\(\int \frac {\sinh ^3(a+b x^2)}{x^2} \, dx\) [20]

Optimal result

Integrand size = 14, antiderivative size = 136 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=-\frac {3}{8} \sqrt {b} e^{-a} \sqrt {\pi } \text {erf}\left (\sqrt {b} x\right )+\frac {1}{8} \sqrt {b} e^{-3 a} \sqrt {3 \pi } \text {erf}\left (\sqrt {3} \sqrt {b} x\right )-\frac {3}{8} \sqrt {b} e^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x\right )+\frac {1}{8} \sqrt {b} e^{3 a} \sqrt {3 \pi } \text {erfi}\left (\sqrt {3} \sqrt {b} x\right )-\frac {\sinh ^3\left (a+b x^2\right )}{x} \] Output:

-3/8*b^(1/2)*Pi^(1/2)*erf(b^(1/2)*x)/exp(a)+1/8*b^(1/2)*3^(1/2)*Pi^(1/2)*e 
rf(3^(1/2)*b^(1/2)*x)/exp(3*a)-3/8*b^(1/2)*exp(a)*Pi^(1/2)*erfi(b^(1/2)*x) 
+1/8*b^(1/2)*exp(3*a)*3^(1/2)*Pi^(1/2)*erfi(3^(1/2)*b^(1/2)*x)-sinh(b*x^2+ 
a)^3/x
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.50 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=\frac {-3 \sqrt {b} \sqrt {\pi } x \cosh (a) \text {erfi}\left (\sqrt {b} x\right )+\sqrt {b} \sqrt {3 \pi } x \cosh (3 a) \text {erfi}\left (\sqrt {3} \sqrt {b} x\right )-3 \sqrt {b} \sqrt {\pi } x \text {erfi}\left (\sqrt {b} x\right ) \sinh (a)+3 \sqrt {b} \sqrt {\pi } x \text {erf}\left (\sqrt {b} x\right ) (-\cosh (a)+\sinh (a))+\sqrt {b} \sqrt {3 \pi } x \text {erf}\left (\sqrt {3} \sqrt {b} x\right ) (\cosh (3 a)-\sinh (3 a))+\sqrt {b} \sqrt {3 \pi } x \text {erfi}\left (\sqrt {3} \sqrt {b} x\right ) \sinh (3 a)+6 \sinh \left (a+b x^2\right )-2 \sinh \left (3 \left (a+b x^2\right )\right )}{8 x} \] Input:

Integrate[Sinh[a + b*x^2]^3/x^2,x]
 

Output:

(-3*Sqrt[b]*Sqrt[Pi]*x*Cosh[a]*Erfi[Sqrt[b]*x] + Sqrt[b]*Sqrt[3*Pi]*x*Cosh 
[3*a]*Erfi[Sqrt[3]*Sqrt[b]*x] - 3*Sqrt[b]*Sqrt[Pi]*x*Erfi[Sqrt[b]*x]*Sinh[ 
a] + 3*Sqrt[b]*Sqrt[Pi]*x*Erf[Sqrt[b]*x]*(-Cosh[a] + Sinh[a]) + Sqrt[b]*Sq 
rt[3*Pi]*x*Erf[Sqrt[3]*Sqrt[b]*x]*(Cosh[3*a] - Sinh[3*a]) + Sqrt[b]*Sqrt[3 
*Pi]*x*Erfi[Sqrt[3]*Sqrt[b]*x]*Sinh[3*a] + 6*Sinh[a + b*x^2] - 2*Sinh[3*(a 
 + b*x^2)])/(8*x)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5853, 6152, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx\)

\(\Big \downarrow \) 5853

\(\displaystyle 6 b \int \cosh \left (b x^2+a\right ) \sinh ^2\left (b x^2+a\right )dx-\frac {\sinh ^3\left (a+b x^2\right )}{x}\)

\(\Big \downarrow \) 6152

\(\displaystyle 6 b \int \left (\frac {1}{4} \cosh \left (3 b x^2+3 a\right )-\frac {1}{4} \cosh \left (b x^2+a\right )\right )dx-\frac {\sinh ^3\left (a+b x^2\right )}{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 6 b \left (-\frac {\sqrt {\pi } e^{-a} \text {erf}\left (\sqrt {b} x\right )}{16 \sqrt {b}}+\frac {\sqrt {\frac {\pi }{3}} e^{-3 a} \text {erf}\left (\sqrt {3} \sqrt {b} x\right )}{16 \sqrt {b}}-\frac {\sqrt {\pi } e^a \text {erfi}\left (\sqrt {b} x\right )}{16 \sqrt {b}}+\frac {\sqrt {\frac {\pi }{3}} e^{3 a} \text {erfi}\left (\sqrt {3} \sqrt {b} x\right )}{16 \sqrt {b}}\right )-\frac {\sinh ^3\left (a+b x^2\right )}{x}\)

Input:

Int[Sinh[a + b*x^2]^3/x^2,x]
 

Output:

6*b*(-1/16*(Sqrt[Pi]*Erf[Sqrt[b]*x])/(Sqrt[b]*E^a) + (Sqrt[Pi/3]*Erf[Sqrt[ 
3]*Sqrt[b]*x])/(16*Sqrt[b]*E^(3*a)) - (E^a*Sqrt[Pi]*Erfi[Sqrt[b]*x])/(16*S 
qrt[b]) + (E^(3*a)*Sqrt[Pi/3]*Erfi[Sqrt[3]*Sqrt[b]*x])/(16*Sqrt[b])) - Sin 
h[a + b*x^2]^3/x
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5853

Int[(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_)]^(p_), x_Symbol] :> Simp[-Sinh[ 
a + b*x^n]^p/((n - 1)*x^(n - 1)), x] + Simp[b*n*(p/(n - 1))   Int[Sinh[a + 
b*x^n]^(p - 1)*Cosh[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IntegersQ[n, 
p] && EqQ[m + n, 0] && GtQ[p, 1] && NeQ[n, 1]
 

rule 6152

Int[Cosh[w_]^(q_.)*Sinh[v_]^(p_.), x_Symbol] :> Int[ExpandTrigReduce[Sinh[v 
]^p*Cosh[w]^q, x], x] /; IGtQ[p, 0] && IGtQ[q, 0] && ((PolynomialQ[v, x] && 
 PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/w], x 
]))
 

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10

method	result	size
risch	\(\frac {{\mathrm e}^{-3 a} {\mathrm e}^{-3 b \,x^{2}}}{8 x}+\frac {{\mathrm e}^{-3 a} \sqrt {b}\, \sqrt {\pi }\, \sqrt {3}\, \operatorname {erf}\left (\sqrt {3}\, \sqrt {b}\, x \right )}{8}-\frac {3 \,{\mathrm e}^{-a} {\mathrm e}^{-b \,x^{2}}}{8 x}-\frac {3 \sqrt {b}\, \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {b}\, x \right ) {\mathrm e}^{-a}}{8}+\frac {3 \,{\mathrm e}^{a} {\mathrm e}^{b \,x^{2}}}{8 x}-\frac {3 \,{\mathrm e}^{a} b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b}\, x \right )}{8 \sqrt {-b}}-\frac {{\mathrm e}^{3 a} {\mathrm e}^{3 b \,x^{2}}}{8 x}+\frac {3 \,{\mathrm e}^{3 a} b \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-3 b}\, x \right )}{8 \sqrt {-3 b}}\)	\(149\)

Input:

int(sinh(b*x^2+a)^3/x^2,x,method=_RETURNVERBOSE)
 

Output:

1/8/exp(a)^3/x*exp(-3*b*x^2)+1/8/exp(a)^3*b^(1/2)*Pi^(1/2)*3^(1/2)*erf(3^( 
1/2)*b^(1/2)*x)-3/8/exp(a)/x*exp(-b*x^2)-3/8*b^(1/2)*Pi^(1/2)*erf(b^(1/2)* 
x)/exp(a)+3/8*exp(a)*exp(b*x^2)/x-3/8*exp(a)*b*Pi^(1/2)/(-b)^(1/2)*erf((-b 
)^(1/2)*x)-1/8*exp(a)^3/x*exp(3*b*x^2)+3/8*exp(a)^3*b*Pi^(1/2)/(-3*b)^(1/2 
)*erf((-3*b)^(1/2)*x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 892 vs. \(2 (98) = 196\).

Time = 0.11 (sec) , antiderivative size = 892, normalized size of antiderivative = 6.56 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=\text {Too large to display} \] Input:

integrate(sinh(b*x^2+a)^3/x^2,x, algorithm="fricas")
 

Output:

-1/8*(cosh(b*x^2 + a)^6 + 6*cosh(b*x^2 + a)*sinh(b*x^2 + a)^5 + sinh(b*x^2 
 + a)^6 + 3*(5*cosh(b*x^2 + a)^2 - 1)*sinh(b*x^2 + a)^4 - 3*cosh(b*x^2 + a 
)^4 + 4*(5*cosh(b*x^2 + a)^3 - 3*cosh(b*x^2 + a))*sinh(b*x^2 + a)^3 + sqrt 
(3)*sqrt(pi)*(x*cosh(b*x^2 + a)^3*cosh(3*a) + x*cosh(b*x^2 + a)^3*sinh(3*a 
) + (x*cosh(3*a) + x*sinh(3*a))*sinh(b*x^2 + a)^3 + 3*(x*cosh(b*x^2 + a)*c 
osh(3*a) + x*cosh(b*x^2 + a)*sinh(3*a))*sinh(b*x^2 + a)^2 + 3*(x*cosh(b*x^ 
2 + a)^2*cosh(3*a) + x*cosh(b*x^2 + a)^2*sinh(3*a))*sinh(b*x^2 + a))*sqrt( 
-b)*erf(sqrt(3)*sqrt(-b)*x) - sqrt(3)*sqrt(pi)*(x*cosh(b*x^2 + a)^3*cosh(3 
*a) - x*cosh(b*x^2 + a)^3*sinh(3*a) + (x*cosh(3*a) - x*sinh(3*a))*sinh(b*x 
^2 + a)^3 + 3*(x*cosh(b*x^2 + a)*cosh(3*a) - x*cosh(b*x^2 + a)*sinh(3*a))* 
sinh(b*x^2 + a)^2 + 3*(x*cosh(b*x^2 + a)^2*cosh(3*a) - x*cosh(b*x^2 + a)^2 
*sinh(3*a))*sinh(b*x^2 + a))*sqrt(b)*erf(sqrt(3)*sqrt(b)*x) - 3*sqrt(pi)*( 
x*cosh(b*x^2 + a)^3*cosh(a) + x*cosh(b*x^2 + a)^3*sinh(a) + (x*cosh(a) + x 
*sinh(a))*sinh(b*x^2 + a)^3 + 3*(x*cosh(b*x^2 + a)*cosh(a) + x*cosh(b*x^2 
+ a)*sinh(a))*sinh(b*x^2 + a)^2 + 3*(x*cosh(b*x^2 + a)^2*cosh(a) + x*cosh( 
b*x^2 + a)^2*sinh(a))*sinh(b*x^2 + a))*sqrt(-b)*erf(sqrt(-b)*x) + 3*sqrt(p 
i)*(x*cosh(b*x^2 + a)^3*cosh(a) - x*cosh(b*x^2 + a)^3*sinh(a) + (x*cosh(a) 
 - x*sinh(a))*sinh(b*x^2 + a)^3 + 3*(x*cosh(b*x^2 + a)*cosh(a) - x*cosh(b* 
x^2 + a)*sinh(a))*sinh(b*x^2 + a)^2 + 3*(x*cosh(b*x^2 + a)^2*cosh(a) - x*c 
osh(b*x^2 + a)^2*sinh(a))*sinh(b*x^2 + a))*sqrt(b)*erf(sqrt(b)*x) + 3*(...
 

Sympy [F]

\[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=\int \frac {\sinh ^{3}{\left (a + b x^{2} \right )}}{x^{2}}\, dx \] Input:

integrate(sinh(b*x**2+a)**3/x**2,x)
 

Output:

Integral(sinh(a + b*x**2)**3/x**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.75 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=\frac {\sqrt {3} \sqrt {b x^{2}} e^{\left (-3 \, a\right )} \Gamma \left (-\frac {1}{2}, 3 \, b x^{2}\right )}{16 \, x} - \frac {\sqrt {3} \sqrt {-b x^{2}} e^{\left (3 \, a\right )} \Gamma \left (-\frac {1}{2}, -3 \, b x^{2}\right )}{16 \, x} - \frac {3 \, \sqrt {b x^{2}} e^{\left (-a\right )} \Gamma \left (-\frac {1}{2}, b x^{2}\right )}{16 \, x} + \frac {3 \, \sqrt {-b x^{2}} e^{a} \Gamma \left (-\frac {1}{2}, -b x^{2}\right )}{16 \, x} \] Input:

integrate(sinh(b*x^2+a)^3/x^2,x, algorithm="maxima")
 

Output:

1/16*sqrt(3)*sqrt(b*x^2)*e^(-3*a)*gamma(-1/2, 3*b*x^2)/x - 1/16*sqrt(3)*sq 
rt(-b*x^2)*e^(3*a)*gamma(-1/2, -3*b*x^2)/x - 3/16*sqrt(b*x^2)*e^(-a)*gamma 
(-1/2, b*x^2)/x + 3/16*sqrt(-b*x^2)*e^a*gamma(-1/2, -b*x^2)/x
 

Giac [F]

\[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=\int { \frac {\sinh \left (b x^{2} + a\right )^{3}}{x^{2}} \,d x } \] Input:

integrate(sinh(b*x^2+a)^3/x^2,x, algorithm="giac")
 

Output:

integrate(sinh(b*x^2 + a)^3/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=\int \frac {{\mathrm {sinh}\left (b\,x^2+a\right )}^3}{x^2} \,d x \] Input:

int(sinh(a + b*x^2)^3/x^2,x)
 

Output:

int(sinh(a + b*x^2)^3/x^2, x)
 

Reduce [F]

\[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x^2} \, dx=\int \frac {\sinh \left (b \,x^{2}+a \right )^{3}}{x^{2}}d x \] Input:

int(sinh(b*x^2+a)^3/x^2,x)
                                                                                    
                                                                                    
 

Output:

int(sinh(a + b*x**2)**3/x**2,x)