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

Optimal. Leaf size=171 \[ \frac {15 \sqrt {\pi } d^{5/2} e^{\frac {b c}{d}-a} \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}-\frac {15 \sqrt {\pi } d^{5/2} e^{a-\frac {b c}{d}} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sinh (a+b x)}{4 b^3}-\frac {5 d (c+d x)^{3/2} \cosh (a+b x)}{2 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x)}{b} \]

[Out]

-5/2*d*(d*x+c)^(3/2)*cosh(b*x+a)/b^2+(d*x+c)^(5/2)*sinh(b*x+a)/b+15/16*d^(5/2)*exp(-a+b*c/d)*erf(b^(1/2)*(d*x+
c)^(1/2)/d^(1/2))*Pi^(1/2)/b^(7/2)-15/16*d^(5/2)*exp(a-b*c/d)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/b^(
7/2)+15/4*d^2*sinh(b*x+a)*(d*x+c)^(1/2)/b^3

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Rubi [A]  time = 0.33, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {15 \sqrt {\pi } d^{5/2} e^{\frac {b c}{d}-a} \text {Erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}-\frac {15 \sqrt {\pi } d^{5/2} e^{a-\frac {b c}{d}} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sinh (a+b x)}{4 b^3}-\frac {5 d (c+d x)^{3/2} \cosh (a+b x)}{2 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)*Cosh[a + b*x],x]

[Out]

(-5*d*(c + d*x)^(3/2)*Cosh[a + b*x])/(2*b^2) + (15*d^(5/2)*E^(-a + (b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x
])/Sqrt[d]])/(16*b^(7/2)) - (15*d^(5/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(16*b^
(7/2)) + (15*d^2*Sqrt[c + d*x]*Sinh[a + b*x])/(4*b^3) + ((c + d*x)^(5/2)*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 (c+d x)^{5/2} \cosh (a+b x) \, dx &=\frac {(c+d x)^{5/2} \sinh (a+b x)}{b}-\frac {(5 d) \int (c+d x)^{3/2} \sinh (a+b x) \, dx}{2 b}\\ &=-\frac {5 d (c+d x)^{3/2} \cosh (a+b x)}{2 b^2}+\frac {(c+d x)^{5/2} \sinh (a+b x)}{b}+\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \cosh (a+b x) \, dx}{4 b^2}\\ &=-\frac {5 d (c+d x)^{3/2} \cosh (a+b x)}{2 b^2}+\frac {15 d^2 \sqrt {c+d x} \sinh (a+b x)}{4 b^3}+\frac {(c+d x)^{5/2} \sinh (a+b x)}{b}-\frac {\left (15 d^3\right ) \int \frac {\sinh (a+b x)}{\sqrt {c+d x}} \, dx}{8 b^3}\\ &=-\frac {5 d (c+d x)^{3/2} \cosh (a+b x)}{2 b^2}+\frac {15 d^2 \sqrt {c+d x} \sinh (a+b x)}{4 b^3}+\frac {(c+d x)^{5/2} \sinh (a+b x)}{b}-\frac {\left (15 d^3\right ) \int \frac {e^{-i (i a+i b x)}}{\sqrt {c+d x}} \, dx}{16 b^3}+\frac {\left (15 d^3\right ) \int \frac {e^{i (i a+i b x)}}{\sqrt {c+d x}} \, dx}{16 b^3}\\ &=-\frac {5 d (c+d x)^{3/2} \cosh (a+b x)}{2 b^2}+\frac {15 d^2 \sqrt {c+d x} \sinh (a+b x)}{4 b^3}+\frac {(c+d x)^{5/2} \sinh (a+b x)}{b}+\frac {\left (15 d^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 )}{8 b^3}-\frac {\left (15 d^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 )}{8 b^3}\\ &=-\frac {5 d (c+d x)^{3/2} \cosh (a+b x)}{2 b^2}+\frac {15 d^{5/2} e^{-a+\frac {b c}{d}} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}-\frac {15 d^{5/2} e^{a-\frac {b c}{d}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \sinh (a+b x)}{4 b^3}+\frac {(c+d x)^{5/2} \sinh (a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 107, normalized size = 0.63 \[ -\frac {d^3 e^{-a-\frac {b c}{d}} \left (e^{2 a} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},-\frac {b (c+d x)}{d}\right )+e^{\frac {2 b c}{d}} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {7}{2},\frac {b (c+d x)}{d}\right )\right )}{2 b^4 \sqrt {c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(d^3*E^(-a - (b*c)/d)*(E^(2*a)*Sqrt[-((b*(c + d*x))/d)]*Gamma[7/2, -((b*(c + d*x))/d)] + E^((2*b*c)/d)*Sq
rt[(b*(c + d*x))/d]*Gamma[7/2, (b*(c + d*x))/d]))/(b^4*Sqrt[c + d*x])

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fricas [B]  time = 0.91, size = 523, normalized size = 3.06 \[ \frac {15 \, \sqrt {\pi } {\left (d^{3} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - d^{3} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d^{3} \cosh \left (-\frac {b c - a d}{d}\right ) - d^{3} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) + 15 \, \sqrt {\pi } {\left (d^{3} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + d^{3} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left (d^{3} \cosh \left (-\frac {b c - a d}{d}\right ) + d^{3} \sinh \left (-\frac {b c - a d}{d}\right )\right )} \sinh \left (b x + a\right )\right )} \sqrt {-\frac {b}{d}} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) - 2 \, {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} + 10 \, b^{2} c d + 15 \, b d^{2} - {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} - 10 \, b^{2} c d + 15 \, b d^{2} + 2 \, {\left (4 \, b^{3} c d - 5 \, b^{2} d^{2}\right )} x\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} - 10 \, b^{2} c d + 15 \, b d^{2} + 2 \, {\left (4 \, b^{3} c d - 5 \, b^{2} d^{2}\right )} x\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (4 \, b^{3} d^{2} x^{2} + 4 \, b^{3} c^{2} - 10 \, b^{2} c d + 15 \, b d^{2} + 2 \, {\left (4 \, b^{3} c d - 5 \, b^{2} d^{2}\right )} x\right )} \sinh \left (b x + a\right )^{2} + 2 \, {\left (4 \, b^{3} c d + 5 \, b^{2} d^{2}\right )} x\right )} \sqrt {d x + c}}{16 \, {\left (b^{4} \cosh \left (b x + a\right ) + b^{4} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(15*sqrt(pi)*(d^3*cosh(b*x + a)*cosh(-(b*c - a*d)/d) - d^3*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + (d^3*cosh
(-(b*c - a*d)/d) - d^3*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/d)) + 15*sqrt(p
i)*(d^3*cosh(b*x + a)*cosh(-(b*c - a*d)/d) + d^3*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + (d^3*cosh(-(b*c - a*d)/d
) + d^3*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(d*x + c)*sqrt(-b/d)) - 2*(4*b^3*d^2*x^2 + 4*b
^3*c^2 + 10*b^2*c*d + 15*b*d^2 - (4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2
)*x)*cosh(b*x + a)^2 - 2*(4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*cos
h(b*x + a)*sinh(b*x + a) - (4*b^3*d^2*x^2 + 4*b^3*c^2 - 10*b^2*c*d + 15*b*d^2 + 2*(4*b^3*c*d - 5*b^2*d^2)*x)*s
inh(b*x + a)^2 + 2*(4*b^3*c*d + 5*b^2*d^2)*x)*sqrt(d*x + c))/(b^4*cosh(b*x + a) + b^4*sinh(b*x + a))

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giac [A]  time = 0.22, size = 232, normalized size = 1.36 \[ -\frac {\frac {15 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (-\frac {\sqrt {b d} \sqrt {d x + c}}{d}\right ) e^{\left (\frac {b c - a d}{d}\right )}}{\sqrt {b d} b^{3}} - \frac {15 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (-\frac {\sqrt {-b d} \sqrt {d x + c}}{d}\right ) e^{\left (-\frac {b c - a d}{d}\right )}}{\sqrt {-b d} b^{3}} - \frac {2 \, {\left (4 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d - 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} + 15 \, \sqrt {d x + c} d^{3}\right )} e^{\left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b^{3}} + \frac {2 \, {\left (4 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d + 10 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{2} + 15 \, \sqrt {d x + c} d^{3}\right )} e^{\left (-\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}}{b^{3}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(15*sqrt(pi)*d^4*erf(-sqrt(b*d)*sqrt(d*x + c)/d)*e^((b*c - a*d)/d)/(sqrt(b*d)*b^3) - 15*sqrt(pi)*d^4*erf
(-sqrt(-b*d)*sqrt(d*x + c)/d)*e^(-(b*c - a*d)/d)/(sqrt(-b*d)*b^3) - 2*(4*(d*x + c)^(5/2)*b^2*d - 10*(d*x + c)^
(3/2)*b*d^2 + 15*sqrt(d*x + c)*d^3)*e^(((d*x + c)*b - b*c + a*d)/d)/b^3 + 2*(4*(d*x + c)^(5/2)*b^2*d + 10*(d*x
+ c)^(3/2)*b*d^2 + 15*sqrt(d*x + c)*d^3)*e^(-((d*x + c)*b - b*c + a*d)/d)/b^3)/d

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{\frac {5}{2}} \cosh \left (b x +a \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 308, normalized size = 1.80 \[ \frac {32 \, {\left (d x + c\right )}^{\frac {7}{2}} \cosh \left (b x + a\right ) - \frac {{\left (\frac {105 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {b}{d}}\right ) e^{\left (a - \frac {b c}{d}\right )}}{b^{4} \sqrt {-\frac {b}{d}}} - \frac {105 \, \sqrt {\pi } d^{4} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {b}{d}}\right ) e^{\left (-a + \frac {b c}{d}\right )}}{b^{4} \sqrt {\frac {b}{d}}} + \frac {2 \, {\left (8 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d e^{\left (\frac {b c}{d}\right )} + 28 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{2} e^{\left (\frac {b c}{d}\right )} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{3} e^{\left (\frac {b c}{d}\right )} + 105 \, \sqrt {d x + c} d^{4} e^{\left (\frac {b c}{d}\right )}\right )} e^{\left (-a - \frac {{\left (d x + c\right )} b}{d}\right )}}{b^{4}} + \frac {2 \, {\left (8 \, {\left (d x + c\right )}^{\frac {7}{2}} b^{3} d e^{a} - 28 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} d^{2} e^{a} + 70 \, {\left (d x + c\right )}^{\frac {3}{2}} b d^{3} e^{a} - 105 \, \sqrt {d x + c} d^{4} e^{a}\right )} e^{\left (\frac {{\left (d x + c\right )} b}{d} - \frac {b c}{d}\right )}}{b^{4}}\right )} b}{d}}{112 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/112*(32*(d*x + c)^(7/2)*cosh(b*x + a) - (105*sqrt(pi)*d^4*erf(sqrt(d*x + c)*sqrt(-b/d))*e^(a - b*c/d)/(b^4*s
qrt(-b/d)) - 105*sqrt(pi)*d^4*erf(sqrt(d*x + c)*sqrt(b/d))*e^(-a + b*c/d)/(b^4*sqrt(b/d)) + 2*(8*(d*x + c)^(7/
2)*b^3*d*e^(b*c/d) + 28*(d*x + c)^(5/2)*b^2*d^2*e^(b*c/d) + 70*(d*x + c)^(3/2)*b*d^3*e^(b*c/d) + 105*sqrt(d*x
+ c)*d^4*e^(b*c/d))*e^(-a - (d*x + c)*b/d)/b^4 + 2*(8*(d*x + c)^(7/2)*b^3*d*e^a - 28*(d*x + c)^(5/2)*b^2*d^2*e
^a + 70*(d*x + c)^(3/2)*b*d^3*e^a - 105*sqrt(d*x + c)*d^4*e^a)*e^((d*x + c)*b/d - b*c/d)/b^4)*b/d)/d

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

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 \[ \int \left (c + d x\right )^{\frac {5}{2}} \cosh {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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