3.47 \(\int \frac{\cosh (a+b x)}{(c+d x)^{7/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac{4 \sqrt{\pi } b^{5/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{4 \sqrt{\pi } b^{5/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}-\frac{8 b^2 \cosh (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}} \]

[Out]

(-2*Cosh[a + b*x])/(5*d*(c + d*x)^(5/2)) - (8*b^2*Cosh[a + b*x])/(15*d^3*Sqrt[c + d*x]) - (4*b^(5/2)*E^(-a + (
b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(15*d^(7/2)) + (4*b^(5/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi
[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(15*d^(7/2)) - (4*b*Sinh[a + b*x])/(15*d^2*(c + d*x)^(3/2))

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Rubi [A]  time = 0.312007, antiderivative size = 174, normalized size of antiderivative = 1., 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 = {3297, 3308, 2180, 2204, 2205} \[ -\frac{4 \sqrt{\pi } b^{5/2} e^{\frac{b c}{d}-a} \text{Erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{4 \sqrt{\pi } b^{5/2} e^{a-\frac{b c}{d}} \text{Erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}-\frac{8 b^2 \cosh (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]/(c + d*x)^(7/2),x]

[Out]

(-2*Cosh[a + b*x])/(5*d*(c + d*x)^(5/2)) - (8*b^2*Cosh[a + b*x])/(15*d^3*Sqrt[c + d*x]) - (4*b^(5/2)*E^(-a + (
b*c)/d)*Sqrt[Pi]*Erf[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(15*d^(7/2)) + (4*b^(5/2)*E^(a - (b*c)/d)*Sqrt[Pi]*Erfi
[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(15*d^(7/2)) - (4*b*Sinh[a + b*x])/(15*d^2*(c + d*x)^(3/2))

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 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]

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]

Rubi steps

\begin{align*} \int \frac{\cosh (a+b x)}{(c+d x)^{7/2}} \, dx &=-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}}+\frac{(2 b) \int \frac{\sinh (a+b x)}{(c+d x)^{5/2}} \, dx}{5 d}\\ &=-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}}-\frac{4 b \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac{\left (4 b^2\right ) \int \frac{\cosh (a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2}\\ &=-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}}-\frac{8 b^2 \cosh (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac{\left (8 b^3\right ) \int \frac{\sinh (a+b x)}{\sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}}-\frac{8 b^2 \cosh (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\frac{\left (4 b^3\right ) \int \frac{e^{-i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{15 d^3}-\frac{\left (4 b^3\right ) \int \frac{e^{i (i a+i b x)}}{\sqrt{c+d x}} \, dx}{15 d^3}\\ &=-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}}-\frac{8 b^2 \cosh (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac{\left (8 b^3\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 )}{15 d^4}+\frac{\left (8 b^3\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 )}{15 d^4}\\ &=-\frac{2 \cosh (a+b x)}{5 d (c+d x)^{5/2}}-\frac{8 b^2 \cosh (a+b x)}{15 d^3 \sqrt{c+d x}}-\frac{4 b^{5/2} e^{-a+\frac{b c}{d}} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}+\frac{4 b^{5/2} e^{a-\frac{b c}{d}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{d}}\right )}{15 d^{7/2}}-\frac{4 b \sinh (a+b x)}{15 d^2 (c+d x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.395199, size = 191, normalized size = 1.1 \[ \frac{e^{-a} \left (2 e^{2 a} \left (-2 b e^{-\frac{b c}{d}} (c+d x) \left (2 d \left (-\frac{b (c+d x)}{d}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{b (c+d x)}{d}\right )+e^{b \left (\frac{c}{d}+x\right )} (2 b (c+d x)+d)\right )-3 d^2 e^{b x}\right )+e^{-b x} \left (8 d^2 e^{b \left (\frac{c}{d}+x\right )} \left (\frac{b (c+d x)}{d}\right )^{5/2} \text{Gamma}\left (\frac{1}{2},\frac{b (c+d x)}{d}\right )-8 b^2 (c+d x)^2+4 b d (c+d x)-6 d^2\right )\right )}{30 d^3 (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]/(c + d*x)^(7/2),x]

[Out]

(2*E^(2*a)*(-3*d^2*E^(b*x) - (2*b*(c + d*x)*(E^(b*(c/d + x))*(d + 2*b*(c + d*x)) + 2*d*(-((b*(c + d*x))/d))^(3
/2)*Gamma[1/2, -((b*(c + d*x))/d)]))/E^((b*c)/d)) + (-6*d^2 + 4*b*d*(c + d*x) - 8*b^2*(c + d*x)^2 + 8*d^2*E^(b
*(c/d + x))*((b*(c + d*x))/d)^(5/2)*Gamma[1/2, (b*(c + d*x))/d])/E^(b*x))/(30*d^3*E^a*(c + d*x)^(5/2))

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{\cosh \left ( bx+a \right ) \left ( dx+c \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)/(d*x+c)^(7/2),x)

[Out]

int(cosh(b*x+a)/(d*x+c)^(7/2),x)

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Maxima [A]  time = 1.18735, size = 155, normalized size = 0.89 \begin{align*} \frac{\frac{{\left (\frac{\left (\frac{{\left (d x + c\right )} b}{d}\right )^{\frac{3}{2}} e^{\left (-a + \frac{b c}{d}\right )} \Gamma \left (-\frac{3}{2}, \frac{{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac{3}{2}}} - \frac{\left (-\frac{{\left (d x + c\right )} b}{d}\right )^{\frac{3}{2}} e^{\left (a - \frac{b c}{d}\right )} \Gamma \left (-\frac{3}{2}, -\frac{{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac{3}{2}}}\right )} b}{d} - \frac{2 \, \cosh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac{5}{2}}}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)/(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

1/5*((((d*x + c)*b/d)^(3/2)*e^(-a + b*c/d)*gamma(-3/2, (d*x + c)*b/d)/(d*x + c)^(3/2) - (-(d*x + c)*b/d)^(3/2)
*e^(a - b*c/d)*gamma(-3/2, -(d*x + c)*b/d)/(d*x + c)^(3/2))*b/d - 2*cosh(b*x + a)/(d*x + c)^(5/2))/d

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Fricas [B]  time = 2.29256, size = 1821, normalized size = 10.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)/(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/15*(4*sqrt(pi)*((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*cosh(-(b*c - a*d)/d
) - (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + ((b^2*d^3*x
^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(-(b*c - a*d)/d) - (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*
c^2*d*x + b^2*c^3)*sinh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(d*x + c)*sqrt(b/d)) + 4*sqrt(pi)*((
b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*cosh(-(b*c - a*d)/d) + (b^2*d^3*x^3 + 3
*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*cosh(b*x + a)*sinh(-(b*c - a*d)/d) + ((b^2*d^3*x^3 + 3*b^2*c*d^2*x^2
+ 3*b^2*c^2*d*x + b^2*c^3)*cosh(-(b*c - a*d)/d) + (b^2*d^3*x^3 + 3*b^2*c*d^2*x^2 + 3*b^2*c^2*d*x + b^2*c^3)*s
inh(-(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(d*x + c)*sqrt(-b/d)) + (4*b^2*d^2*x^2 + 4*b^2*c^2 - 2*
b*c*d + (4*b^2*d^2*x^2 + 4*b^2*c^2 + 2*b*c*d + 3*d^2 + 2*(4*b^2*c*d + b*d^2)*x)*cosh(b*x + a)^2 + 2*(4*b^2*d^2
*x^2 + 4*b^2*c^2 + 2*b*c*d + 3*d^2 + 2*(4*b^2*c*d + b*d^2)*x)*cosh(b*x + a)*sinh(b*x + a) + (4*b^2*d^2*x^2 + 4
*b^2*c^2 + 2*b*c*d + 3*d^2 + 2*(4*b^2*c*d + b*d^2)*x)*sinh(b*x + a)^2 + 3*d^2 + 2*(4*b^2*c*d - b*d^2)*x)*sqrt(
d*x + c))/((d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*cosh(b*x + a) + (d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*
x + c^3*d^3)*sinh(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)/(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)/(d*x+c)^(7/2),x, algorithm="giac")

[Out]

integrate(cosh(b*x + a)/(d*x + c)^(7/2), x)