Integrand size = 18, antiderivative size = 251 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx=-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}+\frac {32 b^{7/2} e^{-2 a+\frac {2 b c}{d}} \sqrt {2 \pi } \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{105 d^{9/2}}+\frac {32 b^{7/2} e^{2 a-\frac {2 b c}{d}} \sqrt {2 \pi } \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{105 d^{9/2}}-\frac {8 b \cosh (a+b x) \sinh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {128 b^3 \cosh (a+b x) \sinh (a+b x)}{105 d^4 \sqrt {c+d x}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {32 b^2 \sinh ^2(a+b x)}{105 d^3 (c+d x)^{3/2}} \]
-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)
Time = 0.36 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.88 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx=\frac {2 \left (-8 b^2 d (c+d x)^2+16 \sqrt {2} b^3 e^{2 a-\frac {2 b c}{d}} (c+d x)^3 \sqrt {-\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},-\frac {2 b (c+d x)}{d}\right )-16 \sqrt {2} b^3 e^{-2 a+\frac {2 b c}{d}} (c+d x)^3 \sqrt {\frac {b (c+d x)}{d}} \Gamma \left (\frac {1}{2},\frac {2 b (c+d x)}{d}\right )-15 d^3 \sinh ^2(a+b x)-16 b^2 d (c+d x)^2 \sinh ^2(a+b x)-6 b d^2 (c+d x) \sinh (2 (a+b x))-32 b^3 (c+d x)^3 \sinh (2 (a+b x))\right )}{105 d^4 (c+d x)^{7/2}} \]
(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))
Time = 0.83 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.21, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {3042, 25, 3795, 17, 25, 3042, 25, 3795, 17, 25, 3042, 25, 3793, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin (i a+i b x)^2}{(c+d x)^{9/2}}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin (i a+i b x)^2}{(c+d x)^{9/2}}dx\) |
| \(\Big \downarrow \) 3795 |
| \(\displaystyle -\frac {16 b^2 \int -\frac {\sinh ^2(a+b x)}{(c+d x)^{5/2}}dx}{35 d^2}+\frac {8 b^2 \int \frac {1}{(c+d x)^{5/2}}dx}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle -\frac {16 b^2 \int -\frac {\sinh ^2(a+b x)}{(c+d x)^{5/2}}dx}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {16 b^2 \int \frac {\sinh ^2(a+b x)}{(c+d x)^{5/2}}dx}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {16 b^2 \int -\frac {\sin (i a+i b x)^2}{(c+d x)^{5/2}}dx}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {16 b^2 \int \frac {\sin (i a+i b x)^2}{(c+d x)^{5/2}}dx}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 3795 |
| \(\displaystyle -\frac {16 b^2 \left (\frac {16 b^2 \int -\frac {\sinh ^2(a+b x)}{\sqrt {c+d x}}dx}{3 d^2}-\frac {8 b^2 \int \frac {1}{\sqrt {c+d x}}dx}{3 d^2}+\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}\right )}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle -\frac {16 b^2 \left (\frac {16 b^2 \int -\frac {\sinh ^2(a+b x)}{\sqrt {c+d x}}dx}{3 d^2}+\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\right )}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {16 b^2 \int \frac {\sinh ^2(a+b x)}{\sqrt {c+d x}}dx}{3 d^2}+\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\right )}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {16 b^2 \left (-\frac {16 b^2 \int -\frac {\sin (i a+i b x)^2}{\sqrt {c+d x}}dx}{3 d^2}+\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\right )}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {16 b^2 \left (\frac {16 b^2 \int \frac {\sin (i a+i b x)^2}{\sqrt {c+d x}}dx}{3 d^2}+\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\right )}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 3793 |
| \(\displaystyle -\frac {16 b^2 \left (\frac {16 b^2 \int \left (\frac {1}{2 \sqrt {c+d x}}-\frac {\cosh (2 a+2 b x)}{2 \sqrt {c+d x}}\right )dx}{3 d^2}+\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\right )}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {16 b^2 \left (\frac {16 b^2 \left (-\frac {\sqrt {\frac {\pi }{2}} e^{\frac {2 b c}{d}-2 a} \text {erf}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}-\frac {\sqrt {\frac {\pi }{2}} e^{2 a-\frac {2 b c}{d}} \text {erfi}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 \sqrt {b} \sqrt {d}}+\frac {\sqrt {c+d x}}{d}\right )}{3 d^2}+\frac {8 b \sinh (a+b x) \cosh (a+b x)}{3 d^2 \sqrt {c+d x}}+\frac {2 \sinh ^2(a+b x)}{3 d (c+d x)^{3/2}}-\frac {16 b^2 \sqrt {c+d x}}{3 d^3}\right )}{35 d^2}-\frac {8 b \sinh (a+b x) \cosh (a+b x)}{35 d^2 (c+d x)^{5/2}}-\frac {2 \sinh ^2(a+b x)}{7 d (c+d x)^{7/2}}-\frac {16 b^2}{105 d^3 (c+d x)^{3/2}}\) |
(-16*b^2)/(105*d^3*(c + d*x)^(3/2)) - (8*b*Cosh[a + b*x]*Sinh[a + b*x])/(3 5*d^2*(c + d*x)^(5/2)) - (2*Sinh[a + b*x]^2)/(7*d*(c + d*x)^(7/2)) - (16*b ^2*((-16*b^2*Sqrt[c + d*x])/(3*d^3) + (16*b^2*(Sqrt[c + d*x]/d - (E^(-2*a + (2*b*c)/d)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]])/(4*S qrt[b]*Sqrt[d]) - (E^(2*a - (2*b*c)/d)*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[b]*Sq rt[c + d*x])/Sqrt[d]])/(4*Sqrt[b]*Sqrt[d])))/(3*d^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))))/(35*d^2)
3.1.52.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[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]))
Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo l] :> Simp[(c + d*x)^(m + 1)*((b*Sin[e + f*x])^n/(d*(m + 1))), 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] + Simp[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] - Simp[f^2*(n^2/(d^2*(m + 1)* (m + 2))) Int[(c + d*x)^(m + 2)*(b*Sin[e + f*x])^n, x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && LtQ[m, -2]
\[\int \frac {\sinh \left (b x +a \right )^{2}}{\left (d x +c \right )^{\frac {9}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 1827 vs. \(2 (199) = 398\).
Time = 0.28 (sec) , antiderivative size = 1827, normalized size of antiderivative = 7.28 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx=\text {Too large to display} \]
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))*sin h(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...
Timed out. \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.47 \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx=-\frac {\frac {14 \, \sqrt {2} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {7}{2}} e^{\left (\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {7}{2}, \frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {7}{2}}} + \frac {14 \, \sqrt {2} \left (-\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {7}{2}} e^{\left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )} \Gamma \left (-\frac {7}{2}, -\frac {2 \, {\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )}^{\frac {7}{2}}} - \frac {1}{{\left (d x + c\right )}^{\frac {7}{2}}}}{7 \, d} \]
-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*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) - 1/(d*x + c)^( 7/2))/d
\[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx=\int { \frac {\sinh \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sinh ^2(a+b x)}{(c+d x)^{9/2}} \, dx=\int \frac {{\mathrm {sinh}\left (a+b\,x\right )}^2}{{\left (c+d\,x\right )}^{9/2}} \,d x \]