\(\int \sinh ((a+b x)^2) \, dx\) [90]

Optimal result

Integrand size = 8, antiderivative size = 37 \[ \int \sinh \left ((a+b x)^2\right ) \, dx=-\frac {\sqrt {\pi } \text {erf}(a+b x)}{4 b}+\frac {\sqrt {\pi } \text {erfi}(a+b x)}{4 b} \] Output:

-1/4*Pi^(1/2)*erf(b*x+a)/b+1/4*Pi^(1/2)*erfi(b*x+a)/b
 

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \sinh \left ((a+b x)^2\right ) \, dx=\frac {\sqrt {\pi } (-\text {erf}(a+b x)+\text {erfi}(a+b x))}{4 b} \] Input:

Integrate[Sinh[(a + b*x)^2],x]
 

Output:

(Sqrt[Pi]*(-Erf[a + b*x] + Erfi[a + b*x]))/(4*b)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5833, 5821, 2633, 2634}

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.

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

\(\Big \downarrow \) 5833

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

\(\Big \downarrow \) 5821

\(\displaystyle \frac {\frac {1}{2} \int e^{(a+b x)^2}d(a+b x)-\frac {1}{2} \int e^{-(a+b x)^2}d(a+b x)}{b}\)

\(\Big \downarrow \) 2633

\(\displaystyle \frac {\frac {1}{4} \sqrt {\pi } \text {erfi}(a+b x)-\frac {1}{2} \int e^{-(a+b x)^2}d(a+b x)}{b}\)

\(\Big \downarrow \) 2634

\(\displaystyle \frac {\frac {1}{4} \sqrt {\pi } \text {erfi}(a+b x)-\frac {1}{4} \sqrt {\pi } \text {erf}(a+b x)}{b}\)

Input:

Int[Sinh[(a + b*x)^2],x]
 

Output:

(-1/4*(Sqrt[Pi]*Erf[a + b*x]) + (Sqrt[Pi]*Erfi[a + b*x])/4)/b
 

Defintions of rubi rules used

rule 2633

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{ 
F, a, b, c, d}, x] && PosQ[b]
 

rule 2634

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr 
eeQ[{F, a, b, c, d}, x] && NegQ[b]
 

rule 5821

Int[Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[1/2   Int[E^(c + d*x^n 
), x], x] - Simp[1/2   Int[E^(-c - d*x^n), x], x] /; FreeQ[{c, d}, x] && IG 
tQ[n, 1]
 

rule 5833

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(u_)^(n_)])^(p_.), x_Symbol] :> Simp[ 
1/Coefficient[u, x, 1]   Subst[Int[(a + b*Sinh[c + d*x^n])^p, x], x, u], x] 
 /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[p] && LinearQ[u, x] && NeQ[u, x]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {\sqrt {\pi }\, \operatorname {erf}\left (b x +a \right )}{4 b}-\frac {i \sqrt {\pi }\, \operatorname {erf}\left (i b x +i a \right )}{4 b}\)	\(36\)

Input:

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

Output:

-1/4*Pi^(1/2)*erf(b*x+a)/b-1/4*I*Pi^(1/2)/b*erf(I*b*x+I*a)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.57 \[ \int \sinh \left ((a+b x)^2\right ) \, dx=-\frac {\sqrt {\pi } \sqrt {-b^{2}} \operatorname {erf}\left (\frac {\sqrt {-b^{2}} {\left (b x + a\right )}}{b}\right ) + \sqrt {\pi } \sqrt {b^{2}} \operatorname {erf}\left (\frac {\sqrt {b^{2}} {\left (b x + a\right )}}{b}\right )}{4 \, b^{2}} \] Input:

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

Output:

-1/4*(sqrt(pi)*sqrt(-b^2)*erf(sqrt(-b^2)*(b*x + a)/b) + sqrt(pi)*sqrt(b^2) 
*erf(sqrt(b^2)*(b*x + a)/b))/b^2
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 477 vs. \(2 (29) = 58\).

Time = 0.28 (sec) , antiderivative size = 477, normalized size of antiderivative = 12.89 \[ \int \sinh \left ((a+b x)^2\right ) \, dx=\frac {1}{2} \, {\left (\frac {{\left (\frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a b^{2} {\left (\operatorname {erf}\left (\frac {\sqrt {{\left (b^{2} x + a b\right )}^{2}}}{b}\right ) - 1\right )}}{\sqrt {{\left (b^{2} x + a b\right )}^{2}} \left (-b^{2}\right )^{\frac {3}{2}}} + \frac {b^{2} e^{\left (-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{\left (-b^{2}\right )^{\frac {3}{2}}}\right )} a}{\sqrt {-b^{2}}} + \frac {{\left (\frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a^{2} b^{3} {\left (\operatorname {erf}\left (\frac {\sqrt {{\left (b^{2} x + a b\right )}^{2}}}{b}\right ) - 1\right )}}{\sqrt {{\left (b^{2} x + a b\right )}^{2}} \left (-b^{2}\right )^{\frac {5}{2}}} - \frac {{\left (b^{2} x + a b\right )}^{3} b^{3} \Gamma \left (\frac {3}{2}, \frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}{{\left ({\left (b^{2} x + a b\right )}^{2}\right )}^{\frac {3}{2}} \left (-b^{2}\right )^{\frac {5}{2}}} + \frac {2 \, a b^{3} e^{\left (-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{\left (-b^{2}\right )^{\frac {5}{2}}}\right )} b}{\sqrt {-b^{2}}} + \frac {a {\left (\frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}\right ) - 1\right )}}{b^{2} \sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}} - \frac {e^{\left (\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{b}\right )}}{b} - \frac {\sqrt {\pi } {\left (b^{2} x + a b\right )} a^{2} {\left (\operatorname {erf}\left (\sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}\right ) - 1\right )}}{b^{3} \sqrt {-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}}} + \frac {2 \, a e^{\left (\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}}{b^{2}} + \frac {{\left (b^{2} x + a b\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )}{b^{5} \left (-\frac {{\left (b^{2} x + a b\right )}^{2}}{b^{2}}\right )^{\frac {3}{2}}}\right )} b + x \sinh \left ({\left (b x + a\right )}^{2}\right ) \] Input:

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

Output:

1/2*((sqrt(pi)*(b^2*x + a*b)*a*b^2*(erf(sqrt((b^2*x + a*b)^2)/b) - 1)/(sqr 
t((b^2*x + a*b)^2)*(-b^2)^(3/2)) + b^2*e^(-(b^2*x + a*b)^2/b^2)/(-b^2)^(3/ 
2))*a/sqrt(-b^2) + (sqrt(pi)*(b^2*x + a*b)*a^2*b^3*(erf(sqrt((b^2*x + a*b) 
^2)/b) - 1)/(sqrt((b^2*x + a*b)^2)*(-b^2)^(5/2)) - (b^2*x + a*b)^3*b^3*gam 
ma(3/2, (b^2*x + a*b)^2/b^2)/(((b^2*x + a*b)^2)^(3/2)*(-b^2)^(5/2)) + 2*a* 
b^3*e^(-(b^2*x + a*b)^2/b^2)/(-b^2)^(5/2))*b/sqrt(-b^2) + a*(sqrt(pi)*(b^2 
*x + a*b)*a*(erf(sqrt(-(b^2*x + a*b)^2/b^2)) - 1)/(b^2*sqrt(-(b^2*x + a*b) 
^2/b^2)) - e^((b^2*x + a*b)^2/b^2)/b)/b - sqrt(pi)*(b^2*x + a*b)*a^2*(erf( 
sqrt(-(b^2*x + a*b)^2/b^2)) - 1)/(b^3*sqrt(-(b^2*x + a*b)^2/b^2)) + 2*a*e^ 
((b^2*x + a*b)^2/b^2)/b^2 + (b^2*x + a*b)^3*gamma(3/2, -(b^2*x + a*b)^2/b^ 
2)/(b^5*(-(b^2*x + a*b)^2/b^2)^(3/2)))*b + x*sinh((b*x + a)^2)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \sinh \left ((a+b x)^2\right ) \, dx=-\frac {i \, \sqrt {\pi } \operatorname {erf}\left (i \, b {\left (x + \frac {a}{b}\right )}\right )}{4 \, b} + \frac {\sqrt {\pi } \operatorname {erf}\left (-b {\left (x + \frac {a}{b}\right )}\right )}{4 \, b} \] Input:

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

Output:

-1/4*I*sqrt(pi)*erf(I*b*(x + a/b))/b + 1/4*sqrt(pi)*erf(-b*(x + a/b))/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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