/Sqrt[a - b + b*Cosh[e + f*x]^2]])/(2*b^(3/2)*f) + (Cosh[e + f*x]*Sqr
t[a - b + b*Cosh[e + f*x]^2])/(2*b*f)
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 388
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{\sinh ^3(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\sqrt{a-b+b x^2}} \, dx,x,\cosh (e+f x)\right )}{f}\\ &=\frac{\cosh (e+f x) \sqrt{a-b+b \cosh ^2(e+f x)}}{2 b f}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-b+b x^2}} \, dx,x,\cosh (e+f x)\right )}{2 b f}\\ &=\frac{\cosh (e+f x) \sqrt{a-b+b \cosh ^2(e+f x)}}{2 b f}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\cosh (e+f x)}{\sqrt{a-b+b \cosh ^2(e+f x)}}\right )}{2 b f}\\ &=-\frac{(a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \cosh (e+f x)}{\sqrt{a-b+b \cosh ^2(e+f x)}}\right )}{2 b^{3/2} f}+\frac{\cosh (e+f x) \sqrt{a-b+b \cosh ^2(e+f x)}}{2 b f}\\ \end{align*}
Mathematica [A] time = 0.269282, size = 98, normalized size = 1.18 \[ \frac{\cosh (e+f x) \sqrt{2 a+b \cosh (2 (e+f x))-b}}{2 \sqrt{2} b f}-\frac{(a+b) \log \left (\sqrt{2 a+b \cosh (2 (e+f x))-b}+\sqrt{2} \sqrt{b} \cosh (e+f x)\right )}{2 b^{3/2} f} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[e + f*x]^3/Sqrt[a + b*Sinh[e + f*x]^2],x]
[Out]
(Cosh[e + f*x]*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])/(2*Sqrt[2]*b*f) - ((a + b)*Log[Sqrt[2]*Sqrt[b]*Cosh[e + f*
x] + Sqrt[2*a - b + b*Cosh[2*(e + f*x)]]])/(2*b^(3/2)*f)
________________________________________________________________________________________
Maple [B] time = 0.109, size = 204, normalized size = 2.5 \begin{align*}{\frac{1}{4\,f\cosh \left ( fx+e \right ) }\sqrt{ \left ( a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,{b}^{3/2}\sqrt{b \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( a-b \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}-{b}^{2}\ln \left ({\frac{1}{2} \left ( 2\,b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{b \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( a-b \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}\sqrt{b}+a-b \right ){\frac{1}{\sqrt{b}}}} \right ) -ba\ln \left ({\frac{1}{2} \left ( 2\,b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}+2\,\sqrt{b \left ( \cosh \left ( fx+e \right ) \right ) ^{4}+ \left ( a-b \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}\sqrt{b}+a-b \right ){\frac{1}{\sqrt{b}}}} \right ) \right ){b}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x)
[Out]
1/4*((a+b*sinh(f*x+e)^2)*cosh(f*x+e)^2)^(1/2)*(2*b^(3/2)*(b*cosh(f*x+e)^4+(a-b)*cosh(f*x+e)^2)^(1/2)-b^2*ln(1/
2*(2*b*cosh(f*x+e)^2+2*(b*cosh(f*x+e)^4+(a-b)*cosh(f*x+e)^2)^(1/2)*b^(1/2)+a-b)/b^(1/2))-b*a*ln(1/2*(2*b*cosh(
f*x+e)^2+2*(b*cosh(f*x+e)^4+(a-b)*cosh(f*x+e)^2)^(1/2)*b^(1/2)+a-b)/b^(1/2)))/b^(5/2)/cosh(f*x+e)/(a+b*sinh(f*
x+e)^2)^(1/2)/f
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (f x + e\right )^{3}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(sinh(f*x + e)^3/sqrt(b*sinh(f*x + e)^2 + a), x)
________________________________________________________________________________________
Fricas [B] time = 2.5622, size = 5576, normalized size = 67.18 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/8*(((a + b)*cosh(f*x + e)^2 + 2*(a + b)*cosh(f*x + e)*sinh(f*x + e) + (a + b)*sinh(f*x + e)^2)*sqrt(b)*log(
(a^2*b*cosh(f*x + e)^8 + 8*a^2*b*cosh(f*x + e)*sinh(f*x + e)^7 + a^2*b*sinh(f*x + e)^8 + 2*(a^3 + a^2*b)*cosh(
f*x + e)^6 + 2*(14*a^2*b*cosh(f*x + e)^2 + a^3 + a^2*b)*sinh(f*x + e)^6 + 4*(14*a^2*b*cosh(f*x + e)^3 + 3*(a^3
+ a^2*b)*cosh(f*x + e))*sinh(f*x + e)^5 + (9*a^2*b - 4*a*b^2 + b^3)*cosh(f*x + e)^4 + (70*a^2*b*cosh(f*x + e)
^4 + 9*a^2*b - 4*a*b^2 + b^3 + 30*(a^3 + a^2*b)*cosh(f*x + e)^2)*sinh(f*x + e)^4 + 4*(14*a^2*b*cosh(f*x + e)^5
+ 10*(a^3 + a^2*b)*cosh(f*x + e)^3 + (9*a^2*b - 4*a*b^2 + b^3)*cosh(f*x + e))*sinh(f*x + e)^3 + b^3 + 2*(3*a*
b^2 - b^3)*cosh(f*x + e)^2 + 2*(14*a^2*b*cosh(f*x + e)^6 + 15*(a^3 + a^2*b)*cosh(f*x + e)^4 + 3*a*b^2 - b^3 +
3*(9*a^2*b - 4*a*b^2 + b^3)*cosh(f*x + e)^2)*sinh(f*x + e)^2 - sqrt(2)*(a^2*cosh(f*x + e)^6 + 6*a^2*cosh(f*x +
e)*sinh(f*x + e)^5 + a^2*sinh(f*x + e)^6 + 3*a^2*cosh(f*x + e)^4 + 3*(5*a^2*cosh(f*x + e)^2 + a^2)*sinh(f*x +
e)^4 + 4*(5*a^2*cosh(f*x + e)^3 + 3*a^2*cosh(f*x + e))*sinh(f*x + e)^3 + (4*a*b - b^2)*cosh(f*x + e)^2 + (15*
a^2*cosh(f*x + e)^4 + 18*a^2*cosh(f*x + e)^2 + 4*a*b - b^2)*sinh(f*x + e)^2 + b^2 + 2*(3*a^2*cosh(f*x + e)^5 +
6*a^2*cosh(f*x + e)^3 + (4*a*b - b^2)*cosh(f*x + e))*sinh(f*x + e))*sqrt(b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(
f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)) + 4*(2*a^2*b*cosh(f
*x + e)^7 + 3*(a^3 + a^2*b)*cosh(f*x + e)^5 + (9*a^2*b - 4*a*b^2 + b^3)*cosh(f*x + e)^3 + (3*a*b^2 - b^3)*cosh
(f*x + e))*sinh(f*x + e))/(cosh(f*x + e)^6 + 6*cosh(f*x + e)^5*sinh(f*x + e) + 15*cosh(f*x + e)^4*sinh(f*x + e
)^2 + 20*cosh(f*x + e)^3*sinh(f*x + e)^3 + 15*cosh(f*x + e)^2*sinh(f*x + e)^4 + 6*cosh(f*x + e)*sinh(f*x + e)^
5 + sinh(f*x + e)^6)) + ((a + b)*cosh(f*x + e)^2 + 2*(a + b)*cosh(f*x + e)*sinh(f*x + e) + (a + b)*sinh(f*x +
e)^2)*sqrt(b)*log(-(b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(a - b)*cosh
(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + a - b)*sinh(f*x + e)^2 - sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sin
h(f*x + e) + sinh(f*x + e)^2 - 1)*sqrt(b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e
)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)) + 4*(b*cosh(f*x + e)^3 + (a - b)*cosh(f*x + e))*sinh(f
*x + e) + b)/(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)) + sqrt(2)*(b*cosh(f*x + e)^2
+ 2*b*cosh(f*x + e)*sinh(f*x + e) + b*sinh(f*x + e)^2 + b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a
- b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/(b^2*f*cosh(f*x + e)^2 + 2*b^2*f*co
sh(f*x + e)*sinh(f*x + e) + b^2*f*sinh(f*x + e)^2), 1/8*(2*((a + b)*cosh(f*x + e)^2 + 2*(a + b)*cosh(f*x + e)*
sinh(f*x + e) + (a + b)*sinh(f*x + e)^2)*sqrt(-b)*arctan(sqrt(2)*(a*cosh(f*x + e)^2 + 2*a*cosh(f*x + e)*sinh(f
*x + e) + a*sinh(f*x + e)^2 + b)*sqrt(-b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e
)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/(a*b*cosh(f*x + e)^4 + 4*a*b*cosh(f*x + e)*sinh(f*x +
e)^3 + a*b*sinh(f*x + e)^4 + (3*a*b - b^2)*cosh(f*x + e)^2 + (6*a*b*cosh(f*x + e)^2 + 3*a*b - b^2)*sinh(f*x +
e)^2 + b^2 + 2*(2*a*b*cosh(f*x + e)^3 + (3*a*b - b^2)*cosh(f*x + e))*sinh(f*x + e))) + 2*((a + b)*cosh(f*x + e
)^2 + 2*(a + b)*cosh(f*x + e)*sinh(f*x + e) + (a + b)*sinh(f*x + e)^2)*sqrt(-b)*arctan(sqrt(2)*(cosh(f*x + e)^
2 + 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2 - 1)*sqrt(-b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2
+ 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/(b*cosh(f*x + e)^4 + 4*b*cosh(
f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(2*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 2*a - b)
*sinh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)) + sqrt(2)*(b*cosh(f*x +
e)^2 + 2*b*cosh(f*x + e)*sinh(f*x + e) + b*sinh(f*x + e)^2 + b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 +
2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/(b^2*f*cosh(f*x + e)^2 + 2*b^2
*f*cosh(f*x + e)*sinh(f*x + e) + b^2*f*sinh(f*x + e)^2)]
________________________________________________________________________________________
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(f*x+e)**3/(a+b*sinh(f*x+e)**2)**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (f x + e\right )^{3}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
integrate(sinh(f*x + e)^3/sqrt(b*sinh(f*x + e)^2 + a), x)