3.43 \(\int \sqrt {c+d x} \cosh (a+b x) \, dx\)

Optimal. Leaf size=123 \[ \frac {\sqrt {\pi } \sqrt {d} e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{3/2}}-\frac {\sqrt {\pi } \sqrt {d} e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{3/2}}+\frac {\sqrt {c+d x} \sinh (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 123, 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 = {3296, 3308, 2180, 2204, 2205} \[ \frac {\sqrt {\pi } \sqrt {d} e^{\frac {b c}{d}-a} \text {Erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{3/2}}-\frac {\sqrt {\pi } \sqrt {d} e^{a-\frac {b c}{d}} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{3/2}}+\frac {\sqrt {c+d x} \sinh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + d*x]*Cosh[a + b*x],x]

[Out]

(Sqrt[d]*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(4*b^(3/2)) - (Sqrt[d]*E^(a - (b*c)/d
)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(4*b^(3/2)) + (Sqrt[c + d*x]*Sinh[a + b*x])/b

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 105, normalized size = 0.85 \[ \frac {\sqrt {c+d x} e^{-a-\frac {b c}{d}} \left (\frac {e^{2 a} \Gamma \left (\frac {3}{2},-\frac {b (c+d x)}{d}\right )}{\sqrt {-\frac {b (c+d x)}{d}}}-\frac {e^{\frac {2 b c}{d}} \Gamma \left (\frac {3}{2},\frac {b (c+d x)}{d}\right )}{\sqrt {\frac {b (c+d x)}{d}}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + d*x]*Cosh[a + b*x],x]

[Out]

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

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fricas [B]  time = 1.02, size = 302, normalized size = 2.46 \[ \frac {\sqrt {\pi } {\left (d \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - d \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d \cosh \left (-\frac {b c - a d}{d}\right ) - d \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) + \sqrt {\pi } {\left (d \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + d \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d \cosh \left (-\frac {b c - a d}{d}\right ) + d \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) + 2 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b\right )} \sqrt {d x + c}}{4 \, {\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 169, normalized size = 1.37 \[ -\frac {\frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c}}{d}\right ) e^{\left (\frac {b c - a d}{d}\right )}}{\sqrt {b d} b} - \frac {\sqrt {\pi } d^{2} \operatorname {erf}\left (-\frac {\sqrt {-b d} \sqrt {d x + c}}{d}\right ) e^{\left (-\frac {b c - a d}{d}\right )}}{\sqrt {-b d} b} - \frac {2 \, \sqrt {d x + c} d e^{\left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b} + \frac {2 \, \sqrt {d x + c} d e^{\left (-\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*(d*x+c)^(1/2),x, algorithm="giac")

[Out]

-1/4*(sqrt(pi)*d^2*erf(-sqrt(b*d)*sqrt(d*x + c)/d)*e^((b*c - a*d)/d)/(sqrt(b*d)*b) - sqrt(pi)*d^2*erf(-sqrt(-b
*d)*sqrt(d*x + c)/d)*e^(-(b*c - a*d)/d)/(sqrt(-b*d)*b) - 2*sqrt(d*x + c)*d*e^(((d*x + c)*b - b*c + a*d)/d)/b +
2*sqrt(d*x + c)*d*e^(-((d*x + c)*b - b*c + a*d)/d)/b)/d

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \cosh \left (b x +a \right ) \sqrt {d x +c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*(d*x+c)^(1/2),x)

[Out]

int(cosh(b*x+a)*(d*x+c)^(1/2),x)

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maxima [B]  time = 0.34, size = 230, normalized size = 1.87 \[ \frac {8 \, {\left (d x + c\right )}^{\frac {3}{2}} \cosh \left (b x + a\right ) - \frac {{\left (\frac {3 \, \sqrt {\pi } d^{2} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (a - \frac {b c}{d}\right )}}{b^{2} \sqrt {-\frac {b}{d}}} - \frac {3 \, \sqrt {\pi } d^{2} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-a + \frac {b c}{d}\right )}}{b^{2} \sqrt {\frac {b}{d}}} + \frac {2 \, {\left (2 \, {\left (d x + c\right )}^{\frac {3}{2}} b d e^{\left (\frac {b c}{d}\right )} + 3 \, \sqrt {d x + c} d^{2} e^{\left (\frac {b c}{d}\right )}\right )} e^{\left (-a - \frac {{\left (d x + c\right )} b}{d}\right )}}{b^{2}} + \frac {2 \, {\left (2 \, {\left (d x + c\right )}^{\frac {3}{2}} b d e^{a} - 3 \, \sqrt {d x + c} d^{2} e^{a}\right )} e^{\left (\frac {{\left (d x + c\right )} b}{d} - \frac {b c}{d}\right )}}{b^{2}}\right )} b}{d}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/12*(8*(d*x + c)^(3/2)*cosh(b*x + a) - (3*sqrt(pi)*d^2*erf(sqrt(d*x + c)*sqrt(-b/d))*e^(a - b*c/d)/(b^2*sqrt(
-b/d)) - 3*sqrt(pi)*d^2*erf(sqrt(d*x + c)*sqrt(b/d))*e^(-a + b*c/d)/(b^2*sqrt(b/d)) + 2*(2*(d*x + c)^(3/2)*b*d
*e^(b*c/d) + 3*sqrt(d*x + c)*d^2*e^(b*c/d))*e^(-a - (d*x + c)*b/d)/b^2 + 2*(2*(d*x + c)^(3/2)*b*d*e^a - 3*sqrt
(d*x + c)*d^2*e^a)*e^((d*x + c)*b/d - b*c/d)/b^2)*b/d)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)*(c + d*x)^(1/2),x)

[Out]

int(cosh(a + b*x)*(c + d*x)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*(d*x+c)**(1/2),x)

[Out]

Integral(sqrt(c + d*x)*cosh(a + b*x), x)

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