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

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

[Out]

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

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Rubi [A]  time = 0.25, 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 = {3297, 3307, 2180, 2204, 2205} \[ \frac {2 \sqrt {\pi } b^{3/2} e^{\frac {b c}{d}-a} \text {Erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {2 \sqrt {\pi } b^{3/2} e^{a-\frac {b c}{d}} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {4 b \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]/(c + d*x)^(5/2),x]

[Out]

(-2*Cosh[a + b*x])/(3*d*(c + d*x)^(3/2)) + (2*b^(3/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sq
rt[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)) -
(4*b*Sinh[a + b*x])/(3*d^2*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 {\cosh (a+b x)}{(c+d x)^{5/2}} \, dx &=-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}+\frac {(2 b) \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx}{3 d}\\ &=-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {4 b \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {\left (4 b^2\right ) \int \frac {\cosh (a+b x)}{\sqrt {c+d x}} \, dx}{3 d^2}\\ &=-\frac {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {4 b \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}+\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 {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}-\frac {4 b \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}+\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 {2 \cosh (a+b x)}{3 d (c+d x)^{3/2}}+\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 {4 b \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.73, size = 150, normalized size = 1.01 \[ \frac {e^{-a} \left (-2 d e^{2 a-\frac {b c}{d}} \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {b (c+d x)}{d}\right )-e^{-b x} \left (2 b \left (e^{2 (a+b x)}-1\right ) (c+d x)+d \left (e^{2 (a+b x)}+1\right )+2 d e^{b \left (\frac {c}{d}+x\right )} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},b \left (\frac {c}{d}+x\right )\right )\right )\right )}{3 d^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((d*(1 + E^(2*(a + b*x))) + 2*b*(-1 + E^(2*(a + b*x)))*(c + d*x) + 2*d*E^(b*(c/d + x))*((b*(c + d*x))/d)^(3/
2)*Gamma[1/2, b*(c/d + x)])/E^(b*x)) - 2*d*E^(2*a - (b*c)/d)*(-((b*(c + d*x))/d))^(3/2)*Gamma[1/2, -((b*(c + d
*x))/d)])/(3*d^2*E^a*(c + d*x)^(3/2))

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fricas [B]  time = 0.91, size = 534, normalized size = 3.58 \[ \frac {2 \, \sqrt {\pi } {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) - 2 \, \sqrt {\pi } {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) + {\left (2 \, b d x - {\left (2 \, b d x + 2 \, b c + d\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (2 \, b d x + 2 \, b c + d\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (2 \, b d x + 2 \, b c + d\right )} \sinh \left (b x + a\right )^{2} + 2 \, b c - d\right )} \sqrt {d x + c}}{3 \, {\left ({\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \cosh \left (b x + a\right ) + {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(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(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*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)*sinh
(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (b x +a \right )}{\left (d x +c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 115, normalized size = 0.77 \[ \frac {\frac {{\left (\frac {\sqrt {\frac {{\left (d x + c\right )} b}{d}} e^{\left (-a + \frac {b c}{d}\right )} \Gamma \left (-\frac {1}{2}, \frac {{\left (d x + c\right )} b}{d}\right )}{\sqrt {d x + c}} - \frac {\sqrt {-\frac {{\left (d x + c\right )} b}{d}} e^{\left (a - \frac {b c}{d}\right )} \Gamma \left (-\frac {1}{2}, -\frac {{\left (d x + c\right )} b}{d}\right )}{\sqrt {d x + c}}\right )} b}{d} - \frac {2 \, \cosh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {3}{2}}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)/(c + d*x)^(5/2),x)

[Out]

int(cosh(a + b*x)/(c + d*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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