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

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

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3314, 32, 3312, 3307, 2180, 2204, 2205} \[ \frac {2 \sqrt {2 \pi } b^{3/2} e^{\frac {2 b c}{d}-2 a} \text {Erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {2 \sqrt {2 \pi } b^{3/2} e^{2 a-\frac {2 b c}{d}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 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]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3314

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(b*Si
n[e + f*x])^n)/(d*(m + 1)), x] + (Dist[(b^2*f^2*n*(n - 1))/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin
[e + f*x])^(n - 2), x], x] - Dist[(f^2*n^2)/(d^2*(m + 1)*(m + 2)), Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x
], x] - Simp[(b*f*n*(c + d*x)^(m + 2)*Cos[e + f*x]*(b*Sin[e + f*x])^(n - 1))/(d^2*(m + 1)*(m + 2)), x]) /; Fre
eQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]

Rubi steps

\begin {align*} \int \frac {\sinh ^2(a+b x)}{(c+d x)^{5/2}} \, dx &=-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac {\left (8 b^2\right ) \int \frac {1}{\sqrt {c+d x}} \, dx}{3 d^2}+\frac {\left (16 b^2\right ) \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx}{3 d^2}\\ &=\frac {16 b^2 \sqrt {c+d x}}{3 d^3}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {\left (16 b^2\right ) \int \left (\frac {1}{2 \sqrt {c+d x}}-\frac {\cosh (2 a+2 b x)}{2 \sqrt {c+d x}}\right ) \, dx}{3 d^2}\\ &=-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac {\left (8 b^2\right ) \int \frac {\cosh (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{3 d^2}\\ &=-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac {\left (4 b^2\right ) \int \frac {e^{-i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{3 d^2}+\frac {\left (4 b^2\right ) \int \frac {e^{i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{3 d^2}\\ &=-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}+\frac {\left (8 b^2\right ) \operatorname {Subst}\left (\int e^{i \left (2 i a-\frac {2 i b c}{d}\right )-\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{3 d^3}+\frac {\left (8 b^2\right ) \operatorname {Subst}\left (\int e^{-i \left (2 i a-\frac {2 i b c}{d}\right )+\frac {2 b x^2}{d}} \, dx,x,\sqrt {c+d x}\right )}{3 d^3}\\ &=\frac {2 b^{3/2} e^{-2 a+\frac {2 b c}{d}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {2 b^{3/2} e^{2 a-\frac {2 b c}{d}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{3 d^2 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.12, size = 156, normalized size = 0.90 \[ -\frac {2 e^{-2 \left (a+\frac {b c}{d}\right )} \left (e^{2 \left (a+\frac {b c}{d}\right )} \left (2 b (c+d x) \sinh (2 (a+b x))+d \sinh ^2(a+b x)\right )+\sqrt {2} e^{4 a} d \left (-\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )+\sqrt {2} d e^{\frac {4 b c}{d}} \left (\frac {b (c+d x)}{d}\right )^{3/2} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )\right )}{3 d^2 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(Sqrt[2]*d*E^(4*a)*(-((b*(c + d*x))/d))^(3/2)*Gamma[1/2, (-2*b*(c + d*x))/d] + Sqrt[2]*d*E^((4*b*c)/d)*((b
*(c + d*x))/d)^(3/2)*Gamma[1/2, (2*b*(c + d*x))/d] + E^(2*(a + (b*c)/d))*(d*Sinh[a + b*x]^2 + 2*b*(c + d*x)*Si
nh[2*(a + b*x)])))/(3*d^2*E^(2*(a + (b*c)/d))*(c + d*x)^(3/2))

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fricas [B]  time = 0.78, size = 864, normalized size = 4.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*sqrt(2)*sqrt(pi)*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) - (b*d^2*x^2 +
2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(-2*(b*c - a
*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((b*d^2*x^2 + 2*b*c*d*x +
b*c^2)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) - (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-2*(b*c - a*
d)/d))*sinh(b*x + a))*sqrt(b/d)*erf(sqrt(2)*sqrt(d*x + c)*sqrt(b/d)) - 4*sqrt(2)*sqrt(pi)*((b*d^2*x^2 + 2*b*c*
d*x + b*c^2)*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)^2*sinh(-2*
(b*c - a*d)/d) + ((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(-2*(b*c - a*d)/d) + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*sin
h(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d)
+ (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*cosh(b*x + a)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(sqrt(2)
*sqrt(d*x + c)*sqrt(-b/d)) - ((4*b*d*x + 4*b*c + d)*cosh(b*x + a)^4 + 4*(4*b*d*x + 4*b*c + d)*cosh(b*x + a)*si
nh(b*x + a)^3 + (4*b*d*x + 4*b*c + d)*sinh(b*x + a)^4 - 4*b*d*x - 2*d*cosh(b*x + a)^2 + 2*(3*(4*b*d*x + 4*b*c
+ d)*cosh(b*x + a)^2 - d)*sinh(b*x + a)^2 - 4*b*c + 4*((4*b*d*x + 4*b*c + d)*cosh(b*x + a)^3 - d*cosh(b*x + a)
)*sinh(b*x + a) + d)*sqrt(d*x + c))/((d^4*x^2 + 2*c*d^3*x + c^2*d^2)*cosh(b*x + a)^2 + 2*(d^4*x^2 + 2*c*d^3*x
+ c^2*d^2)*cosh(b*x + a)*sinh(b*x + a) + (d^4*x^2 + 2*c*d^3*x + c^2*d^2)*sinh(b*x + a)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}\left (b x +a \right )}{\left (d x +c \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 118, normalized size = 0.68 \[ -\frac {\frac {3 \, \sqrt {2} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {3}{2}, \frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} + \frac {3 \, \sqrt {2} \left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {3}{2}} e^{\left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {3}{2}, -\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {3}{2}}} - \frac {2}{{\left (d x + c\right )}^{\frac {3}{2}}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*sqrt(2)*((d*x + c)*b/d)^(3/2)*e^(2*(b*c - a*d)/d)*gamma(-3/2, 2*(d*x + c)*b/d)/(d*x + c)^(3/2) + 3*sqr
t(2)*(-(d*x + c)*b/d)^(3/2)*e^(-2*(b*c - a*d)/d)*gamma(-3/2, -2*(d*x + c)*b/d)/(d*x + c)^(3/2) - 2/(d*x + c)^(
3/2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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