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

Optimal. Leaf size=118 \[ \frac{\sqrt{\pi } \sqrt{b} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{\sqrt{\pi } \sqrt{b} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{2 \sinh (a+b x)}{d \sqrt{c+d x}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.200403, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3297, 3307, 2180, 2204, 2205} \[ \frac{\sqrt{\pi } \sqrt{b} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{\sqrt{\pi } \sqrt{b} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{2 \sinh (a+b x)}{d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]/(c + d*x)^(3/2),x]

[Out]

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

Rule 3297

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] - Dist[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 3307

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

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[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] &&  !$UseGamma === True

Rule 2204

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 2205

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] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{\sinh (a+b x)}{(c+d x)^{3/2}} \, dx &=-\frac{2 \sinh (a+b x)}{d \sqrt{c+d x}}+\frac{(2 b) \int \frac{\cosh (a+b x)}{\sqrt{c+d x}} \, dx}{d}\\ &=-\frac{2 \sinh (a+b x)}{d \sqrt{c+d x}}+\frac{b \int \frac{e^{-i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{d}+\frac{b \int \frac{e^{i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{d}\\ &=-\frac{2 \sinh (a+b x)}{d \sqrt{c+d x}}+\frac{(2 b) \operatorname{Subst}\left (\int e^{i \left (i a-\frac{i b c}{d}\right )-\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^2}+\frac{(2 b) \operatorname{Subst}\left (\int e^{-i \left (i a-\frac{i b c}{d}\right )+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{d^2}\\ &=\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}}\\ \end{align*}

Mathematica [A]  time = 0.148855, size = 120, normalized size = 1.02 \[ \frac{e^{-a-\frac{b c}{d}} \left (e^{2 a} \sqrt{-\frac{b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},-\frac{b (c+d x)}{d}\right )-e^{\frac{2 b c}{d}} \sqrt{\frac{b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},\frac{b (c+d x)}{d}\right )-2 e^{a+\frac{b c}{d}} \sinh (a+b x)\right )}{d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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])

________________________________________________________________________________________

Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int{\sinh \left ( bx+a \right ) \left ( dx+c \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.11129, size = 139, normalized size = 1.18 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.76847, size = 814, normalized size = 6.9 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)*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(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (a + b x \right )}}{\left (c + d x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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