\(\int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx\) [44]

Optimal result

Integrand size = 16, antiderivative size = 174 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {4 b \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac {4 b^{5/2} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {4 b^{5/2} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}-\frac {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}}-\frac {8 b^2 \sinh (a+b x)}{15 d^3 \sqrt {c+d x}} \] Output:

-4/15*b*cosh(b*x+a)/d^2/(d*x+c)^(3/2)+4/15*b^(5/2)*exp(-a+b*c/d)*Pi^(1/2)* 
erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(7/2)+4/15*b^(5/2)*exp(a-b*c/d)*Pi^(1 
/2)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))/d^(7/2)-2/5*sinh(b*x+a)/d/(d*x+c)^ 
(5/2)-8/15*b^2*sinh(b*x+a)/d^3/(d*x+c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.97 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx=\frac {2 \left (-b (c+d x) \left (e^{a-\frac {b c}{d}} \left (e^{b \left (\frac {c}{d}+x\right )} (d+2 b (c+d x))+2 d \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {b (c+d x)}{d}\right )\right )+e^{-a-b x} \left (d-2 b (c+d x)+2 d e^{b \left (\frac {c}{d}+x\right )} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {b (c+d x)}{d}\right )\right )\right )-3 d^2 \sinh (a+b x)\right )}{15 d^3 (c+d x)^{5/2}} \] Input:

Integrate[Sinh[a + b*x]/(c + d*x)^(7/2),x]
 

Output:

(2*(-(b*(c + d*x)*(E^(a - (b*c)/d)*(E^(b*(c/d + x))*(d + 2*b*(c + d*x)) + 
2*d*(-((b*(c + d*x))/d))^(3/2)*Gamma[1/2, -((b*(c + d*x))/d)]) + E^(-a - b 
*x)*(d - 2*b*(c + d*x) + 2*d*E^(b*(c/d + x))*((b*(c + d*x))/d)^(3/2)*Gamma 
[1/2, (b*(c + d*x))/d]))) - 3*d^2*Sinh[a + b*x]))/(15*d^3*(c + d*x)^(5/2))
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.80 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.19, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.938, Rules used = {3042, 26, 3778, 3042, 3778, 26, 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.

\(\displaystyle \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \sin (i a+i b x)}{(c+d x)^{7/2}}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\sin (i a+i b x)}{(c+d x)^{7/2}}dx\)

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}+\frac {2 i b \int -\frac {i \sinh (a+b x)}{(c+d x)^{3/2}}dx}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -i \left (\frac {2 i b \left (\frac {2 b \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}}dx}{3 d}-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}+\frac {2 b \int -\frac {i \sin (i a+i b x)}{(c+d x)^{3/2}}dx}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \int \frac {\sin (i a+i b x)}{(c+d x)^{3/2}}dx}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 3778

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \left (\frac {2 i b \int \frac {\cosh (a+b x)}{\sqrt {c+d x}}dx}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \left (\frac {2 i b \int \frac {\sin \left (i a+i b x+\frac {\pi }{2}\right )}{\sqrt {c+d x}}dx}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 3788

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \left (\frac {2 i b \left (\frac {1}{2} i \int -\frac {i e^{a+b x}}{\sqrt {c+d x}}dx-\frac {1}{2} i \int \frac {i e^{-a-b x}}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 26

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \left (\frac {2 i b \left (\frac {1}{2} \int \frac {e^{-a-b x}}{\sqrt {c+d x}}dx+\frac {1}{2} \int \frac {e^{a+b x}}{\sqrt {c+d x}}dx\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 2611

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \left (\frac {2 i b \left (\frac {\int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}+\frac {\int e^{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d}\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 2633

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \left (\frac {2 i b \left (\frac {\int e^{-a-\frac {b (c+d x)}{d}+\frac {b c}{d}}d\sqrt {c+d x}}{d}+\frac {\sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

\(\Big \downarrow \) 2634

\(\displaystyle -i \left (\frac {2 i b \left (-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {2 i b \left (\frac {2 i b \left (\frac {\sqrt {\pi } e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}+\frac {\sqrt {\pi } e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{2 \sqrt {b} \sqrt {d}}\right )}{d}-\frac {2 i \sinh (a+b x)}{d \sqrt {c+d x}}\right )}{3 d}\right )}{5 d}-\frac {2 i \sinh (a+b x)}{5 d (c+d x)^{5/2}}\right )\)

Input:

Int[Sinh[a + b*x]/(c + d*x)^(7/2),x]
 

Output:

(-I)*((((-2*I)/5)*Sinh[a + b*x])/(d*(c + d*x)^(5/2)) + (((2*I)/5)*b*((-2*C 
osh[a + b*x])/(3*d*(c + d*x)^(3/2)) - (((2*I)/3)*b*(((2*I)*b*((E^(-a + (b* 
c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[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*Sqrt[c + d*x])))/d))/d)
 

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]

\[\int \frac {\sinh \left (b x +a \right )}{\left (d x +c \right )^{\frac {7}{2}}}d x\]

Input:

int(sinh(b*x+a)/(d*x+c)^(7/2),x)
 

Output:

int(sinh(b*x+a)/(d*x+c)^(7/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 855 vs. \(2 (132) = 264\).

Time = 0.14 (sec) , antiderivative size = 855, normalized size of antiderivative = 4.91 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx =\text {Too large to display} \] Input:

integrate(sinh(b*x+a)/(d*x+c)^(7/2),x, algorithm="fricas")
 

Output:

1/15*(4*sqrt(pi)*((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3 
)*cosh(b*x + a)*cosh(-(b*c - a*d)/d) - (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3* 
b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + ((b^2*d^3*x^3 
+ 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(-(b*c - a*d)/d) - (b^2*d 
^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*sinh(-(b*c - a*d)/d))* 
sinh(b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/d)) - 4*sqrt(pi)*((b^2*d 
^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*cosh(-(b 
*c - a*d)/d) + (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*c 
osh(b*x + a)*sinh(-(b*c - a*d)/d) + ((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^ 
2*c^2*d*x + b^2*c^3)*cosh(-(b*c - a*d)/d) + (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 
 + 3*b^2*c^2*d*x + b^2*c^3)*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d 
)*erf(sqrt(d*x + c)*sqrt(-b/d)) + (4*b^2*d^2*x^2 + 4*b^2*c^2 - 2*b*c*d - ( 
4*b^2*d^2*x^2 + 4*b^2*c^2 + 2*b*c*d + 3*d^2 + 2*(4*b^2*c*d + b*d^2)*x)*cos 
h(b*x + a)^2 - 2*(4*b^2*d^2*x^2 + 4*b^2*c^2 + 2*b*c*d + 3*d^2 + 2*(4*b^2*c 
*d + b*d^2)*x)*cosh(b*x + a)*sinh(b*x + a) - (4*b^2*d^2*x^2 + 4*b^2*c^2 + 
2*b*c*d + 3*d^2 + 2*(4*b^2*c*d + b*d^2)*x)*sinh(b*x + a)^2 + 3*d^2 + 2*(4* 
b^2*c*d - b*d^2)*x)*sqrt(d*x + c))/((d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + 
 c^3*d^3)*cosh(b*x + a) + (d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)* 
sinh(b*x + a))
 

Sympy [F]

\[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\sinh {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(sinh(b*x+a)/(d*x+c)**(7/2),x)
 

Output:

Integral(sinh(a + b*x)/(c + d*x)**(7/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.66 \[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx=-\frac {\frac {{\left (\frac {\left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (-a + \frac {b c}{d}\right )} \Gamma \left (-\frac {3}{2}, \frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} + \frac {\left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (a - \frac {b c}{d}\right )} \Gamma \left (-\frac {3}{2}, -\frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )} b}{d} + \frac {2 \, \sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {5}{2}}}}{5 \, d} \] Input:

integrate(sinh(b*x+a)/(d*x+c)^(7/2),x, algorithm="maxima")
 

Output:

-1/5*((((d*x + c)*b/d)^(3/2)*e^(-a + b*c/d)*gamma(-3/2, (d*x + c)*b/d)/(d* 
x + c)^(3/2) + (-(d*x + c)*b/d)^(3/2)*e^(a - b*c/d)*gamma(-3/2, -(d*x + c) 
*b/d)/(d*x + c)^(3/2))*b/d + 2*sinh(b*x + a)/(d*x + c)^(5/2))/d
 

Giac [F]

\[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx=\int { \frac {\sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate(sinh(b*x+a)/(d*x+c)^(7/2),x, algorithm="giac")
 

Output:

integrate(sinh(b*x + a)/(d*x + c)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\mathrm {sinh}\left (a+b\,x\right )}{{\left (c+d\,x\right )}^{7/2}} \,d x \] Input:

int(sinh(a + b*x)/(c + d*x)^(7/2),x)
 

Output:

int(sinh(a + b*x)/(c + d*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx=\int \frac {\sinh \left (b x +a \right )}{\sqrt {d x +c}\, c^{3}+3 \sqrt {d x +c}\, c^{2} d x +3 \sqrt {d x +c}\, c \,d^{2} x^{2}+\sqrt {d x +c}\, d^{3} x^{3}}d x \] Input:

int(sinh(b*x+a)/(d*x+c)^(7/2),x)
 

Output:

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