Integrand size = 26, antiderivative size = 269 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx=\frac {3 b^2 (b B d-2 A c d+A b e) (b d+(2 c d-b e) x) \sqrt {b x+c x^2}}{128 d^3 (c d-b e)^3 (d+e x)^2}-\frac {(b B d-2 A c d+A b e) (b d+(2 c d-b e) x) \left (b x+c x^2\right )^{3/2}}{16 d^2 (c d-b e)^2 (d+e x)^4}+\frac {(B d-A e) \left (b x+c x^2\right )^{5/2}}{5 d (c d-b e) (d+e x)^5}-\frac {3 b^4 (b B d-2 A c d+A b e) \text {arctanh}\left (\frac {b d+(2 c d-b e) x}{2 \sqrt {d} \sqrt {c d-b e} \sqrt {b x+c x^2}}\right )}{256 d^{7/2} (c d-b e)^{7/2}} \]
-1/16*(A*b*e-2*A*c*d+B*b*d)*(b*d+(-b*e+2*c*d)*x)*(c*x^2+b*x)^(3/2)/d^2/(-b *e+c*d)^2/(e*x+d)^4+1/5*(-A*e+B*d)*(c*x^2+b*x)^(5/2)/d/(-b*e+c*d)/(e*x+d)^ 5-3/256*b^4*(A*b*e-2*A*c*d+B*b*d)*arctanh(1/2*(b*d+(-b*e+2*c*d)*x)/d^(1/2) /(-b*e+c*d)^(1/2)/(c*x^2+b*x)^(1/2))/d^(7/2)/(-b*e+c*d)^(7/2)+3/128*b^2*(A *b*e-2*A*c*d+B*b*d)*(b*d+(-b*e+2*c*d)*x)*(c*x^2+b*x)^(1/2)/d^3/(-b*e+c*d)^ 3/(e*x+d)^2
Time = 10.63 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx=\frac {(x (b+c x))^{3/2} \left (\frac {2 (-B d+A e) x^{5/2} (b+c x)}{(d+e x)^5}+\frac {5 (b B d-2 A c d+A b e) \left (-\frac {2 x^{3/2} (b+c x)^{5/2}}{(d+e x)^4}+\frac {b \sqrt {x} (b+c x)^{5/2}}{(c d-b e) (d+e x)^3}-\frac {b^2 \sqrt {x} \sqrt {b+c x} (5 b d+2 c d x+3 b e x)}{8 d^2 (c d-b e) (d+e x)^2}-\frac {3 b^4 \text {arctanh}\left (\frac {\sqrt {c d-b e} \sqrt {x}}{\sqrt {d} \sqrt {b+c x}}\right )}{8 d^{5/2} (c d-b e)^{3/2}}\right )}{8 (-c d+b e) (b+c x)^{3/2}}\right )}{10 d (-c d+b e) x^{3/2}} \]
((x*(b + c*x))^(3/2)*((2*(-(B*d) + A*e)*x^(5/2)*(b + c*x))/(d + e*x)^5 + ( 5*(b*B*d - 2*A*c*d + A*b*e)*((-2*x^(3/2)*(b + c*x)^(5/2))/(d + e*x)^4 + (b *Sqrt[x]*(b + c*x)^(5/2))/((c*d - b*e)*(d + e*x)^3) - (b^2*Sqrt[x]*Sqrt[b + c*x]*(5*b*d + 2*c*d*x + 3*b*e*x))/(8*d^2*(c*d - b*e)*(d + e*x)^2) - (3*b ^4*ArcTanh[(Sqrt[c*d - b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(8*d^(5/2)* (c*d - b*e)^(3/2))))/(8*(-(c*d) + b*e)*(b + c*x)^(3/2))))/(10*d*(-(c*d) + b*e)*x^(3/2))
Time = 0.42 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1228, 1152, 1152, 1154, 219}
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.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx\) |
\(\Big \downarrow \) 1228 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{5/2} (B d-A e)}{5 d (d+e x)^5 (c d-b e)}-\frac {(A b e-2 A c d+b B d) \int \frac {\left (c x^2+b x\right )^{3/2}}{(d+e x)^5}dx}{2 d (c d-b e)}\) |
\(\Big \downarrow \) 1152 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{5/2} (B d-A e)}{5 d (d+e x)^5 (c d-b e)}-\frac {(A b e-2 A c d+b B d) \left (\frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \int \frac {\sqrt {c x^2+b x}}{(d+e x)^3}dx}{16 d (c d-b e)}\right )}{2 d (c d-b e)}\) |
\(\Big \downarrow \) 1152 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{5/2} (B d-A e)}{5 d (d+e x)^5 (c d-b e)}-\frac {(A b e-2 A c d+b B d) \left (\frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \left (\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac {b^2 \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{8 d (c d-b e)}\right )}{16 d (c d-b e)}\right )}{2 d (c d-b e)}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{5/2} (B d-A e)}{5 d (d+e x)^5 (c d-b e)}-\frac {(A b e-2 A c d+b B d) \left (\frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \left (\frac {b^2 \int \frac {1}{4 d (c d-b e)-\frac {(b d+(2 c d-b e) x)^2}{c x^2+b x}}d\left (-\frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}\right )}{4 d (c d-b e)}+\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}\right )}{16 d (c d-b e)}\right )}{2 d (c d-b e)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (b x+c x^2\right )^{5/2} (B d-A e)}{5 d (d+e x)^5 (c d-b e)}-\frac {(A b e-2 A c d+b B d) \left (\frac {\left (b x+c x^2\right )^{3/2} (x (2 c d-b e)+b d)}{8 d (d+e x)^4 (c d-b e)}-\frac {3 b^2 \left (\frac {\sqrt {b x+c x^2} (x (2 c d-b e)+b d)}{4 d (d+e x)^2 (c d-b e)}-\frac {b^2 \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{8 d^{3/2} (c d-b e)^{3/2}}\right )}{16 d (c d-b e)}\right )}{2 d (c d-b e)}\) |
((B*d - A*e)*(b*x + c*x^2)^(5/2))/(5*d*(c*d - b*e)*(d + e*x)^5) - ((b*B*d - 2*A*c*d + A*b*e)*(((b*d + (2*c*d - b*e)*x)*(b*x + c*x^2)^(3/2))/(8*d*(c* d - b*e)*(d + e*x)^4) - (3*b^2*(((b*d + (2*c*d - b*e)*x)*Sqrt[b*x + c*x^2] )/(4*d*(c*d - b*e)*(d + e*x)^2) - (b^2*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2* Sqrt[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(8*d^(3/2)*(c*d - b*e)^(3/2)) ))/(16*d*(c*d - b*e))))/(2*d*(c*d - b*e))
3.12.80.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-(d + e*x)^(m + 1))*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b *x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a *c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] && GtQ[p, 0]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Time = 0.76 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.44
method | result | size |
pseudoelliptic | \(-\frac {3 \left (\left (\left (-2 A c +B b \right ) d +A b e \right ) \left (e x +d \right )^{5} b^{4} \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )+\sqrt {d \left (b e -c d \right )}\, \left (\left (b^{4} B -2 c \left (\frac {B x}{3}+A \right ) b^{3}+\frac {4 c^{2} x \left (\frac {2 B x}{5}+A \right ) b^{2}}{3}+16 c^{3} x^{2} \left (\frac {11 B x}{15}+A \right ) b +\frac {32 c^{4} \left (\frac {4 B x}{5}+A \right ) x^{3}}{3}\right ) d^{5}+\left (\left (A +\frac {14 B x}{3}\right ) b^{4}-10 c x \left (\frac {23 B x}{75}+A \right ) b^{3}-\frac {668 c^{2} x^{2} \left (\frac {128 B x}{167}+A \right ) b^{2}}{15}-\frac {112 c^{3} x^{3} \left (B x +A \right ) b}{5}+\frac {32 A \,c^{4} x^{4}}{15}\right ) e \,d^{4}+\frac {14 x \left (\left (\frac {64 B x}{35}+A \right ) b^{3}+\frac {233 c x \left (B x +A \right ) b^{2}}{35}+\frac {46 c^{2} x^{2} \left (\frac {62 B x}{23}+A \right ) b}{35}-\frac {32 A \,c^{3} x^{3}}{35}\right ) e^{2} b \,d^{3}}{3}-\frac {128 x^{2} e^{3} \left (\left (\frac {35 B x}{64}+A \right ) b^{2}-\frac {47 c \left (-\frac {5 B x}{47}+A \right ) x b}{64}-\frac {3 A \,c^{2} x^{2}}{32}\right ) b^{2} d^{2}}{15}-\frac {14 x^{3} \left (\left (\frac {3 B x}{14}+A \right ) b -\frac {2 A c x}{7}\right ) e^{4} b^{3} d}{3}-A \,b^{4} e^{5} x^{4}\right ) \sqrt {x \left (c x +b \right )}\right )}{128 \sqrt {d \left (b e -c d \right )}\, \left (e x +d \right )^{5} \left (b e -c d \right )^{3} d^{3}}\) | \(387\) |
default | \(\text {Expression too large to display}\) | \(9004\) |
-3/128/(d*(b*e-c*d))^(1/2)*(((-2*A*c+B*b)*d+A*b*e)*(e*x+d)^5*b^4*arctan((x *(c*x+b))^(1/2)/x*d/(d*(b*e-c*d))^(1/2))+(d*(b*e-c*d))^(1/2)*((b^4*B-2*c*( 1/3*B*x+A)*b^3+4/3*c^2*x*(2/5*B*x+A)*b^2+16*c^3*x^2*(11/15*B*x+A)*b+32/3*c ^4*(4/5*B*x+A)*x^3)*d^5+((A+14/3*B*x)*b^4-10*c*x*(23/75*B*x+A)*b^3-668/15* c^2*x^2*(128/167*B*x+A)*b^2-112/5*c^3*x^3*(B*x+A)*b+32/15*A*c^4*x^4)*e*d^4 +14/3*x*((64/35*B*x+A)*b^3+233/35*c*x*(B*x+A)*b^2+46/35*c^2*x^2*(62/23*B*x +A)*b-32/35*A*c^3*x^3)*e^2*b*d^3-128/15*x^2*e^3*((35/64*B*x+A)*b^2-47/64*c *(-5/47*B*x+A)*x*b-3/32*A*c^2*x^2)*b^2*d^2-14/3*x^3*((3/14*B*x+A)*b-2/7*A* c*x)*e^4*b^3*d-A*b^4*e^5*x^4)*(x*(c*x+b))^(1/2))/(e*x+d)^5/(b*e-c*d)^3/d^3
Leaf count of result is larger than twice the leaf count of optimal. 1163 vs. \(2 (243) = 486\).
Time = 0.59 (sec) , antiderivative size = 2338, normalized size of antiderivative = 8.69 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx=\text {Too large to display} \]
[-1/1280*(15*(A*b^5*d^5*e + (B*b^5 - 2*A*b^4*c)*d^6 + (A*b^5*e^6 + (B*b^5 - 2*A*b^4*c)*d*e^5)*x^5 + 5*(A*b^5*d*e^5 + (B*b^5 - 2*A*b^4*c)*d^2*e^4)*x^ 4 + 10*(A*b^5*d^2*e^4 + (B*b^5 - 2*A*b^4*c)*d^3*e^3)*x^3 + 10*(A*b^5*d^3*e ^3 + (B*b^5 - 2*A*b^4*c)*d^4*e^2)*x^2 + 5*(A*b^5*d^4*e^2 + (B*b^5 - 2*A*b^ 4*c)*d^5*e)*x)*sqrt(c*d^2 - b*d*e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d ^2 - b*d*e)*sqrt(c*x^2 + b*x))/(e*x + d)) + 2*(15*A*b^5*d^5*e^2 - 15*(B*b^ 4*c - 2*A*b^3*c^2)*d^7 + 15*(B*b^5 - 3*A*b^4*c)*d^6*e - (128*B*c^5*d^7 + 1 5*A*b^5*d*e^6 - 16*(29*B*b*c^4 - 2*A*c^5)*d^6*e + 8*(73*B*b^2*c^3 - 12*A*b *c^4)*d^5*e^2 - 2*(129*B*b^3*c^2 - 38*A*b^2*c^3)*d^4*e^3 - (5*B*b^4*c - 8* A*b^3*c^2)*d^3*e^4 + 5*(3*B*b^5 - 7*A*b^4*c)*d^2*e^5)*x^4 - 2*(35*A*b^5*d^ 2*e^5 + 8*(11*B*b*c^4 + 10*A*c^5)*d^7 - 8*(43*B*b^2*c^3 + 31*A*b*c^4)*d^6* e + (489*B*b^3*c^2 + 214*A*b^2*c^3)*d^5*e^2 - (268*B*b^4*c - A*b^3*c^2)*d^ 4*e^3 + (35*B*b^5 - 82*A*b^4*c)*d^3*e^4)*x^3 - 2*(64*A*b^5*d^3*e^4 + 4*(B* b^2*c^3 + 30*A*b*c^4)*d^7 - (27*B*b^3*c^2 + 454*A*b^2*c^3)*d^6*e + 3*(29*B *b^4*c + 189*A*b^3*c^2)*d^5*e^2 - (64*B*b^5 + 297*A*b^4*c)*d^4*e^3)*x^2 + 10*(7*A*b^5*d^4*e^3 + (B*b^3*c^2 - 2*A*b^2*c^3)*d^7 - (8*B*b^4*c - 17*A*b^ 3*c^2)*d^6*e + (7*B*b^5 - 22*A*b^4*c)*d^5*e^2)*x)*sqrt(c*x^2 + b*x))/(c^4* d^13 - 4*b*c^3*d^12*e + 6*b^2*c^2*d^11*e^2 - 4*b^3*c*d^10*e^3 + b^4*d^9*e^ 4 + (c^4*d^8*e^5 - 4*b*c^3*d^7*e^6 + 6*b^2*c^2*d^6*e^7 - 4*b^3*c*d^5*e^8 + b^4*d^4*e^9)*x^5 + 5*(c^4*d^9*e^4 - 4*b*c^3*d^8*e^5 + 6*b^2*c^2*d^7*e^...
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 4306 vs. \(2 (243) = 486\).
Time = 0.39 (sec) , antiderivative size = 4306, normalized size of antiderivative = 16.01 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx=\text {Too large to display} \]
-3/128*(B*b^5*d - 2*A*b^4*c*d + A*b^5*e)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/((c^3*d^6 - 3*b*c^2*d^5*e + 3 *b^2*c*d^4*e^2 - b^3*d^3*e^3)*sqrt(-c*d^2 + b*d*e)) + 1/640*(1280*(sqrt(c) *x - sqrt(c*x^2 + b*x))^9*B*c^5*d^6*e^4 - 3840*(sqrt(c)*x - sqrt(c*x^2 + b *x))^9*B*b*c^4*d^5*e^5 + 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*B*b^2*c^3* d^4*e^6 - 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*B*b^3*c^2*d^3*e^7 + 15*(s qrt(c)*x - sqrt(c*x^2 + b*x))^9*B*b^5*d*e^9 - 30*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*A*b^4*c*d*e^9 + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^9*A*b^5*e^10 + 5120*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*c^(11/2)*d^7*e^3 - 12800*(sqrt(c )*x - sqrt(c*x^2 + b*x))^8*B*b*c^(9/2)*d^6*e^4 + 1280*(sqrt(c)*x - sqrt(c* x^2 + b*x))^8*A*c^(11/2)*d^6*e^4 + 7680*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8* B*b^2*c^(7/2)*d^5*e^5 - 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*b*c^(9/2) *d^5*e^5 + 2560*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*B*b^3*c^(5/2)*d^4*e^6 + 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*b^2*c^(7/2)*d^4*e^6 - 2560*(sqrt( c)*x - sqrt(c*x^2 + b*x))^8*B*b^4*c^(3/2)*d^3*e^7 - 1280*(sqrt(c)*x - sqrt (c*x^2 + b*x))^8*A*b^3*c^(5/2)*d^3*e^7 + 135*(sqrt(c)*x - sqrt(c*x^2 + b*x ))^8*B*b^5*sqrt(c)*d^2*e^8 - 270*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*b^4*c ^(3/2)*d^2*e^8 + 135*(sqrt(c)*x - sqrt(c*x^2 + b*x))^8*A*b^5*sqrt(c)*d*e^9 + 10240*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*c^6*d^8*e^2 - 21760*(sqrt(c)* x - sqrt(c*x^2 + b*x))^7*B*b*c^5*d^7*e^3 + 2560*(sqrt(c)*x - sqrt(c*x^2...
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{(d+e x)^6} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{{\left (d+e\,x\right )}^6} \,d x \]