\(\int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx\) [42]

Optimal result

Integrand size = 16, antiderivative size = 118 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {b} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {\sqrt {b} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sinh (a+b x)}{d \sqrt {c+d x}} \] Output:

b^(1/2)*exp(-a+b*c/d)*Pi^(1/2)*erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(3/2)+ 
b^(1/2)*exp(a-b*c/d)*Pi^(1/2)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(3/2)- 
2*sinh(b*x+a)/d/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {e^{-a-\frac {b c}{d}} \left (e^{2 a} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {b (c+d x)}{d}\right )-e^{\frac {2 b c}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {b (c+d x)}{d}\right )-2 e^{a+\frac {b c}{d}} \sinh (a+b x)\right )}{d \sqrt {c+d x}} \] Input:

Integrate[Sinh[a + b*x]/(c + d*x)^(3/2),x]
 

Output:

(E^(-a - (b*c)/d)*(E^(2*a)*Sqrt[-((b*(c + d*x))/d)]*Gamma[1/2, -((b*(c + d 
*x))/d)] - E^((2*b*c)/d)*Sqrt[(b*(c + d*x))/d]*Gamma[1/2, (b*(c + d*x))/d] 
 - 2*E^(a + (b*c)/d)*Sinh[a + b*x]))/(d*Sqrt[c + d*x])
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 26, 3778, 3042, 3788, 26, 2611, 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.

\(\displaystyle \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \sin (i a+i b x)}{(c+d x)^{3/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\sin (i a+i b x)}{(c+d x)^{3/2}}dx\)

\(\Big \downarrow \) 3778

\(\displaystyle -i \left (\frac {2 i b \int \frac {\cosh (a+b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (\frac {2 i b \int \frac {\sin \left (i a+i b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )\)

\(\Big \downarrow \) 3788

\(\displaystyle -i \left (\frac {2 i b \left (\frac {1}{2} i \int -\frac {i e^{a+b x}}{\sqrt {c+d x}}dx-\frac {1}{2} i \int \frac {i e^{-a-b x}}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -i \left (\frac {2 i b \left (\frac {1}{2} \int \frac {e^{-a-b x}}{\sqrt {c+d x}}dx+\frac {1}{2} \int \frac {e^{a+b x}}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )\)

\(\Big \downarrow \) 2611

\(\displaystyle -i \left (\frac {2 i b \left (\frac {\int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}+\frac {\int e^{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )\)

\(\Big \downarrow \) 2633

\(\displaystyle -i \left (\frac {2 i b \left (\frac {\int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}+\frac {\sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )\)

\(\Big \downarrow \) 2634

\(\displaystyle -i \left (\frac {2 i b \left (\frac {\sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}+\frac {\sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )\)

Input:

Int[Sinh[a + b*x]/(c + d*x)^(3/2),x]
 

Output:

(-I)*(((2*I)*b*((E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqr 
t[d]])/(2*Sqrt[b]*Sqrt[d]) + (E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[ 
c + d*x])/Sqrt[d]])/(2*Sqrt[b]*Sqrt[d])))/d - ((2*I)*Sinh[a + b*x])/(d*Sqr 
t[c + d*x]))
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2611

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] : 
> Simp[2/d   Subst[Int[F^(g*(e - c*(f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d 
*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3778

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

rule 3788

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol 
] :> Simp[I/2   Int[(c + d*x)^m/(E^(I*k*Pi)*E^(I*(e + f*x))), x], x] - Simp 
[I/2   Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e 
, f, m}, x] && IntegerQ[2*k]
 

Maple [F]

\[\int \frac {\sinh \left (b x +a \right )}{\left (d x +c \right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (90) = 180\).

Time = 0.10 (sec) , antiderivative size = 339, normalized size of antiderivative = 2.87 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {\sqrt {\pi } {\left ({\left (d x + c\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (d x + c\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (d x + c\right )} \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (d x + c\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) - \sqrt {\pi } {\left ({\left (d x + c\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (d x + c\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (d x + c\right )} \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (d x + c\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) - \sqrt {d x + c} {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}}{{\left (d^{2} x + c d\right )} \cosh \left (b x + a\right ) + {\left (d^{2} x + c d\right )} \sinh \left (b x + a\right )} \] Input:

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

Output:

(sqrt(pi)*((d*x + c)*cosh(b*x + a)*cosh(-(b*c - a*d)/d) - (d*x + c)*cosh(b 
*x + a)*sinh(-(b*c - a*d)/d) + ((d*x + c)*cosh(-(b*c - a*d)/d) - (d*x + c) 
*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/d 
)) - sqrt(pi)*((d*x + c)*cosh(b*x + a)*cosh(-(b*c - a*d)/d) + (d*x + c)*co 
sh(b*x + a)*sinh(-(b*c - a*d)/d) + ((d*x + c)*cosh(-(b*c - a*d)/d) + (d*x 
+ c)*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(d*x + c)*sqr 
t(-b/d)) - sqrt(d*x + c)*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) 
+ sinh(b*x + a)^2 - 1))/((d^2*x + c*d)*cosh(b*x + a) + (d^2*x + c*d)*sinh( 
b*x + a))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx=\frac {\frac {{\left (\frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (a - \frac {b c}{d}\right )}}{\sqrt {-\frac {b}{d}}} + \frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-a + \frac {b c}{d}\right )}}{\sqrt {\frac {b}{d}}}\right )} b}{d} - \frac {2 \, \sinh \left (b x + a\right )}{\sqrt {d x + c}}}{d} \] Input:

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

Output:

((sqrt(pi)*erf(sqrt(d*x + c)*sqrt(-b/d))*e^(a - b*c/d)/sqrt(-b/d) + sqrt(p 
i)*erf(sqrt(d*x + c)*sqrt(b/d))*e^(-a + b*c/d)/sqrt(b/d))*b/d - 2*sinh(b*x 
 + a)/sqrt(d*x + c))/d
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sinh(a + b*x)/(sqrt(c + d*x)*c + sqrt(c + d*x)*d*x),x)