3.1.43 \(\int \frac {\sinh (a+b x)}{(c+d x)^{5/2}} \, dx\) [43]

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

[Out]

-2/3*sinh(b*x+a)/d/(d*x+c)^(3/2)-2/3*b^(3/2)*exp(-a+b*c/d)*erf(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/d^(5/2)
+2/3*b^(3/2)*exp(a-b*c/d)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/d^(5/2)-4/3*b*cosh(b*x+a)/d^2/(d*x+c)^(
1/2)

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Rubi [A]
time = 0.17, antiderivative size = 149, normalized size of antiderivative = 1.00, 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 = {3378, 3389, 2211, 2235, 2236} \begin {gather*} -\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}} \end {gather*}

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 2211

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] &&  !TrueQ[$UseGamma]

Rule 2235

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 2236

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 3378

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 3389

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]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.53, size = 161, normalized size = 1.08 \begin {gather*} \frac {2 b \left (\frac {e^a \left (-e^{b x}+e^{-\frac {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}}+\frac {e^{-a-b x} \left (-1+e^{b \left (\frac {c}{d}+x\right )} \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}}\right )}{3 d}-\frac {2 \sinh (a+b x)}{3 d (c+d x)^{3/2}} \end {gather*}

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 = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\sinh \left (b x +a \right )}{\left (d x +c \right )^{\frac {5}{2}}}\, dx\]

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 = 0.34, size = 114, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (111) = 222\).
time = 0.39, size = 532, normalized size = 3.57 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {5}{2}}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {sinh}\left (a+b\,x\right )}{{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)/(c + d*x)^(5/2),x)

[Out]

int(sinh(a + b*x)/(c + d*x)^(5/2), x)

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