\(\int \frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}} \, dx\) [49]

Optimal result

Integrand size = 18, antiderivative size = 142 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {\sqrt {b} e^{-2 a+\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}+\frac {\sqrt {b} e^{2 a-\frac {2 b c}{d}} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.07 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}} \, dx=\frac {e^{-2 \left (a+b \left (\frac {c}{d}+x\right )\right )} \left (\sqrt {2} e^{4 a+2 b x} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )+e^{\frac {2 b c}{d}} \left (-\left (-1+e^{2 (a+b x)}\right )^2+\sqrt {2} e^{\frac {2 b (c+d x)}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )\right )\right )}{2 d \sqrt {c+d x}} \] Input:

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

Output:

(Sqrt[2]*E^(4*a + 2*b*x)*Sqrt[-((b*(c + d*x))/d)]*Gamma[1/2, (-2*b*(c + d* 
x))/d] + E^((2*b*c)/d)*(-(-1 + E^(2*(a + b*x)))^2 + Sqrt[2]*E^((2*b*(c + d 
*x))/d)*Sqrt[(b*(c + d*x))/d]*Gamma[1/2, (2*b*(c + d*x))/d]))/(2*d*E^(2*(a 
 + b*(c/d + x)))*Sqrt[c + d*x])
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.59 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 25, 3794, 27, 3042, 26, 3789, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3794

\(\displaystyle -\frac {2 \sinh ^2(a+b x)}{d \sqrt {c+d x}}-\frac {4 i b \int \frac {i \sinh (2 a+2 b x)}{2 \sqrt {c+d x}}dx}{d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3789

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

\(\Big \downarrow \) 2611

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

\(\Big \downarrow \) 2633

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

\(\Big \downarrow \) 2634

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

Input:

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

Output:

((-2*I)*b*(((-1/2*I)*E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[b]* 
Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d]) + ((I/2)*E^(2*a - (2*b*c)/d)*Sq 
rt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(Sqrt[b]*Sqrt[d])) 
)/d - (2*Sinh[a + b*x]^2)/(d*Sqrt[c + d*x])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 3789

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

rule 3794

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 
)))   Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n 
- 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & 
& LtQ[m, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (109) = 218\).

Time = 0.10 (sec) , antiderivative size = 571, normalized size of antiderivative = 4.02 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(sqrt(2)*sqrt(pi)*((d*x + c)*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) - 
 (d*x + c)*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d) + ((d*x + c)*cosh(-2*(b* 
c - a*d)/d) - (d*x + c)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((d*x 
+ c)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) - (d*x + c)*cosh(b*x + a)*sinh(- 
2*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(2)*sqrt(d*x + c)*sqrt( 
b/d)) + sqrt(2)*sqrt(pi)*((d*x + c)*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) 
 + (d*x + c)*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d) + ((d*x + c)*cosh(-2*( 
b*c - a*d)/d) + (d*x + c)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((d* 
x + c)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) + (d*x + c)*cosh(b*x + a)*sinh 
(-2*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(2)*sqrt(d*x + c)*sq 
rt(-b/d)) + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x 
+ a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4 
*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*sqrt(d*x + c))/((d^2 
*x + c*d)*cosh(b*x + a)^2 + 2*(d^2*x + c*d)*cosh(b*x + a)*sinh(b*x + a) + 
(d^2*x + c*d)*sinh(b*x + a)^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.82 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{3/2}} \, dx=-\frac {\frac {\sqrt {2} \sqrt {\frac {{\left (d x + c\right )} b}{d}} e^{\left (\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {1}{2}, \frac {2 \, {\left (d x + c\right )} b}{d}\right )}{\sqrt {d x + c}} + \frac {\sqrt {2} \sqrt {-\frac {{\left (d x + c\right )} b}{d}} e^{\left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {1}{2}, -\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{\sqrt {d x + c}} - \frac {4}{\sqrt {d x + c}}}{4 \, d} \] Input:

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

Output:

-1/4*(sqrt(2)*sqrt((d*x + c)*b/d)*e^(2*(b*c - a*d)/d)*gamma(-1/2, 2*(d*x + 
 c)*b/d)/sqrt(d*x + c) + sqrt(2)*sqrt(-(d*x + c)*b/d)*e^(-2*(b*c - a*d)/d) 
*gamma(-1/2, -2*(d*x + c)*b/d)/sqrt(d*x + c) - 4/sqrt(d*x + c))/d
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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