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]

exp(-a+b*c/d)*erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*b^(1/2)*Pi^(1/2)/d^(3/2)+exp(a-b*c/d)*erfi(b^(1/2)*(d*x+c)^(1
/2)/d^(1/2))*b^(1/2)*Pi^(1/2)/d^(3/2)-2*sinh(b*x+a)/d/(d*x+c)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.14, size = 120, normalized size = 1.02 \[ \frac {e^{-a-\frac {b c}{d}} \left (-2 e^{a+\frac {b c}{d}} \sinh (a+b x)+e^{2 a} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {b (c+d x)}{d}\right )-e^{\frac {2 b c}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {b (c+d x)}{d}\right )\right )}{d \sqrt {c+d x}} \]

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])

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fricas [B]  time = 0.46, size = 339, normalized size = 2.87 \[ \frac {\sqrt {\pi } {\left ({\left (d x + c\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (d x + c\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (d x + c\right )} \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (d x + c\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) - \sqrt {\pi } {\left ({\left (d x + c\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (d x + c\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (d x + c\right )} \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (d x + c\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) - \sqrt {d x + c} {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )}}{{\left (d^{2} x + c d\right )} \cosh \left (b x + a\right ) + {\left (d^{2} x + c d\right )} \sinh \left (b x + a\right )} \]

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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x +a \right )}{\left (d x +c \right )^{\frac {3}{2}}}\, dx \]

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)

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maxima [A]  time = 0.62, size = 103, normalized size = 0.87 \[ \frac {\frac {{\left (\frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (a - \frac {b c}{d}\right )}}{\sqrt {-\frac {b}{d}}} + \frac {\sqrt {\pi } \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-a + \frac {b c}{d}\right )}}{\sqrt {\frac {b}{d}}}\right )} b}{d} - \frac {2 \, \sinh \left (b x + a\right )}{\sqrt {d x + c}}}{d} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\left (a+b\,x\right )}{{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)/(c + d*x)^(3/2),x)

[Out]

int(sinh(a + b*x)/(c + d*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \]

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)

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