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

Optimal. Leaf size=146 \[ -\frac{3 \sqrt{\pi } d^{3/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{5/2}}+\frac{3 \sqrt{\pi } d^{3/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{5/2}}-\frac{3 d \sqrt{c+d x} \sinh (a+b x)}{2 b^2}+\frac{(c+d x)^{3/2} \cosh (a+b x)}{b} \]

[Out]

((c + d*x)^(3/2)*Cosh[a + b*x])/b - (3*d^(3/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])
/(8*b^(5/2)) + (3*d^(3/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(8*b^(5/2)) - (3*d*S
qrt[c + d*x]*Sinh[a + b*x])/(2*b^2)

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Rubi [A]  time = 0.248011, antiderivative size = 146, 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 = {3296, 3308, 2180, 2204, 2205} \[ -\frac{3 \sqrt{\pi } d^{3/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{5/2}}+\frac{3 \sqrt{\pi } d^{3/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{5/2}}-\frac{3 d \sqrt{c+d x} \sinh (a+b x)}{2 b^2}+\frac{(c+d x)^{3/2} \cosh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)*Sinh[a + b*x],x]

[Out]

((c + d*x)^(3/2)*Cosh[a + b*x])/b - (3*d^(3/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])
/(8*b^(5/2)) + (3*d^(3/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(8*b^(5/2)) - (3*d*S
qrt[c + d*x]*Sinh[a + b*x])/(2*b^2)

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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 (c+d x)^{3/2} \sinh (a+b x) \, dx &=\frac{(c+d x)^{3/2} \cosh (a+b x)}{b}-\frac{(3 d) \int \sqrt{c+d x} \cosh (a+b x) \, dx}{2 b}\\ &=\frac{(c+d x)^{3/2} \cosh (a+b x)}{b}-\frac{3 d \sqrt{c+d x} \sinh (a+b x)}{2 b^2}+\frac{\left (3 d^2\right ) \int \frac{\sinh (a+b x)}{\sqrt{c+d x}} \, dx}{4 b^2}\\ &=\frac{(c+d x)^{3/2} \cosh (a+b x)}{b}-\frac{3 d \sqrt{c+d x} \sinh (a+b x)}{2 b^2}+\frac{\left (3 d^2\right ) \int \frac{e^{-i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{8 b^2}-\frac{\left (3 d^2\right ) \int \frac{e^{i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{8 b^2}\\ &=\frac{(c+d x)^{3/2} \cosh (a+b x)}{b}-\frac{3 d \sqrt{c+d x} \sinh (a+b x)}{2 b^2}-\frac{(3 d) \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 )}{4 b^2}+\frac{(3 d) \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 )}{4 b^2}\\ &=\frac{(c+d x)^{3/2} \cosh (a+b x)}{b}-\frac{3 d^{3/2} e^{-a+\frac{b c}{d}} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{5/2}}+\frac{3 d^{3/2} e^{a-\frac{b c}{d}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{8 b^{5/2}}-\frac{3 d \sqrt{c+d x} \sinh (a+b x)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0952787, size = 106, normalized size = 0.73 \[ \frac{d \sqrt{c+d x} e^{-a-\frac{b c}{d}} \left (\frac{e^{\frac{2 b c}{d}} \text{Gamma}\left (\frac{5}{2},\frac{b (c+d x)}{d}\right )}{\sqrt{\frac{b (c+d x)}{d}}}-\frac{e^{2 a} \text{Gamma}\left (\frac{5}{2},-\frac{b (c+d x)}{d}\right )}{\sqrt{-\frac{b (c+d x)}{d}}}\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*E^(-a - (b*c)/d)*Sqrt[c + d*x]*(-((E^(2*a)*Gamma[5/2, -((b*(c + d*x))/d)])/Sqrt[-((b*(c + d*x))/d)]) + (E^(
(2*b*c)/d)*Gamma[5/2, (b*(c + d*x))/d])/Sqrt[(b*(c + d*x))/d]))/(2*b^2)

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Maple [F]  time = 0.026, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{{\frac{3}{2}}}\sinh \left ( bx+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.19344, size = 362, normalized size = 2.48 \begin{align*} \frac{16 \,{\left (d x + c\right )}^{\frac{5}{2}} \sinh \left (b x + a\right ) + \frac{{\left (\frac{15 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (\sqrt{d x + c} \sqrt{-\frac{b}{d}}\right ) e^{\left (a - \frac{b c}{d}\right )}}{b^{3} \sqrt{-\frac{b}{d}}} - \frac{15 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (\sqrt{d x + c} \sqrt{\frac{b}{d}}\right ) e^{\left (-a + \frac{b c}{d}\right )}}{b^{3} \sqrt{\frac{b}{d}}} + \frac{2 \,{\left (4 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{2} d e^{\left (\frac{b c}{d}\right )} + 10 \,{\left (d x + c\right )}^{\frac{3}{2}} b d^{2} e^{\left (\frac{b c}{d}\right )} + 15 \, \sqrt{d x + c} d^{3} e^{\left (\frac{b c}{d}\right )}\right )} e^{\left (-a - \frac{{\left (d x + c\right )} b}{d}\right )}}{b^{3}} - \frac{2 \,{\left (4 \,{\left (d x + c\right )}^{\frac{5}{2}} b^{2} d e^{a} - 10 \,{\left (d x + c\right )}^{\frac{3}{2}} b d^{2} e^{a} + 15 \, \sqrt{d x + c} d^{3} e^{a}\right )} e^{\left (\frac{{\left (d x + c\right )} b}{d} - \frac{b c}{d}\right )}}{b^{3}}\right )} b}{d}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(16*(d*x + c)^(5/2)*sinh(b*x + a) + (15*sqrt(pi)*d^3*erf(sqrt(d*x + c)*sqrt(-b/d))*e^(a - b*c/d)/(b^3*sqr
t(-b/d)) - 15*sqrt(pi)*d^3*erf(sqrt(d*x + c)*sqrt(b/d))*e^(-a + b*c/d)/(b^3*sqrt(b/d)) + 2*(4*(d*x + c)^(5/2)*
b^2*d*e^(b*c/d) + 10*(d*x + c)^(3/2)*b*d^2*e^(b*c/d) + 15*sqrt(d*x + c)*d^3*e^(b*c/d))*e^(-a - (d*x + c)*b/d)/
b^3 - 2*(4*(d*x + c)^(5/2)*b^2*d*e^a - 10*(d*x + c)^(3/2)*b*d^2*e^a + 15*sqrt(d*x + c)*d^3*e^a)*e^((d*x + c)*b
/d - b*c/d)/b^3)*b/d)/d

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Fricas [B]  time = 2.64267, size = 894, normalized size = 6.12 \begin{align*} -\frac{3 \, \sqrt{\pi }{\left (d^{2} \cosh \left (b x + a\right ) \cosh \left (-\frac{b c - a d}{d}\right ) - d^{2} \cosh \left (b x + a\right ) \sinh \left (-\frac{b c - a d}{d}\right ) +{\left (d^{2} \cosh \left (-\frac{b c - a d}{d}\right ) - d^{2} \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 ) + 3 \, \sqrt{\pi }{\left (d^{2} \cosh \left (b x + a\right ) \cosh \left (-\frac{b c - a d}{d}\right ) + d^{2} \cosh \left (b x + a\right ) \sinh \left (-\frac{b c - a d}{d}\right ) +{\left (d^{2} \cosh \left (-\frac{b c - a d}{d}\right ) + d^{2} \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 \,{\left (2 \, b^{2} d x + 2 \, b^{2} c +{\left (2 \, b^{2} d x + 2 \, b^{2} c - 3 \, b d\right )} \cosh \left (b x + a\right )^{2} + 2 \,{\left (2 \, b^{2} d x + 2 \, b^{2} c - 3 \, b d\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (2 \, b^{2} d x + 2 \, b^{2} c - 3 \, b d\right )} \sinh \left (b x + a\right )^{2} + 3 \, b d\right )} \sqrt{d x + c}}{8 \,{\left (b^{3} \cosh \left (b x + a\right ) + b^{3} \sinh \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*sqrt(pi)*(d^2*cosh(b*x + a)*cosh(-(b*c - a*d)/d) - d^2*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + (d^2*cosh(
-(b*c - a*d)/d) - d^2*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/d)) + 3*sqrt(pi)
*(d^2*cosh(b*x + a)*cosh(-(b*c - a*d)/d) + d^2*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + (d^2*cosh(-(b*c - a*d)/d)
+ d^2*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(d*x + c)*sqrt(-b/d)) - 2*(2*b^2*d*x + 2*b^2*c +
(2*b^2*d*x + 2*b^2*c - 3*b*d)*cosh(b*x + a)^2 + 2*(2*b^2*d*x + 2*b^2*c - 3*b*d)*cosh(b*x + a)*sinh(b*x + a) +
(2*b^2*d*x + 2*b^2*c - 3*b*d)*sinh(b*x + a)^2 + 3*b*d)*sqrt(d*x + c))/(b^3*cosh(b*x + a) + b^3*sinh(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.33518, size = 273, normalized size = 1.87 \begin{align*} \frac{\frac{3 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (-\frac{\sqrt{b d} \sqrt{d x + c}}{d}\right ) e^{\left (\frac{b c - a d}{d}\right )}}{\sqrt{b d} b^{2}} - \frac{3 \, \sqrt{\pi } d^{3} \operatorname{erf}\left (-\frac{\sqrt{-b d} \sqrt{d x + c}}{d}\right ) e^{\left (-\frac{b c - a d}{d}\right )}}{\sqrt{-b d} b^{2}} + \frac{2 \,{\left (2 \,{\left (d x + c\right )}^{\frac{3}{2}} b d - 3 \, \sqrt{d x + c} d^{2}\right )} e^{\left (\frac{{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b^{2}} + \frac{2 \,{\left (2 \,{\left (d x + c\right )}^{\frac{3}{2}} b d + 3 \, \sqrt{d x + c} d^{2}\right )} e^{\left (-\frac{{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*sqrt(pi)*d^3*erf(-sqrt(b*d)*sqrt(d*x + c)/d)*e^((b*c - a*d)/d)/(sqrt(b*d)*b^2) - 3*sqrt(pi)*d^3*erf(-sq
rt(-b*d)*sqrt(d*x + c)/d)*e^(-(b*c - a*d)/d)/(sqrt(-b*d)*b^2) + 2*(2*(d*x + c)^(3/2)*b*d - 3*sqrt(d*x + c)*d^2
)*e^(((d*x + c)*b - b*c + a*d)/d)/b^2 + 2*(2*(d*x + c)^(3/2)*b*d + 3*sqrt(d*x + c)*d^2)*e^(-((d*x + c)*b - b*c
+ a*d)/d)/b^2)/d