Integrand size = 28, antiderivative size = 494 \[ \int (f+g x) \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {b d^2 g x \sqrt {d+c^2 d x^2}}{7 c \sqrt {1+c^2 x^2}}-\frac {25 b c d^2 f x^2 \sqrt {d+c^2 d x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {b c d^2 g x^3 \sqrt {d+c^2 d x^2}}{7 \sqrt {1+c^2 x^2}}-\frac {5 b c^3 d^2 f x^4 \sqrt {d+c^2 d x^2}}{96 \sqrt {1+c^2 x^2}}-\frac {3 b c^3 d^2 g x^5 \sqrt {d+c^2 d x^2}}{35 \sqrt {1+c^2 x^2}}-\frac {b c^5 d^2 g x^7 \sqrt {d+c^2 d x^2}}{49 \sqrt {1+c^2 x^2}}-\frac {b d^2 f \left (1+c^2 x^2\right )^{5/2} \sqrt {d+c^2 d x^2}}{36 c}+\frac {5}{16} d^2 f x \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))+\frac {5}{24} d^2 f x \left (1+c^2 x^2\right ) \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))+\frac {1}{6} d^2 f x \left (1+c^2 x^2\right )^2 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))+\frac {d^2 g \left (1+c^2 x^2\right )^3 \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))}{7 c^2}+\frac {5 d^2 f \sqrt {d+c^2 d x^2} (a+b \text {arcsinh}(c x))^2}{32 b c \sqrt {1+c^2 x^2}} \]
-1/36*b*d^2*f*(c^2*x^2+1)^(5/2)*(c^2*d*x^2+d)^(1/2)/c+5/16*d^2*f*x*(a+b*ar csinh(c*x))*(c^2*d*x^2+d)^(1/2)+5/24*d^2*f*x*(c^2*x^2+1)*(a+b*arcsinh(c*x) )*(c^2*d*x^2+d)^(1/2)+1/6*d^2*f*x*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))*(c^2*d* x^2+d)^(1/2)+1/7*d^2*g*(c^2*x^2+1)^3*(a+b*arcsinh(c*x))*(c^2*d*x^2+d)^(1/2 )/c^2-1/7*b*d^2*g*x*(c^2*d*x^2+d)^(1/2)/c/(c^2*x^2+1)^(1/2)-25/96*b*c*d^2* f*x^2*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/7*b*c*d^2*g*x^3*(c^2*d*x^2+d )^(1/2)/(c^2*x^2+1)^(1/2)-5/96*b*c^3*d^2*f*x^4*(c^2*d*x^2+d)^(1/2)/(c^2*x^ 2+1)^(1/2)-3/35*b*c^3*d^2*g*x^5*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)-1/49 *b*c^5*d^2*g*x^7*(c^2*d*x^2+d)^(1/2)/(c^2*x^2+1)^(1/2)+5/32*d^2*f*(a+b*arc sinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/b/c/(c^2*x^2+1)^(1/2)
Time = 0.89 (sec) , antiderivative size = 390, normalized size of antiderivative = 0.79 \[ \int (f+g x) \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {-d^3 \left (1+c^2 x^2\right ) \left (-1680 a \sqrt {1+c^2 x^2} \left (48 g \left (1+c^2 x^2\right )^3+7 c^2 f x \left (33+26 c^2 x^2+8 c^4 x^4\right )\right )+b c \left (2304 g x \left (35+35 c^2 x^2+21 c^4 x^4+5 c^6 x^6\right )+245 f \left (299+792 c^2 x^2+312 c^4 x^4+64 c^6 x^6\right )\right )\right )+88200 b c d^3 f \left (1+c^2 x^2\right ) \text {arcsinh}(c x)^2+176400 a c d^{5/2} f \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )+420 b d^3 \left (1+c^2 x^2\right ) \text {arcsinh}(c x) \left (192 g \sqrt {1+c^2 x^2}+576 c^2 g x^2 \sqrt {1+c^2 x^2}+576 c^4 g x^4 \sqrt {1+c^2 x^2}+192 c^6 g x^6 \sqrt {1+c^2 x^2}+315 c f \sinh (2 \text {arcsinh}(c x))+63 c f \sinh (4 \text {arcsinh}(c x))+7 c f \sinh (6 \text {arcsinh}(c x))\right )}{564480 c^2 \sqrt {1+c^2 x^2} \sqrt {d+c^2 d x^2}} \]
(-(d^3*(1 + c^2*x^2)*(-1680*a*Sqrt[1 + c^2*x^2]*(48*g*(1 + c^2*x^2)^3 + 7* c^2*f*x*(33 + 26*c^2*x^2 + 8*c^4*x^4)) + b*c*(2304*g*x*(35 + 35*c^2*x^2 + 21*c^4*x^4 + 5*c^6*x^6) + 245*f*(299 + 792*c^2*x^2 + 312*c^4*x^4 + 64*c^6* x^6)))) + 88200*b*c*d^3*f*(1 + c^2*x^2)*ArcSinh[c*x]^2 + 176400*a*c*d^(5/2 )*f*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2 *d*x^2]] + 420*b*d^3*(1 + c^2*x^2)*ArcSinh[c*x]*(192*g*Sqrt[1 + c^2*x^2] + 576*c^2*g*x^2*Sqrt[1 + c^2*x^2] + 576*c^4*g*x^4*Sqrt[1 + c^2*x^2] + 192*c ^6*g*x^6*Sqrt[1 + c^2*x^2] + 315*c*f*Sinh[2*ArcSinh[c*x]] + 63*c*f*Sinh[4* ArcSinh[c*x]] + 7*c*f*Sinh[6*ArcSinh[c*x]]))/(564480*c^2*Sqrt[1 + c^2*x^2] *Sqrt[d + c^2*d*x^2])
Time = 0.69 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.50, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6260, 6253, 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.
\(\displaystyle \int \left (c^2 d x^2+d\right )^{5/2} (f+g x) (a+b \text {arcsinh}(c x)) \, dx\) |
\(\Big \downarrow \) 6260 |
\(\displaystyle \frac {d^2 \sqrt {c^2 d x^2+d} \int (f+g x) \left (c^2 x^2+1\right )^{5/2} (a+b \text {arcsinh}(c x))dx}{\sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 6253 |
\(\displaystyle \frac {d^2 \sqrt {c^2 d x^2+d} \int \left (f (a+b \text {arcsinh}(c x)) \left (c^2 x^2+1\right )^{5/2}+g x (a+b \text {arcsinh}(c x)) \left (c^2 x^2+1\right )^{5/2}\right )dx}{\sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d^2 \sqrt {c^2 d x^2+d} \left (\frac {1}{6} f x \left (c^2 x^2+1\right )^{5/2} (a+b \text {arcsinh}(c x))+\frac {5}{24} f x \left (c^2 x^2+1\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {5}{16} f x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))+\frac {g \left (c^2 x^2+1\right )^{7/2} (a+b \text {arcsinh}(c x))}{7 c^2}+\frac {5 f (a+b \text {arcsinh}(c x))^2}{32 b c}-\frac {1}{49} b c^5 g x^7-\frac {5}{96} b c^3 f x^4-\frac {3}{35} b c^3 g x^5-\frac {b f \left (c^2 x^2+1\right )^3}{36 c}-\frac {25}{96} b c f x^2-\frac {1}{7} b c g x^3-\frac {b g x}{7 c}\right )}{\sqrt {c^2 x^2+1}}\) |
(d^2*Sqrt[d + c^2*d*x^2]*(-1/7*(b*g*x)/c - (25*b*c*f*x^2)/96 - (b*c*g*x^3) /7 - (5*b*c^3*f*x^4)/96 - (3*b*c^3*g*x^5)/35 - (b*c^5*g*x^7)/49 - (b*f*(1 + c^2*x^2)^3)/(36*c) + (5*f*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/16 + (5*f*x*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/24 + (f*x*(1 + c^2*x^2)^ (5/2)*(a + b*ArcSinh[c*x]))/6 + (g*(1 + c^2*x^2)^(7/2)*(a + b*ArcSinh[c*x] ))/(7*c^2) + (5*f*(a + b*ArcSinh[c*x])^2)/(32*b*c)))/Sqrt[1 + c^2*x^2]
3.1.45.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d _) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n , 0] && ((EqQ[n, 1] && GtQ[p, -1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] && LtQ[p, -2]))
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d _) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[Simp[(d + e*x^2)^p/(1 + c^2*x^2) ^p] Int[(f + g*x)^m*(1 + c^2*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && IntegerQ [p - 1/2] && !GtQ[d, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1678\) vs. \(2(430)=860\).
Time = 1.00 (sec) , antiderivative size = 1679, normalized size of antiderivative = 3.40
method | result | size |
default | \(\text {Expression too large to display}\) | \(1679\) |
parts | \(\text {Expression too large to display}\) | \(1679\) |
1/6*a*f*x*(c^2*d*x^2+d)^(5/2)+5/24*a*f*d*x*(c^2*d*x^2+d)^(3/2)+5/16*a*f*d^ 2*x*(c^2*d*x^2+d)^(1/2)+5/16*a*f*d^3*ln(c^2*d*x/(c^2*d)^(1/2)+(c^2*d*x^2+d )^(1/2))/(c^2*d)^(1/2)+1/7*a*g*(c^2*d*x^2+d)^(7/2)/c^2/d+b*(5/32*(d*(c^2*x ^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c*f*arcsinh(c*x)^2*d^2+1/6272*(d*(c^2*x^2+1 ))^(1/2)*(64*c^8*x^8+64*c^7*x^7*(c^2*x^2+1)^(1/2)+144*c^6*x^6+112*c^5*x^5* (c^2*x^2+1)^(1/2)+104*c^4*x^4+56*c^3*x^3*(c^2*x^2+1)^(1/2)+25*c^2*x^2+7*c* x*(c^2*x^2+1)^(1/2)+1)*g*(-1+7*arcsinh(c*x))*d^2/c^2/(c^2*x^2+1)+1/2304*(d *(c^2*x^2+1))^(1/2)*(32*c^7*x^7+32*c^6*x^6*(c^2*x^2+1)^(1/2)+64*c^5*x^5+48 *c^4*x^4*(c^2*x^2+1)^(1/2)+38*c^3*x^3+18*c^2*x^2*(c^2*x^2+1)^(1/2)+6*c*x+( c^2*x^2+1)^(1/2))*f*(-1+6*arcsinh(c*x))*d^2/c/(c^2*x^2+1)+1/640*(d*(c^2*x^ 2+1))^(1/2)*(16*c^6*x^6+16*c^5*x^5*(c^2*x^2+1)^(1/2)+28*c^4*x^4+20*c^3*x^3 *(c^2*x^2+1)^(1/2)+13*c^2*x^2+5*c*x*(c^2*x^2+1)^(1/2)+1)*g*(-1+5*arcsinh(c *x))*d^2/c^2/(c^2*x^2+1)+3/512*(d*(c^2*x^2+1))^(1/2)*(8*c^5*x^5+8*c^4*x^4* (c^2*x^2+1)^(1/2)+12*c^3*x^3+8*c^2*x^2*(c^2*x^2+1)^(1/2)+4*c*x+(c^2*x^2+1) ^(1/2))*f*(-1+4*arcsinh(c*x))*d^2/c/(c^2*x^2+1)+1/128*(d*(c^2*x^2+1))^(1/2 )*(4*c^4*x^4+4*c^3*x^3*(c^2*x^2+1)^(1/2)+5*c^2*x^2+3*c*x*(c^2*x^2+1)^(1/2) +1)*g*(-1+3*arcsinh(c*x))*d^2/c^2/(c^2*x^2+1)+15/256*(d*(c^2*x^2+1))^(1/2) *(2*c^3*x^3+2*c^2*x^2*(c^2*x^2+1)^(1/2)+2*c*x+(c^2*x^2+1)^(1/2))*f*(-1+2*a rcsinh(c*x))*d^2/c/(c^2*x^2+1)+5/128*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c ^2*x^2+1)^(1/2)+1)*g*(-1+arcsinh(c*x))*d^2/c^2/(c^2*x^2+1)+5/128*(d*(c^...
\[ \int (f+g x) \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (c^{2} d x^{2} + d\right )}^{\frac {5}{2}} {\left (g x + f\right )} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} \,d x } \]
integral((a*c^4*d^2*g*x^5 + a*c^4*d^2*f*x^4 + 2*a*c^2*d^2*g*x^3 + 2*a*c^2* d^2*f*x^2 + a*d^2*g*x + a*d^2*f + (b*c^4*d^2*g*x^5 + b*c^4*d^2*f*x^4 + 2*b *c^2*d^2*g*x^3 + 2*b*c^2*d^2*f*x^2 + b*d^2*g*x + b*d^2*f)*arcsinh(c*x))*sq rt(c^2*d*x^2 + d), x)
\[ \int (f+g x) \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\int \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \left (f + g x\right )\, dx \]
Exception generated. \[ \int (f+g x) \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int (f+g x) \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve cteur & l) Error: Bad Argument Value
Timed out. \[ \int (f+g x) \left (d+c^2 d x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\int \left (f+g\,x\right )\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d\,c^2\,x^2+d\right )}^{5/2} \,d x \]