3.52 \(\int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx\)

Optimal. Leaf size=251 \[ \frac {32 \sqrt {2 \pi } b^{7/2} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{105 d^{9/2}}+\frac {32 \sqrt {2 \pi } b^{7/2} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{105 d^{9/2}}-\frac {128 b^3 \sinh (a+b x) \cosh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}} \]

[Out]

-16/105*b^2/d^3/(d*x+c)^(3/2)-8/35*b*cosh(b*x+a)*sinh(b*x+a)/d^2/(d*x+c)^(5/2)-2/7*sinh(b*x+a)^2/d/(d*x+c)^(7/
2)-32/105*b^2*sinh(b*x+a)^2/d^3/(d*x+c)^(3/2)+32/105*b^(7/2)*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^(9/2)+32/105*b^(7/2)*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^(9/2)-128/105*b^3*cosh(b*x+a)*sinh(b*x+a)/d^4/(d*x+c)^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 11, 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} \[ \frac {32 \sqrt {2 \pi } b^{7/2} e^{\frac {2 b c}{d}-2 a} \text {Erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{105 d^{9/2}}+\frac {32 \sqrt {2 \pi } b^{7/2} e^{2 a-\frac {2 b c}{d}} \text {Erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{105 d^{9/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}-\frac {128 b^3 \sinh (a+b x) \cosh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^2/(c + d*x)^(9/2),x]

[Out]

(-16*b^2)/(105*d^3*(c + d*x)^(3/2)) + (32*b^(7/2)*E^(-2*a + (2*b*c)/d)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[b]*Sqrt[c
+ d*x])/Sqrt[d]])/(105*d^(9/2)) + (32*b^(7/2)*E^(2*a - (2*b*c)/d)*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*
x])/Sqrt[d]])/(105*d^(9/2)) - (8*b*Cosh[a + b*x]*Sinh[a + b*x])/(35*d^2*(c + d*x)^(5/2)) - (128*b^3*Cosh[a + b
*x]*Sinh[a + b*x])/(105*d^4*Sqrt[c + d*x]) - (2*Sinh[a + b*x]^2)/(7*d*(c + d*x)^(7/2)) - (32*b^2*Sinh[a + b*x]
^2)/(105*d^3*(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)^{9/2}} \, dx &=-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}+\frac {\left (8 b^2\right ) \int \frac {1}{(c+d x)^{5/2}} \, dx}{35 d^2}+\frac {\left (16 b^2\right ) \int \frac {\sinh ^2(a+b x)}{(c+d x)^{5/2}} \, dx}{35 d^2}\\ &=-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {128 b^3 \cosh (a+b x) \sinh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}+\frac {\left (128 b^4\right ) \int \frac {1}{\sqrt {c+d x}} \, dx}{105 d^4}+\frac {\left (256 b^4\right ) \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}} \, dx}{105 d^4}\\ &=-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}+\frac {256 b^4 \sqrt {c+d x}}{105 d^5}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {128 b^3 \cosh (a+b x) \sinh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}-\frac {\left (256 b^4\right ) \int \left (\frac {1}{2 \sqrt {c+d x}}-\frac {\cosh (2 a+2 b x)}{2 \sqrt {c+d x}}\right ) \, dx}{105 d^4}\\ &=-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {128 b^3 \cosh (a+b x) \sinh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}+\frac {\left (128 b^4\right ) \int \frac {\cosh (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{105 d^4}\\ &=-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {128 b^3 \cosh (a+b x) \sinh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}+\frac {\left (64 b^4\right ) \int \frac {e^{-i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{105 d^4}+\frac {\left (64 b^4\right ) \int \frac {e^{i (2 i a+2 i b x)}}{\sqrt {c+d x}} \, dx}{105 d^4}\\ &=-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {128 b^3 \cosh (a+b x) \sinh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}+\frac {\left (128 b^4\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 )}{105 d^5}+\frac {\left (128 b^4\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 )}{105 d^5}\\ &=-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}+\frac {32 b^{7/2} 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 )}{105 d^{9/2}}+\frac {32 b^{7/2} 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 )}{105 d^{9/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {128 b^3 \cosh (a+b x) \sinh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 222, normalized size = 0.88 \[ \frac {2 \left (-32 b^3 (c+d x)^3 \sinh (2 (a+b x))+16 \sqrt {2} b^3 (c+d x)^3 e^{2 a-\frac {2 b c}{d}} \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )-16 \sqrt {2} b^3 (c+d x)^3 e^{\frac {2 b c}{d}-2 a} \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )-16 b^2 d (c+d x)^2 \sinh ^2(a+b x)-6 b d^2 (c+d x) \sinh (2 (a+b x))-15 d^3 \sinh ^2(a+b x)-8 b^2 d (c+d x)^2\right )}{105 d^4 (c+d x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^2/(c + d*x)^(9/2),x]

[Out]

(2*(-8*b^2*d*(c + d*x)^2 + 16*Sqrt[2]*b^3*E^(2*a - (2*b*c)/d)*(c + d*x)^3*Sqrt[-((b*(c + d*x))/d)]*Gamma[1/2,
(-2*b*(c + d*x))/d] - 16*Sqrt[2]*b^3*E^(-2*a + (2*b*c)/d)*(c + d*x)^3*Sqrt[(b*(c + d*x))/d]*Gamma[1/2, (2*b*(c
+ d*x))/d] - 15*d^3*Sinh[a + b*x]^2 - 16*b^2*d*(c + d*x)^2*Sinh[a + b*x]^2 - 6*b*d^2*(c + d*x)*Sinh[2*(a + b*
x)] - 32*b^3*(c + d*x)^3*Sinh[2*(a + b*x)]))/(105*d^4*(c + d*x)^(7/2))

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fricas [B]  time = 0.58, size = 1827, normalized size = 7.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(64*sqrt(2)*sqrt(pi)*((b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^3*c^3*d*x + b^3*c^4)*cosh
(b*x + a)^2*cosh(-2*(b*c - a*d)/d) - (b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^3*c^3*d*x + b^3*
c^4)*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d) + ((b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^3*c^3*
d*x + b^3*c^4)*cosh(-2*(b*c - a*d)/d) - (b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^3*c^3*d*x + b
^3*c^4)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^
3*c^3*d*x + b^3*c^4)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) - (b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2
+ 4*b^3*c^3*d*x + b^3*c^4)*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)) - 64*sqrt(2)*sqrt(pi)*((b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^3*c^3*d*x +
b^3*c^4)*cosh(b*x + a)^2*cosh(-2*(b*c - a*d)/d) + (b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^3*c
^3*d*x + b^3*c^4)*cosh(b*x + a)^2*sinh(-2*(b*c - a*d)/d) + ((b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2
+ 4*b^3*c^3*d*x + b^3*c^4)*cosh(-2*(b*c - a*d)/d) + (b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^
3*c^3*d*x + b^3*c^4)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a)^2 + 2*((b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d
^2*x^2 + 4*b^3*c^3*d*x + b^3*c^4)*cosh(b*x + a)*cosh(-2*(b*c - a*d)/d) + (b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^
3*c^2*d^2*x^2 + 4*b^3*c^3*d*x + b^3*c^4)*cosh(b*x + a)*sinh(-2*(b*c - a*d)/d))*sinh(b*x + a))*sqrt(-b/d)*erf(s
qrt(2)*sqrt(d*x + c)*sqrt(-b/d)) + (64*b^3*d^3*x^3 + 64*b^3*c^3 - 16*b^2*c^2*d + 30*d^3*cosh(b*x + a)^2 - (64*
b^3*d^3*x^3 + 64*b^3*c^3 + 16*b^2*c^2*d + 12*b*c*d^2 + 15*d^3 + 16*(12*b^3*c*d^2 + b^2*d^3)*x^2 + 4*(48*b^3*c^
2*d + 8*b^2*c*d^2 + 3*b*d^3)*x)*cosh(b*x + a)^4 - 4*(64*b^3*d^3*x^3 + 64*b^3*c^3 + 16*b^2*c^2*d + 12*b*c*d^2 +
15*d^3 + 16*(12*b^3*c*d^2 + b^2*d^3)*x^2 + 4*(48*b^3*c^2*d + 8*b^2*c*d^2 + 3*b*d^3)*x)*cosh(b*x + a)*sinh(b*x
+ a)^3 - (64*b^3*d^3*x^3 + 64*b^3*c^3 + 16*b^2*c^2*d + 12*b*c*d^2 + 15*d^3 + 16*(12*b^3*c*d^2 + b^2*d^3)*x^2
+ 4*(48*b^3*c^2*d + 8*b^2*c*d^2 + 3*b*d^3)*x)*sinh(b*x + a)^4 + 12*b*c*d^2 - 15*d^3 + 16*(12*b^3*c*d^2 - b^2*d
^3)*x^2 + 6*(5*d^3 - (64*b^3*d^3*x^3 + 64*b^3*c^3 + 16*b^2*c^2*d + 12*b*c*d^2 + 15*d^3 + 16*(12*b^3*c*d^2 + b^
2*d^3)*x^2 + 4*(48*b^3*c^2*d + 8*b^2*c*d^2 + 3*b*d^3)*x)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*(48*b^3*c^2*d -
8*b^2*c*d^2 + 3*b*d^3)*x + 4*(15*d^3*cosh(b*x + a) - (64*b^3*d^3*x^3 + 64*b^3*c^3 + 16*b^2*c^2*d + 12*b*c*d^2
+ 15*d^3 + 16*(12*b^3*c*d^2 + b^2*d^3)*x^2 + 4*(48*b^3*c^2*d + 8*b^2*c*d^2 + 3*b*d^3)*x)*cosh(b*x + a)^3)*sinh
(b*x + a))*sqrt(d*x + c))/((d^8*x^4 + 4*c*d^7*x^3 + 6*c^2*d^6*x^2 + 4*c^3*d^5*x + c^4*d^4)*cosh(b*x + a)^2 + 2
*(d^8*x^4 + 4*c*d^7*x^3 + 6*c^2*d^6*x^2 + 4*c^3*d^5*x + c^4*d^4)*cosh(b*x + a)*sinh(b*x + a) + (d^8*x^4 + 4*c*
d^7*x^3 + 6*c^2*d^6*x^2 + 4*c^3*d^5*x + c^4*d^4)*sinh(b*x + a)^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x + a)^2/(d*x + c)^(9/2), x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{2}\left (b x +a \right )}{\left (d x +c \right )^{\frac {9}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^2/(d*x+c)^(9/2),x)

[Out]

int(sinh(b*x+a)^2/(d*x+c)^(9/2),x)

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maxima [A]  time = 0.52, size = 118, normalized size = 0.47 \[ -\frac {\frac {14 \, \sqrt {2} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {7}{2}} e^{\left (\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {7}{2}, \frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {7}{2}}} + \frac {14 \, \sqrt {2} \left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {7}{2}} e^{\left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {7}{2}, -\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {7}{2}}} - \frac {1}{{\left (d x + c\right )}^{\frac {7}{2}}}}{7 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*(14*sqrt(2)*((d*x + c)*b/d)^(7/2)*e^(2*(b*c - a*d)/d)*gamma(-7/2, 2*(d*x + c)*b/d)/(d*x + c)^(7/2) + 14*s
qrt(2)*(-(d*x + c)*b/d)^(7/2)*e^(-2*(b*c - a*d)/d)*gamma(-7/2, -2*(d*x + c)*b/d)/(d*x + c)^(7/2) - 1/(d*x + c)
^(7/2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^2/(c + d*x)^(9/2),x)

[Out]

int(sinh(a + b*x)^2/(c + d*x)^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)**2/(d*x+c)**(9/2),x)

[Out]

Timed out

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