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

Optimal. Leaf size=149 \[ -\frac{2 \sqrt{\pi } b^{3/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{2 \sqrt{\pi } b^{3/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 b \cosh (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sinh (a+b x)}{3 d (c+d x)^{3/2}} \]

[Out]

(-4*b*Cosh[a + b*x])/(3*d^2*Sqrt[c + d*x]) - (2*b^(3/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/
Sqrt[d]])/(3*d^(5/2)) + (2*b^(3/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(3*d^(5/2))
- (2*Sinh[a + b*x])/(3*d*(c + d*x)^(3/2))

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Rubi [A]  time = 0.250753, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3297, 3308, 2180, 2204, 2205} \[ -\frac{2 \sqrt{\pi } b^{3/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}+\frac{2 \sqrt{\pi } b^{3/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{3 d^{5/2}}-\frac{4 b \cosh (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sinh (a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]/(c + d*x)^(5/2),x]

[Out]

(-4*b*Cosh[a + b*x])/(3*d^2*Sqrt[c + d*x]) - (2*b^(3/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/
Sqrt[d]])/(3*d^(5/2)) + (2*b^(3/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(3*d^(5/2))
- (2*Sinh[a + b*x])/(3*d*(c + d*x)^(3/2))

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 3308

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

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)^{5/2}} \, dx &=-\frac{2 \sinh (a+b x)}{3 d (c+d x)^{3/2}}+\frac{(2 b) \int \frac{\cosh (a+b x)}{(c+d x)^{3/2}} \, dx}{3 d}\\ &=-\frac{4 b \cosh (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sinh (a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (4 b^2\right ) \int \frac{\sinh (a+b x)}{\sqrt{c+d x}} \, dx}{3 d^2}\\ &=-\frac{4 b \cosh (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sinh (a+b x)}{3 d (c+d x)^{3/2}}+\frac{\left (2 b^2\right ) \int \frac{e^{-i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{3 d^2}-\frac{\left (2 b^2\right ) \int \frac{e^{i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{3 d^2}\\ &=-\frac{4 b \cosh (a+b x)}{3 d^2 \sqrt{c+d x}}-\frac{2 \sinh (a+b x)}{3 d (c+d x)^{3/2}}-\frac{\left (4 b^2\right ) \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 )}{3 d^3}+\frac{\left (4 b^2\right ) \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 )}{3 d^3}\\ &=-\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}}\\ \end{align*}

Mathematica [A]  time = 0.754084, size = 161, normalized size = 1.08 \[ \frac{2 b \left (\frac{e^a \left (e^{-\frac{b c}{d}} \sqrt{-\frac{b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},-\frac{b (c+d x)}{d}\right )-e^{b x}\right )}{d \sqrt{c+d x}}+\frac{e^{-a-b x} \left (e^{b \left (\frac{c}{d}+x\right )} \sqrt{\frac{b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},\frac{b (c+d x)}{d}\right )-1\right )}{d \sqrt{c+d x}}\right )}{3 d}-\frac{2 \sinh (a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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))

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{\sinh \left ( bx+a \right ) \left ( dx+c \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.33558, size = 154, normalized size = 1.03 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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Fricas [B]  time = 2.7866, size = 1226, normalized size = 8.23 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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(sqrt(d*x + c)*sqrt(b/d)) + 2*sqrt(p
i)*((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)*sin
h(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))

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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