Integrand size = 30, antiderivative size = 208 \[ \int (d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {2 (b d-a e)^3 (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^4 (a+b x)}+\frac {2 b (b d-a e)^2 (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x)}-\frac {6 b^2 (b d-a e) (d+e x)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 e^4 (a+b x)}+\frac {2 b^3 (d+e x)^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 e^4 (a+b x)} \]
-2/7*(-a*e+b*d)^3*(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/3*b*(-a*e+ b*d)^2*(e*x+d)^(9/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)-6/11*b^2*(-a*e+b*d)*(e* x+d)^(11/2)*((b*x+a)^2)^(1/2)/e^4/(b*x+a)+2/13*b^3*(e*x+d)^(13/2)*((b*x+a) ^2)^(1/2)/e^4/(b*x+a)
Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.58 \[ \int (d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \sqrt {(a+b x)^2} (d+e x)^{7/2} \left (429 a^3 e^3+143 a^2 b e^2 (-2 d+7 e x)+13 a b^2 e \left (8 d^2-28 d e x+63 e^2 x^2\right )+b^3 \left (-16 d^3+56 d^2 e x-126 d e^2 x^2+231 e^3 x^3\right )\right )}{3003 e^4 (a+b x)} \]
(2*Sqrt[(a + b*x)^2]*(d + e*x)^(7/2)*(429*a^3*e^3 + 143*a^2*b*e^2*(-2*d + 7*e*x) + 13*a*b^2*e*(8*d^2 - 28*d*e*x + 63*e^2*x^2) + b^3*(-16*d^3 + 56*d^ 2*e*x - 126*d*e^2*x^2 + 231*e^3*x^3)))/(3003*e^4*(a + b*x))
Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1102, 27, 53, 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.
\(\displaystyle \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (d+e x)^{5/2} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b^3 (a+b x)^3 (d+e x)^{5/2}dx}{b^3 (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x)^3 (d+e x)^{5/2}dx}{a+b x}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^3 (d+e x)^{11/2}}{e^3}-\frac {3 b^2 (b d-a e) (d+e x)^{9/2}}{e^3}+\frac {3 b (b d-a e)^2 (d+e x)^{7/2}}{e^3}+\frac {(a e-b d)^3 (d+e x)^{5/2}}{e^3}\right )dx}{a+b x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {6 b^2 (d+e x)^{11/2} (b d-a e)}{11 e^4}+\frac {2 b (d+e x)^{9/2} (b d-a e)^2}{3 e^4}-\frac {2 (d+e x)^{7/2} (b d-a e)^3}{7 e^4}+\frac {2 b^3 (d+e x)^{13/2}}{13 e^4}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^3*(d + e*x)^(7/2))/(7*e^4) + (2*b*(b*d - a*e)^2*(d + e*x)^(9/2))/(3*e^4) - (6*b^2*(b*d - a*e)*(d + e *x)^(11/2))/(11*e^4) + (2*b^3*(d + e*x)^(13/2))/(13*e^4)))/(a + b*x)
3.17.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.51 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.63
method | result | size |
gosper | \(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (231 e^{3} x^{3} b^{3}+819 x^{2} a \,b^{2} e^{3}-126 x^{2} b^{3} d \,e^{2}+1001 a^{2} b \,e^{3} x -364 x a \,b^{2} d \,e^{2}+56 b^{3} d^{2} e x +429 a^{3} e^{3}-286 a^{2} b d \,e^{2}+104 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{3003 e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
default | \(\frac {2 \left (e x +d \right )^{\frac {7}{2}} \left (231 e^{3} x^{3} b^{3}+819 x^{2} a \,b^{2} e^{3}-126 x^{2} b^{3} d \,e^{2}+1001 a^{2} b \,e^{3} x -364 x a \,b^{2} d \,e^{2}+56 b^{3} d^{2} e x +429 a^{3} e^{3}-286 a^{2} b d \,e^{2}+104 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{3003 e^{4} \left (b x +a \right )^{3}}\) | \(132\) |
risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (231 b^{3} e^{6} x^{6}+819 a \,b^{2} e^{6} x^{5}+567 b^{3} d \,e^{5} x^{5}+1001 a^{2} b \,e^{6} x^{4}+2093 a \,b^{2} d \,e^{5} x^{4}+371 b^{3} d^{2} e^{4} x^{4}+429 a^{3} e^{6} x^{3}+2717 a^{2} b d \,e^{5} x^{3}+1469 a \,b^{2} d^{2} e^{4} x^{3}+5 b^{3} d^{3} e^{3} x^{3}+1287 a^{3} d \,e^{5} x^{2}+2145 a^{2} b \,d^{2} e^{4} x^{2}+39 a \,b^{2} d^{3} e^{3} x^{2}-6 b^{3} d^{4} e^{2} x^{2}+1287 a^{3} d^{2} e^{4} x +143 a^{2} b \,d^{3} e^{3} x -52 a \,b^{2} d^{4} e^{2} x +8 b^{3} d^{5} e x +429 a^{3} d^{3} e^{3}-286 a^{2} b \,d^{4} e^{2}+104 a \,b^{2} d^{5} e -16 b^{3} d^{6}\right ) \sqrt {e x +d}}{3003 \left (b x +a \right ) e^{4}}\) | \(302\) |
2/3003*(e*x+d)^(7/2)*(231*b^3*e^3*x^3+819*a*b^2*e^3*x^2-126*b^3*d*e^2*x^2+ 1001*a^2*b*e^3*x-364*a*b^2*d*e^2*x+56*b^3*d^2*e*x+429*a^3*e^3-286*a^2*b*d* e^2+104*a*b^2*d^2*e-16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3
Time = 0.59 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.29 \[ \int (d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (231 \, b^{3} e^{6} x^{6} - 16 \, b^{3} d^{6} + 104 \, a b^{2} d^{5} e - 286 \, a^{2} b d^{4} e^{2} + 429 \, a^{3} d^{3} e^{3} + 63 \, {\left (9 \, b^{3} d e^{5} + 13 \, a b^{2} e^{6}\right )} x^{5} + 7 \, {\left (53 \, b^{3} d^{2} e^{4} + 299 \, a b^{2} d e^{5} + 143 \, a^{2} b e^{6}\right )} x^{4} + {\left (5 \, b^{3} d^{3} e^{3} + 1469 \, a b^{2} d^{2} e^{4} + 2717 \, a^{2} b d e^{5} + 429 \, a^{3} e^{6}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{4} e^{2} - 13 \, a b^{2} d^{3} e^{3} - 715 \, a^{2} b d^{2} e^{4} - 429 \, a^{3} d e^{5}\right )} x^{2} + {\left (8 \, b^{3} d^{5} e - 52 \, a b^{2} d^{4} e^{2} + 143 \, a^{2} b d^{3} e^{3} + 1287 \, a^{3} d^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{3003 \, e^{4}} \]
2/3003*(231*b^3*e^6*x^6 - 16*b^3*d^6 + 104*a*b^2*d^5*e - 286*a^2*b*d^4*e^2 + 429*a^3*d^3*e^3 + 63*(9*b^3*d*e^5 + 13*a*b^2*e^6)*x^5 + 7*(53*b^3*d^2*e ^4 + 299*a*b^2*d*e^5 + 143*a^2*b*e^6)*x^4 + (5*b^3*d^3*e^3 + 1469*a*b^2*d^ 2*e^4 + 2717*a^2*b*d*e^5 + 429*a^3*e^6)*x^3 - 3*(2*b^3*d^4*e^2 - 13*a*b^2* d^3*e^3 - 715*a^2*b*d^2*e^4 - 429*a^3*d*e^5)*x^2 + (8*b^3*d^5*e - 52*a*b^2 *d^4*e^2 + 143*a^2*b*d^3*e^3 + 1287*a^3*d^2*e^4)*x)*sqrt(e*x + d)/e^4
\[ \int (d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int \left (d + e x\right )^{\frac {5}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.29 \[ \int (d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (231 \, b^{3} e^{6} x^{6} - 16 \, b^{3} d^{6} + 104 \, a b^{2} d^{5} e - 286 \, a^{2} b d^{4} e^{2} + 429 \, a^{3} d^{3} e^{3} + 63 \, {\left (9 \, b^{3} d e^{5} + 13 \, a b^{2} e^{6}\right )} x^{5} + 7 \, {\left (53 \, b^{3} d^{2} e^{4} + 299 \, a b^{2} d e^{5} + 143 \, a^{2} b e^{6}\right )} x^{4} + {\left (5 \, b^{3} d^{3} e^{3} + 1469 \, a b^{2} d^{2} e^{4} + 2717 \, a^{2} b d e^{5} + 429 \, a^{3} e^{6}\right )} x^{3} - 3 \, {\left (2 \, b^{3} d^{4} e^{2} - 13 \, a b^{2} d^{3} e^{3} - 715 \, a^{2} b d^{2} e^{4} - 429 \, a^{3} d e^{5}\right )} x^{2} + {\left (8 \, b^{3} d^{5} e - 52 \, a b^{2} d^{4} e^{2} + 143 \, a^{2} b d^{3} e^{3} + 1287 \, a^{3} d^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{3003 \, e^{4}} \]
2/3003*(231*b^3*e^6*x^6 - 16*b^3*d^6 + 104*a*b^2*d^5*e - 286*a^2*b*d^4*e^2 + 429*a^3*d^3*e^3 + 63*(9*b^3*d*e^5 + 13*a*b^2*e^6)*x^5 + 7*(53*b^3*d^2*e ^4 + 299*a*b^2*d*e^5 + 143*a^2*b*e^6)*x^4 + (5*b^3*d^3*e^3 + 1469*a*b^2*d^ 2*e^4 + 2717*a^2*b*d*e^5 + 429*a^3*e^6)*x^3 - 3*(2*b^3*d^4*e^2 - 13*a*b^2* d^3*e^3 - 715*a^2*b*d^2*e^4 - 429*a^3*d*e^5)*x^2 + (8*b^3*d^5*e - 52*a*b^2 *d^4*e^2 + 143*a^2*b*d^3*e^3 + 1287*a^3*d^2*e^4)*x)*sqrt(e*x + d)/e^4
Leaf count of result is larger than twice the leaf count of optimal. 953 vs. \(2 (148) = 296\).
Time = 0.36 (sec) , antiderivative size = 953, normalized size of antiderivative = 4.58 \[ \int (d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\text {Too large to display} \]
2/15015*(15015*sqrt(e*x + d)*a^3*d^3*sgn(b*x + a) + 15015*((e*x + d)^(3/2) - 3*sqrt(e*x + d)*d)*a^3*d^2*sgn(b*x + a) + 15015*((e*x + d)^(3/2) - 3*sq rt(e*x + d)*d)*a^2*b*d^3*sgn(b*x + a)/e + 3003*(3*(e*x + d)^(5/2) - 10*(e* x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a^3*d*sgn(b*x + a) + 3003*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a*b^2*d^3*sgn(b*x + a)/e^2 + 9009*(3*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d + 15*sqrt(e*x + d)*d^2)*a^2*b*d^2*sgn(b*x + a)/e + 429*(5*(e*x + d)^(7/2) - 21*(e*x + d)^ (5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a^3*sgn(b*x + a) + 429*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*b^3*d^3*sgn(b*x + a)/e^3 + 3861*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a* b^2*d^2*sgn(b*x + a)/e^2 + 3861*(5*(e*x + d)^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d)*d^3)*a^2*b*d*sgn(b*x + a)/e + 143*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*b^3*d^2*sgn(b*x + a)/e^ 3 + 429*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)* d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*a*b^2*d*sgn(b*x + a )/e^2 + 143*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7/2)*d + 378*(e*x + d)^(5 /2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*a^2*b*sgn(b*x + a)/e + 65*(63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/2)*d + 990*(e*x + d)...
Timed out. \[ \int (d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int {\left (d+e\,x\right )}^{5/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]