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

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

[Out]

-4/15*b*cosh(b*x+a)/d^2/(d*x+c)^(3/2)-2/5*sinh(b*x+a)/d/(d*x+c)^(5/2)+4/15*b^(5/2)*exp(-a+b*c/d)*erf(b^(1/2)*(
d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/d^(7/2)+4/15*b^(5/2)*exp(a-b*c/d)*erfi(b^(1/2)*(d*x+c)^(1/2)/d^(1/2))*Pi^(1/2)/
d^(7/2)-8/15*b^2*sinh(b*x+a)/d^3/(d*x+c)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 174, 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 = {3297, 3307, 2180, 2204, 2205} \[ \frac {4 \sqrt {\pi } b^{5/2} e^{\frac {b c}{d}-a} \text {Erf}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}+\frac {4 \sqrt {\pi } b^{5/2} e^{a-\frac {b c}{d}} \text {Erfi}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{15 d^{7/2}}-\frac {8 b^2 \sinh (a+b x)}{15 d^3 \sqrt {c+d x}}-\frac {4 b \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]/(c + d*x)^(7/2),x]

[Out]

(-4*b*Cosh[a + b*x])/(15*d^2*(c + d*x)^(3/2)) + (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)) - (2*Sinh[a + b*x])/(5*d*(c + d*x)^(5/2)) - (8*b^2*Sinh[a + b*x])/(15*d^3*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 {\sinh (a+b x)}{(c+d x)^{7/2}} \, dx &=-\frac {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}}+\frac {(2 b) \int \frac {\cosh (a+b x)}{(c+d x)^{5/2}} \, dx}{5 d}\\ &=-\frac {4 b \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}}+\frac {\left (4 b^2\right ) \int \frac {\sinh (a+b x)}{(c+d x)^{3/2}} \, dx}{15 d^2}\\ &=-\frac {4 b \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}}-\frac {8 b^2 \sinh (a+b x)}{15 d^3 \sqrt {c+d x}}+\frac {\left (8 b^3\right ) \int \frac {\cosh (a+b x)}{\sqrt {c+d x}} \, dx}{15 d^3}\\ &=-\frac {4 b \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}}-\frac {8 b^2 \sinh (a+b x)}{15 d^3 \sqrt {c+d x}}+\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 {4 b \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}-\frac {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}}-\frac {8 b^2 \sinh (a+b x)}{15 d^3 \sqrt {c+d x}}+\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 {4 b \cosh (a+b x)}{15 d^2 (c+d x)^{3/2}}+\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 {2 \sinh (a+b x)}{5 d (c+d x)^{5/2}}-\frac {8 b^2 \sinh (a+b x)}{15 d^3 \sqrt {c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 168, normalized size = 0.97 \[ \frac {2 \left (-b (c+d x) \left (e^{a-\frac {b c}{d}} \left (e^{b \left (\frac {c}{d}+x\right )} (2 b (c+d x)+d)+2 d \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {b (c+d x)}{d}\right )\right )+e^{-a-b x} \left (-2 b (c+d x)+2 d e^{b \left (\frac {c}{d}+x\right )} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {b (c+d x)}{d}\right )+d\right )\right )-3 d^2 \sinh (a+b x)\right )}{15 d^3 (c+d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-(b*(c + d*x)*(E^(a - (b*c)/d)*(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^(-a - b*x)*(d - 2*b*(c + d*x) + 2*d*E^(b*(c/d + x))*((b*(c + d*x))/d)^(3/2)*Gam
ma[1/2, (b*(c + d*x))/d]))) - 3*d^2*Sinh[a + b*x]))/(15*d^3*(c + d*x)^(5/2))

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fricas [B]  time = 0.51, size = 855, normalized size = 4.91 \[ \frac {4 \, \sqrt {\pi } {\left ({\left (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}\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (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}\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (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}\right )} \cosh \left (-\frac {b c - a d}{d}\right ) - {\left (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}\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 ) - 4 \, \sqrt {\pi } {\left ({\left (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}\right )} \cosh \left (b x + a\right ) \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (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}\right )} \cosh \left (b x + a\right ) \sinh \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (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}\right )} \cosh \left (-\frac {b c - a d}{d}\right ) + {\left (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}\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 (4 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c^{2} - 2 \, b c d - {\left (4 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c^{2} + 2 \, b c d + 3 \, d^{2} + 2 \, {\left (4 \, b^{2} c d + b d^{2}\right )} x\right )} \cosh \left (b x + a\right )^{2} - 2 \, {\left (4 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c^{2} + 2 \, b c d + 3 \, d^{2} + 2 \, {\left (4 \, b^{2} c d + b d^{2}\right )} x\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (4 \, b^{2} d^{2} x^{2} + 4 \, b^{2} c^{2} + 2 \, b c d + 3 \, d^{2} + 2 \, {\left (4 \, b^{2} c d + b d^{2}\right )} x\right )} \sinh \left (b x + a\right )^{2} + 3 \, d^{2} + 2 \, {\left (4 \, b^{2} c d - b d^{2}\right )} x\right )} \sqrt {d x + c}}{15 \, {\left ({\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )} \cosh \left (b x + a\right ) + {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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)*si
nh(-(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x +a \right )}{\left (d x +c \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.78, size = 114, normalized size = 0.66 \[ -\frac {\frac {{\left (\frac {\left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (-a + \frac {b c}{d}\right )} \Gamma \left (-\frac {3}{2}, \frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} + \frac {\left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (a - \frac {b c}{d}\right )} \Gamma \left (-\frac {3}{2}, -\frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}}\right )} b}{d} + \frac {2 \, \sinh \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {5}{2}}}}{5 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\left (a+b\,x\right )}{{\left (c+d\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)/(c + d*x)^(7/2),x)

[Out]

int(sinh(a + b*x)/(c + d*x)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x)/(c + d*x)**(7/2), x)

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