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

Optimal. Leaf size=142 \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{\frac{2 b c}{d}-2 a} \text{Erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{2 a-\frac{2 b c}{d}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{2 \sinh ^2(a+b x)}{d \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.253019, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3313, 12, 3308, 2180, 2204, 2205} \[ -\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{\frac{2 b c}{d}-2 a} \text{Erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \sqrt{b} e^{2 a-\frac{2 b c}{d}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{d^{3/2}}-\frac{2 \sinh ^2(a+b x)}{d \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3313

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x]^
n)/(d*(m + 1)), x] - Dist[(f*n)/(d*(m + 1)), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [B]  time = 4.85797, size = 570, normalized size = 4.01 \[ \frac{e^{-\frac{2 b (c+d x)}{d}} \left (\sqrt{2} \sqrt{d} e^{\frac{2 b (c+d x)}{d}} \sqrt{-\frac{b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},-\frac{2 b (c+d x)}{d}\right ) \left (\cosh \left (2 a-\frac{2 b c}{d}\right )+\sinh (2 a) \cosh \left (\frac{2 b c}{d}\right )\right )+\sqrt{2} \sqrt{d} e^{\frac{2 b (c+d x)}{d}} \sqrt{\frac{b (c+d x)}{d}} \text{Gamma}\left (\frac{1}{2},\frac{2 b (c+d x)}{d}\right ) \left (\cosh (2 a) \cosh \left (\frac{2 b c}{d}\right )-\sinh (2 a) \left (\sinh \left (\frac{2 b c}{d}\right )+\cosh \left (\frac{2 b c}{d}\right )\right )\right )-\sqrt{2 \pi } \sqrt{b} \cosh (2 a) \sqrt{c+d x} e^{\frac{2 b (c+d x)}{d}} \sinh \left (\frac{2 b c}{d}\right ) \text{Erf}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )-\sqrt{2 \pi } \sqrt{b} \cosh (2 a) \sqrt{c+d x} e^{\frac{2 b (c+d x)}{d}} \sinh \left (\frac{2 b c}{d}\right ) \text{Erfi}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )+\sqrt{d} \sinh (2 a) e^{\frac{4 b (c+d x)}{d}} \sinh \left (\frac{2 b c}{d}\right )-\sqrt{d} \cosh (2 a) e^{\frac{4 b (c+d x)}{d}} \cosh \left (\frac{2 b c}{d}\right )-\sqrt{d} \sinh (2 a) e^{\frac{4 b (c+d x)}{d}} \cosh \left (\frac{2 b c}{d}\right )+\sqrt{d} \cosh (2 a) e^{\frac{4 b (c+d x)}{d}} \sinh \left (\frac{2 b c}{d}\right )+\sqrt{d} \sinh (2 a) \sinh \left (\frac{2 b c}{d}\right )-\sqrt{d} \cosh (2 a) \cosh \left (\frac{2 b c}{d}\right )+\sqrt{d} \sinh (2 a) \cosh \left (\frac{2 b c}{d}\right )-\sqrt{d} \cosh (2 a) \sinh \left (\frac{2 b c}{d}\right )+2 \sqrt{d} e^{\frac{2 b (c+d x)}{d}}\right )}{2 d^{3/2} \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d]*E^((2*b*(c + d*x))/d) - Sqrt[d]*Cosh[2*a]*Cosh[(2*b*c)/d] - Sqrt[d]*E^((4*b*(c + d*x))/d)*Cosh[2*a]
*Cosh[(2*b*c)/d] + Sqrt[d]*Cosh[(2*b*c)/d]*Sinh[2*a] - Sqrt[d]*E^((4*b*(c + d*x))/d)*Cosh[(2*b*c)/d]*Sinh[2*a]
+ Sqrt[2]*Sqrt[d]*E^((2*b*(c + d*x))/d)*Sqrt[-((b*(c + d*x))/d)]*Gamma[1/2, (-2*b*(c + d*x))/d]*(Cosh[2*a - (
2*b*c)/d] + Cosh[(2*b*c)/d]*Sinh[2*a]) - Sqrt[d]*Cosh[2*a]*Sinh[(2*b*c)/d] + Sqrt[d]*E^((4*b*(c + d*x))/d)*Cos
h[2*a]*Sinh[(2*b*c)/d] - Sqrt[b]*E^((2*b*(c + d*x))/d)*Sqrt[2*Pi]*Sqrt[c + d*x]*Cosh[2*a]*Erf[(Sqrt[2]*Sqrt[b]
*Sqrt[c + d*x])/Sqrt[d]]*Sinh[(2*b*c)/d] - Sqrt[b]*E^((2*b*(c + d*x))/d)*Sqrt[2*Pi]*Sqrt[c + d*x]*Cosh[2*a]*Er
fi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]*Sinh[(2*b*c)/d] + Sqrt[d]*Sinh[2*a]*Sinh[(2*b*c)/d] + Sqrt[d]*E^((
4*b*(c + d*x))/d)*Sinh[2*a]*Sinh[(2*b*c)/d] + Sqrt[2]*Sqrt[d]*E^((2*b*(c + d*x))/d)*Sqrt[(b*(c + d*x))/d]*Gamm
a[1/2, (2*b*(c + d*x))/d]*(Cosh[2*a]*Cosh[(2*b*c)/d] - Sinh[2*a]*(Cosh[(2*b*c)/d] + Sinh[(2*b*c)/d])))/(2*d^(3
/2)*E^((2*b*(c + d*x))/d)*Sqrt[c + d*x])

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \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)^2/(d*x+c)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.29964, size = 157, normalized size = 1.11 \begin{align*} -\frac{\frac{\sqrt{2} \sqrt{\frac{{\left (d x + c\right )} b}{d}} e^{\left (\frac{2 \,{\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac{1}{2}, \frac{2 \,{\left (d x + c\right )} b}{d}\right )}{\sqrt{d x + c}} + \frac{\sqrt{2} \sqrt{-\frac{{\left (d x + c\right )} b}{d}} e^{\left (-\frac{2 \,{\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac{1}{2}, -\frac{2 \,{\left (d x + c\right )} b}{d}\right )}{\sqrt{d x + c}} - \frac{4}{\sqrt{d x + c}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(sqrt(2)*sqrt((d*x + c)*b/d)*e^(2*(b*c - a*d)/d)*gamma(-1/2, 2*(d*x + c)*b/d)/sqrt(d*x + c) + sqrt(2)*sqr
t(-(d*x + c)*b/d)*e^(-2*(b*c - a*d)/d)*gamma(-1/2, -2*(d*x + c)*b/d)/sqrt(d*x + c) - 4/sqrt(d*x + c))/d

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Fricas [B]  time = 2.85584, size = 1442, normalized size = 10.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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