3.1.43 \(\int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx\) [43]

3.1.43.1 Optimal result

Integrand size = 16, antiderivative size = 149 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {4 b \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 b^{3/2} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {2 b^{3/2} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {2 \sinh (a+b x)}{3 d (c+d x)^{3/2}} \]

output

-2/3*sinh(b*x+a)/d/(d*x+c)^(3/2)-2/3*b^(3/2)*exp(-a+b*c/d)*erf(b^(1/2)*(d* 
x+c)^(1/2)/d^(1/2))*Pi^(1/2)/d^(5/2)+2/3*b^(3/2)*exp(a-b*c/d)*erfi(b^(1/2) 
*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/d^(5/2)-4/3*b*cosh(b*x+a)/d^2/(d*x+c)^(1/ 
2)
 

3.1.43.2 Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.08 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx=\frac {2 b \left (\frac {e^a \left (-e^{b x}+e^{-\frac {b c}{d}} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {b (c+d x)}{d}\right )\right )}{d \sqrt {c+d x}}+\frac {e^{-a-b x} \left (-1+e^{b \left (\frac {c}{d}+x\right )} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {b (c+d x)}{d}\right )\right )}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 \sinh (a+b x)}{3 d (c+d x)^{3/2}} \]

input

Integrate[Sinh[a + b*x]/(c + d*x)^(5/2),x]
 

output

(2*b*((E^a*(-E^(b*x) + (Sqrt[-((b*(c + d*x))/d)]*Gamma[1/2, -((b*(c + d*x) 
)/d)])/E^((b*c)/d)))/(d*Sqrt[c + d*x]) + (E^(-a - b*x)*(-1 + E^(b*(c/d + x 
))*Sqrt[(b*(c + d*x))/d]*Gamma[1/2, (b*(c + d*x))/d]))/(d*Sqrt[c + d*x]))) 
/(3*d) - (2*Sinh[a + b*x])/(3*d*(c + d*x)^(3/2))
 

3.1.43.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.63 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.18, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 26, 3778, 3042, 3778, 26, 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.

\(\displaystyle \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \sin (i a+i b x)}{(c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\sin (i a+i b x)}{(c+d x)^{5/2}}dx\)

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{d \sqrt {c+d x}}+\frac {2 i b \int -\frac {i \sinh (a+b x)}{\sqrt {c+d x}}dx}{d}\right )}{3 d}-\frac {2 i \sinh (a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -i \left (\frac {2 i b \left (\frac {2 b \int \frac {\sinh (a+b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 \cosh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}-\frac {2 i \sinh (a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3789

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{d \sqrt {c+d x}}-\frac {2 i b \left (\frac {1}{2} i \int \frac {e^{a+b x}}{\sqrt {c+d x}}dx-\frac {1}{2} i \int \frac {e^{-a-b x}}{\sqrt {c+d x}}dx\right )}{d}\right )}{3 d}-\frac {2 i \sinh (a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 2611

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{d \sqrt {c+d x}}-\frac {2 i b \left (\frac {i \int e^{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d}-\frac {i \int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{d}\right )}{3 d}-\frac {2 i \sinh (a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 2633

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{d \sqrt {c+d x}}-\frac {2 i b \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{d}\right )}{3 d}-\frac {2 i \sinh (a+b x)}{3 d (c+d x)^{3/2}}\right )\)

\(\Big \downarrow \) 2634

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{d \sqrt {c+d x}}-\frac {2 i b \left (\frac {i \sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}-\frac {i \sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d}\right )}{3 d}-\frac {2 i \sinh (a+b x)}{3 d (c+d x)^{3/2}}\right )\)

input

Int[Sinh[a + b*x]/(c + d*x)^(5/2),x]
 

output

(-I)*((((2*I)/3)*b*((-2*Cosh[a + b*x])/(d*Sqrt[c + d*x]) - ((2*I)*b*(((-1/ 
2*I)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(Sqrt 
[b]*Sqrt[d]) + ((I/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x] 
)/Sqrt[d]])/(Sqrt[b]*Sqrt[d])))/d))/d - (((2*I)/3)*Sinh[a + b*x])/(d*(c + 
d*x)^(3/2)))
 

3.1.43.3.1 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 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]
 

3.1.43.4 Maple [F]

\[\int \frac {\sinh \left (b x +a \right )}{\left (d x +c \right )^{\frac {5}{2}}}d x\]

input

int(sinh(b*x+a)/(d*x+c)^(5/2),x)
 

output

int(sinh(b*x+a)/(d*x+c)^(5/2),x)
 

3.1.43.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (111) = 222\).

Time = 0.27 (sec) , antiderivative size = 532, normalized size of antiderivative = 3.57 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {2 \, \sqrt {\pi } {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) + 2 \, \sqrt {\pi } {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) + {\left (2 \, b d x + {\left (2 \, b d x + 2 \, b c + d\right )} \cosh \left (b x + a\right )^{2} + 2 \, {\left (2 \, b d x + 2 \, b c + d\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (2 \, b d x + 2 \, b c + d\right )} \sinh \left (b x + a\right )^{2} + 2 \, b c - d\right )} \sqrt {d x + c}}{3 \, {\left ({\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \cosh \left (b x + a\right ) + {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \sinh \left (b x + a\right )\right )}} \]

input

integrate(sinh(b*x+a)/(d*x+c)^(5/2),x, algorithm="fricas")
 

output

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

3.1.43.6 Sympy [F]

\[ \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\sinh {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \]

input

integrate(sinh(b*x+a)/(d*x+c)**(5/2),x)
 

output

Integral(sinh(a + b*x)/(c + d*x)**(5/2), x)
 

3.1.43.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.77 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx=-\frac {\frac {{\left (\frac {\sqrt {\frac {{\left (d x + c\right )} b}{d}} e^{\left (-a + \frac {b c}{d}\right )} \Gamma \left (-\frac {1}{2}, \frac {{\left (d x + c\right )} b}{d}\right )}{\sqrt {d x + c}} + \frac {\sqrt {-\frac {{\left (d x + c\right )} b}{d}} e^{\left (a - \frac {b c}{d}\right )} \Gamma \left (-\frac {1}{2}, -\frac {{\left (d x + c\right )} b}{d}\right )}{\sqrt {d x + c}}\right )} b}{d} + \frac {2 \, \sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {3}{2}}}}{3 \, d} \]

input

integrate(sinh(b*x+a)/(d*x+c)^(5/2),x, algorithm="maxima")
 

output

-1/3*((sqrt((d*x + c)*b/d)*e^(-a + b*c/d)*gamma(-1/2, (d*x + c)*b/d)/sqrt( 
d*x + c) + sqrt(-(d*x + c)*b/d)*e^(a - b*c/d)*gamma(-1/2, -(d*x + c)*b/d)/ 
sqrt(d*x + c))*b/d + 2*sinh(b*x + a)/(d*x + c)^(3/2))/d
 

3.1.43.8 Giac [F]

\[ \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx=\int { \frac {\sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(sinh(b*x+a)/(d*x+c)^(5/2),x, algorithm="giac")
 

output

integrate(sinh(b*x + a)/(d*x + c)^(5/2), x)
 

3.1.43.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx=\int \frac {\mathrm {sinh}\left (a+b\,x\right )}{{\left (c+d\,x\right )}^{5/2}} \,d x \]

input

int(sinh(a + b*x)/(c + d*x)^(5/2),x)
 

output

int(sinh(a + b*x)/(c + d*x)^(5/2), x)