3.10.66 \(\int (-\frac {3 d^2 e^{a+b x}}{4 (b^2-\frac {9 d^2}{4}) \sqrt {\sinh (c+d x)}}+e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)) \, dx\) [966]

Optimal. Leaf size=73 \[ -\frac {6 d e^{a+b x} \cosh (c+d x) \sqrt {\sinh (c+d x)}}{4 b^2-9 d^2}+\frac {4 b e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)}{4 b^2-9 d^2} \]

[Out]

4*b*exp(b*x+a)*sinh(d*x+c)^(3/2)/(4*b^2-9*d^2)-6*d*exp(b*x+a)*cosh(d*x+c)*sinh(d*x+c)^(1/2)/(4*b^2-9*d^2)

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Rubi [A]
time = 0.40, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {5590, 2285, 2284, 2283, 5584} \begin {gather*} \frac {4 b e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)}{4 b^2-9 d^2}-\frac {6 d e^{a+b x} \sqrt {\sinh (c+d x)} \cosh (c+d x)}{4 b^2-9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*d^2*E^(a + b*x))/(4*(b^2 - (9*d^2)/4)*Sqrt[Sinh[c + d*x]]) + E^(a + b*x)*Sinh[c + d*x]^(3/2),x]

[Out]

(-6*d*E^(a + b*x)*Cosh[c + d*x]*Sqrt[Sinh[c + d*x]])/(4*b^2 - 9*d^2) + (4*b*E^(a + b*x)*Sinh[c + d*x]^(3/2))/(
4*b^2 - 9*d^2)

Rule 2283

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[a^p*(G^(h*(f + g*x))/(g*h*Log[G]))*Hypergeometric2F1[-p, g*h*(Log[G]/(d*e*Log[F])), g*h*(Log[G]/(d*e*Log[F]))
+ 1, Simplify[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] || G
tQ[a, 0])

Rule 2284

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Dist
[(a + b*F^(e*(c + d*x)))^p/(1 + (b/a)*F^(e*(c + d*x)))^p, Int[G^(h*(f + g*x))*(1 + (b/a)*F^(e*(c + d*x)))^p, x
], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 2285

Int[((a_) + (b_.)*(F_)^((e_.)*(v_)))^(p_)*(G_)^((h_.)*(u_)), x_Symbol] :> Int[G^(h*ExpandToSum[u, x])*(a + b*F
^(e*ExpandToSum[v, x]))^p, x] /; FreeQ[{F, G, a, b, e, h, p}, x] && LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v
}, x]

Rule 5584

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(-b)*c*Log[F]*F^(c*(a +
b*x))*(Sinh[d + e*x]^n/(e^2*n^2 - b^2*c^2*Log[F]^2)), x] + (-Dist[n*(n - 1)*(e^2/(e^2*n^2 - b^2*c^2*Log[F]^2)
), Int[F^(c*(a + b*x))*Sinh[d + e*x]^(n - 2), x], x] + Simp[e*n*F^(c*(a + b*x))*Cosh[d + e*x]*(Sinh[d + e*x]^(
n - 1)/(e^2*n^2 - b^2*c^2*Log[F]^2)), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 - b^2*c^2*Log[F]^2, 0
] && GtQ[n, 1]

Rule 5590

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Dist[E^(n*(d + e*x))*(Sinh[d
+ e*x]^n/(-1 + E^(2*(d + e*x)))^n), Int[F^(c*(a + b*x))*((-1 + E^(2*(d + e*x)))^n/E^(n*(d + e*x))), x], x] /;
FreeQ[{F, a, b, c, d, e, n}, x] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \left (-\frac {3 d^2 e^{a+b x}}{4 \left (b^2-\frac {9 d^2}{4}\right ) \sqrt {\sinh (c+d x)}}+e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)\right ) \, dx &=-\frac {\left (3 d^2\right ) \int \frac {e^{a+b x}}{\sqrt {\sinh (c+d x)}} \, dx}{4 b^2-9 d^2}+\int e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x) \, dx\\ &=-\frac {6 d e^{a+b x} \cosh (c+d x) \sqrt {\sinh (c+d x)}}{4 b^2-9 d^2}+\frac {4 b e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)}{4 b^2-9 d^2}+\frac {\left (3 d^2\right ) \int \frac {e^{a+b x}}{\sqrt {\sinh (c+d x)}} \, dx}{4 b^2-9 d^2}-\frac {\left (3 d^2 e^{\frac {1}{2} (-c-d x)} \sqrt {-1+e^{2 (c+d x)}}\right ) \int \frac {e^{a+b x+\frac {1}{2} (c+d x)}}{\sqrt {-1+e^{2 (c+d x)}}} \, dx}{\left (4 b^2-9 d^2\right ) \sqrt {\sinh (c+d x)}}\\ &=-\frac {6 d e^{a+b x} \cosh (c+d x) \sqrt {\sinh (c+d x)}}{4 b^2-9 d^2}+\frac {4 b e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)}{4 b^2-9 d^2}-\frac {\left (3 d^2 e^{\frac {1}{2} (-c-d x)} \sqrt {-1+e^{2 (c+d x)}}\right ) \int \frac {e^{\frac {1}{2} (2 a+c)+\frac {1}{2} (2 b+d) x}}{\sqrt {-1+e^{2 (c+d x)}}} \, dx}{\left (4 b^2-9 d^2\right ) \sqrt {\sinh (c+d x)}}+\frac {\left (3 d^2 e^{\frac {1}{2} (-c-d x)} \sqrt {-1+e^{2 (c+d x)}}\right ) \int \frac {e^{a+b x+\frac {1}{2} (c+d x)}}{\sqrt {-1+e^{2 (c+d x)}}} \, dx}{\left (4 b^2-9 d^2\right ) \sqrt {\sinh (c+d x)}}\\ &=-\frac {6 d e^{a+b x} \cosh (c+d x) \sqrt {\sinh (c+d x)}}{4 b^2-9 d^2}+\frac {4 b e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)}{4 b^2-9 d^2}-\frac {\left (3 d^2 e^{\frac {1}{2} (-c-d x)} \sqrt {1-e^{2 (c+d x)}}\right ) \int \frac {e^{\frac {1}{2} (2 a+c)+\frac {1}{2} (2 b+d) x}}{\sqrt {1-e^{2 (c+d x)}}} \, dx}{\left (4 b^2-9 d^2\right ) \sqrt {\sinh (c+d x)}}+\frac {\left (3 d^2 e^{\frac {1}{2} (-c-d x)} \sqrt {-1+e^{2 (c+d x)}}\right ) \int \frac {e^{\frac {1}{2} (2 a+c)+\frac {1}{2} (2 b+d) x}}{\sqrt {-1+e^{2 (c+d x)}}} \, dx}{\left (4 b^2-9 d^2\right ) \sqrt {\sinh (c+d x)}}\\ &=-\frac {6 d^2 \exp \left (\frac {1}{2} (2 a+c)+\frac {1}{2} (2 b+d) x+\frac {1}{2} (-c-d x)\right ) \sqrt {1-e^{2 (c+d x)}} \, _2F_1\left (\frac {1}{2},\frac {2 b+d}{4 d};\frac {1}{4} \left (5+\frac {2 b}{d}\right );e^{2 (c+d x)}\right )}{(2 b+d) \left (4 b^2-9 d^2\right ) \sqrt {\sinh (c+d x)}}-\frac {6 d e^{a+b x} \cosh (c+d x) \sqrt {\sinh (c+d x)}}{4 b^2-9 d^2}+\frac {4 b e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)}{4 b^2-9 d^2}+\frac {\left (3 d^2 e^{\frac {1}{2} (-c-d x)} \sqrt {1-e^{2 (c+d x)}}\right ) \int \frac {e^{\frac {1}{2} (2 a+c)+\frac {1}{2} (2 b+d) x}}{\sqrt {1-e^{2 (c+d x)}}} \, dx}{\left (4 b^2-9 d^2\right ) \sqrt {\sinh (c+d x)}}\\ &=-\frac {6 d e^{a+b x} \cosh (c+d x) \sqrt {\sinh (c+d x)}}{4 b^2-9 d^2}+\frac {4 b e^{a+b x} \sinh ^{\frac {3}{2}}(c+d x)}{4 b^2-9 d^2}\\ \end {align*}

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Mathematica [A]
time = 1.26, size = 51, normalized size = 0.70 \begin {gather*} \frac {2 e^{a+b x} \sqrt {\sinh (c+d x)} (-3 d \cosh (c+d x)+2 b \sinh (c+d x))}{4 b^2-9 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*d^2*E^(a + b*x))/(4*(b^2 - (9*d^2)/4)*Sqrt[Sinh[c + d*x]]) + E^(a + b*x)*Sinh[c + d*x]^(3/2),x]

[Out]

(2*E^(a + b*x)*Sqrt[Sinh[c + d*x]]*(-3*d*Cosh[c + d*x] + 2*b*Sinh[c + d*x]))/(4*b^2 - 9*d^2)

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Maple [F]
time = 4.05, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{b x +a} \left (\sinh ^{\frac {3}{2}}\left (d x +c \right )\right )-\frac {3 d^{2} {\mathrm e}^{b x +a}}{4 \left (b^{2}-\frac {9 d^{2}}{4}\right ) \sqrt {\sinh \left (d x +c \right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*sinh(d*x+c)^(3/2)-3/4*d^2*exp(b*x+a)/(b^2-9/4*d^2)/sinh(d*x+c)^(1/2),x)

[Out]

int(exp(b*x+a)*sinh(d*x+c)^(3/2)-3/4*d^2*exp(b*x+a)/(b^2-9/4*d^2)/sinh(d*x+c)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sinh(d*x+c)^(3/2)-3/4*d^2*exp(b*x+a)/(b^2-9/4*d^2)/sinh(d*x+c)^(1/2),x, algorithm="maxima
")

[Out]

integrate(e^(b*x + a)*sinh(d*x + c)^(3/2) - 3*d^2*e^(b*x + a)/((4*b^2 - 9*d^2)*sqrt(sinh(d*x + c))), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sinh(d*x+c)^(3/2)-3/4*d^2*exp(b*x+a)/(b^2-9/4*d^2)/sinh(d*x+c)^(1/2),x, algorithm="fricas
")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\left (\int 4 b^{2} e^{b x} \sinh ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \left (- \frac {3 d^{2} e^{b x}}{\sqrt {\sinh {\left (c + d x \right )}}}\right )\, dx + \int \left (- 9 d^{2} e^{b x} \sinh ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx\right ) e^{a}}{\left (2 b - 3 d\right ) \left (2 b + 3 d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sinh(d*x+c)**(3/2)-3/4*d**2*exp(b*x+a)/(b**2-9/4*d**2)/sinh(d*x+c)**(1/2),x)

[Out]

(Integral(4*b**2*exp(b*x)*sinh(c + d*x)**(3/2), x) + Integral(-3*d**2*exp(b*x)/sqrt(sinh(c + d*x)), x) + Integ
ral(-9*d**2*exp(b*x)*sinh(c + d*x)**(3/2), x))*exp(a)/((2*b - 3*d)*(2*b + 3*d))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sinh(d*x+c)^(3/2)-3/4*d^2*exp(b*x+a)/(b^2-9/4*d^2)/sinh(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate(e^(b*x + a)*sinh(d*x + c)^(3/2) - 3*d^2*e^(b*x + a)/((4*b^2 - 9*d^2)*sqrt(sinh(d*x + c))), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{a+b\,x}\,{\mathrm {sinh}\left (c+d\,x\right )}^{3/2}-\frac {3\,d^2\,{\mathrm {e}}^{a+b\,x}}{4\,\sqrt {\mathrm {sinh}\left (c+d\,x\right )}\,\left (b^2-\frac {9\,d^2}{4}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)*sinh(c + d*x)^(3/2) - (3*d^2*exp(a + b*x))/(4*sinh(c + d*x)^(1/2)*(b^2 - (9*d^2)/4)),x)

[Out]

int(exp(a + b*x)*sinh(c + d*x)^(3/2) - (3*d^2*exp(a + b*x))/(4*sinh(c + d*x)^(1/2)*(b^2 - (9*d^2)/4)), x)

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