Integrand size = 30, antiderivative size = 250 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {35 e^3 \sqrt {d+e x}}{64 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 (d+e x)^{3/2}}{96 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (d+e x)^{5/2}}{24 b^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^4 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{9/2} \sqrt {b d-a e} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-35/96*e^2*(e*x+d)^(3/2)/b^3/(b*x+a)/((b*x+a)^2)^(1/2)-7/24*e*(e*x+d)^(5/2 )/b^2/(b*x+a)^2/((b*x+a)^2)^(1/2)-1/4*(e*x+d)^(7/2)/b/(b*x+a)^3/((b*x+a)^2 )^(1/2)-35/64*e^4*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/ b^(9/2)/(-a*e+b*d)^(1/2)/((b*x+a)^2)^(1/2)-35/64*e^3*(e*x+d)^(1/2)/b^4/((b *x+a)^2)^(1/2)
Time = 0.81 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^4 (a+b x)^5 \left (-\frac {\sqrt {b} \sqrt {d+e x} \left (105 a^3 e^3+35 a^2 b e^2 (2 d+11 e x)+7 a b^2 e \left (8 d^2+36 d e x+73 e^2 x^2\right )+b^3 \left (48 d^3+200 d^2 e x+326 d e^2 x^2+279 e^3 x^3\right )\right )}{e^4 (a+b x)^4}+\frac {105 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {-b d+a e}}\right )}{192 b^{9/2} \left ((a+b x)^2\right )^{5/2}} \]
(e^4*(a + b*x)^5*(-((Sqrt[b]*Sqrt[d + e*x]*(105*a^3*e^3 + 35*a^2*b*e^2*(2* d + 11*e*x) + 7*a*b^2*e*(8*d^2 + 36*d*e*x + 73*e^2*x^2) + b^3*(48*d^3 + 20 0*d^2*e*x + 326*d*e^2*x^2 + 279*e^3*x^3)))/(e^4*(a + b*x)^4)) + (105*ArcTa n[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/Sqrt[-(b*d) + a*e]))/(192*b ^(9/2)*((a + b*x)^2)^(5/2))
Time = 0.29 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.77, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1102, 27, 51, 51, 51, 51, 73, 221}
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.
\(\displaystyle \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1102 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {(d+e x)^{7/2}}{b^5 (a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{7/2}}{(a+b x)^5}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(a+b x) \left (\frac {7 e \int \frac {(d+e x)^{5/2}}{(a+b x)^4}dx}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{(a+b x)^3}dx}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {3 e \left (\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {3 e \left (\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {5 e \left (\frac {3 e \left (-\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}-\frac {\sqrt {d+e x}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}\right )}{6 b}-\frac {(d+e x)^{5/2}}{3 b (a+b x)^3}\right )}{8 b}-\frac {(d+e x)^{7/2}}{4 b (a+b x)^4}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/4*(d + e*x)^(7/2)/(b*(a + b*x)^4) + (7*e*(-1/3*(d + e*x)^(5 /2)/(b*(a + b*x)^3) + (5*e*(-1/2*(d + e*x)^(3/2)/(b*(a + b*x)^2) + (3*e*(- (Sqrt[d + e*x]/(b*(a + b*x))) - (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b* d - a*e]])/(b^(3/2)*Sqrt[b*d - a*e])))/(4*b)))/(6*b)))/(8*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.18.23.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(466\) vs. \(2(167)=334\).
Time = 2.61 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.87
method | result | size |
default | \(-\frac {\left (-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{4} e^{4} x^{4}-420 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{3} e^{4} x^{3}+279 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {7}{2}} b^{3}-630 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b^{2} e^{4} x^{2}+511 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e -511 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d -420 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} b \,e^{4} x +385 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} a^{2} b \,e^{2}-770 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} a \,b^{2} d e +385 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} b^{3} d^{2}-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{4} e^{4}+105 \sqrt {e x +d}\, a^{3} e^{3} \sqrt {\left (a e -b d \right ) b}-315 \sqrt {e x +d}\, a^{2} d \,e^{2} b \sqrt {\left (a e -b d \right ) b}+315 \sqrt {e x +d}\, a \,d^{2} e \,b^{2} \sqrt {\left (a e -b d \right ) b}-105 \sqrt {e x +d}\, d^{3} b^{3} \sqrt {\left (a e -b d \right ) b}\right ) \left (b x +a \right )}{192 \sqrt {\left (a e -b d \right ) b}\, b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(467\) |
-1/192*(-105*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^4*e^4*x^4-420*a rctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^3*e^4*x^3+279*((a*e-b*d)*b) ^(1/2)*(e*x+d)^(7/2)*b^3-630*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a ^2*b^2*e^4*x^2+511*(e*x+d)^(5/2)*((a*e-b*d)*b)^(1/2)*a*b^2*e-511*(e*x+d)^( 5/2)*((a*e-b*d)*b)^(1/2)*b^3*d-420*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1 /2))*a^3*b*e^4*x+385*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a^2*b*e^2-770*((a*e -b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b^2*d*e+385*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/ 2)*b^3*d^2-105*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^4*e^4+105*(e* x+d)^(1/2)*a^3*e^3*((a*e-b*d)*b)^(1/2)-315*(e*x+d)^(1/2)*a^2*d*e^2*b*((a*e -b*d)*b)^(1/2)+315*(e*x+d)^(1/2)*a*d^2*e*b^2*((a*e-b*d)*b)^(1/2)-105*(e*x+ d)^(1/2)*d^3*b^3*((a*e-b*d)*b)^(1/2))*(b*x+a)/((a*e-b*d)*b)^(1/2)/b^4/((b* x+a)^2)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (167) = 334\).
Time = 0.31 (sec) , antiderivative size = 765, normalized size of antiderivative = 3.06 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [\frac {105 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (48 \, b^{5} d^{4} + 8 \, a b^{4} d^{3} e + 14 \, a^{2} b^{3} d^{2} e^{2} + 35 \, a^{3} b^{2} d e^{3} - 105 \, a^{4} b e^{4} + 279 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (326 \, b^{5} d^{2} e^{2} + 185 \, a b^{4} d e^{3} - 511 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (200 \, b^{5} d^{3} e + 52 \, a b^{4} d^{2} e^{2} + 133 \, a^{2} b^{3} d e^{3} - 385 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{384 \, {\left (a^{4} b^{6} d - a^{5} b^{5} e + {\left (b^{10} d - a b^{9} e\right )} x^{4} + 4 \, {\left (a b^{9} d - a^{2} b^{8} e\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d - a^{3} b^{7} e\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d - a^{4} b^{6} e\right )} x\right )}}, \frac {105 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (48 \, b^{5} d^{4} + 8 \, a b^{4} d^{3} e + 14 \, a^{2} b^{3} d^{2} e^{2} + 35 \, a^{3} b^{2} d e^{3} - 105 \, a^{4} b e^{4} + 279 \, {\left (b^{5} d e^{3} - a b^{4} e^{4}\right )} x^{3} + {\left (326 \, b^{5} d^{2} e^{2} + 185 \, a b^{4} d e^{3} - 511 \, a^{2} b^{3} e^{4}\right )} x^{2} + {\left (200 \, b^{5} d^{3} e + 52 \, a b^{4} d^{2} e^{2} + 133 \, a^{2} b^{3} d e^{3} - 385 \, a^{3} b^{2} e^{4}\right )} x\right )} \sqrt {e x + d}}{192 \, {\left (a^{4} b^{6} d - a^{5} b^{5} e + {\left (b^{10} d - a b^{9} e\right )} x^{4} + 4 \, {\left (a b^{9} d - a^{2} b^{8} e\right )} x^{3} + 6 \, {\left (a^{2} b^{8} d - a^{3} b^{7} e\right )} x^{2} + 4 \, {\left (a^{3} b^{7} d - a^{4} b^{6} e\right )} x\right )}}\right ] \]
[1/384*(105*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e ^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2* d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(48*b^5*d^4 + 8*a*b^4*d^3*e + 14* a^2*b^3*d^2*e^2 + 35*a^3*b^2*d*e^3 - 105*a^4*b*e^4 + 279*(b^5*d*e^3 - a*b^ 4*e^4)*x^3 + (326*b^5*d^2*e^2 + 185*a*b^4*d*e^3 - 511*a^2*b^3*e^4)*x^2 + ( 200*b^5*d^3*e + 52*a*b^4*d^2*e^2 + 133*a^2*b^3*d*e^3 - 385*a^3*b^2*e^4)*x) *sqrt(e*x + d))/(a^4*b^6*d - a^5*b^5*e + (b^10*d - a*b^9*e)*x^4 + 4*(a*b^9 *d - a^2*b^8*e)*x^3 + 6*(a^2*b^8*d - a^3*b^7*e)*x^2 + 4*(a^3*b^7*d - a^4*b ^6*e)*x), 1/192*(105*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)* sqrt(e*x + d)/(b*e*x + b*d)) - (48*b^5*d^4 + 8*a*b^4*d^3*e + 14*a^2*b^3*d^ 2*e^2 + 35*a^3*b^2*d*e^3 - 105*a^4*b*e^4 + 279*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (326*b^5*d^2*e^2 + 185*a*b^4*d*e^3 - 511*a^2*b^3*e^4)*x^2 + (200*b^5*d^ 3*e + 52*a*b^4*d^2*e^2 + 133*a^2*b^3*d*e^3 - 385*a^3*b^2*e^4)*x)*sqrt(e*x + d))/(a^4*b^6*d - a^5*b^5*e + (b^10*d - a*b^9*e)*x^4 + 4*(a*b^9*d - a^2*b ^8*e)*x^3 + 6*(a^2*b^8*d - a^3*b^7*e)*x^2 + 4*(a^3*b^7*d - a^4*b^6*e)*x)]
Timed out. \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {35 \, e^{4} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{64 \, \sqrt {-b^{2} d + a b e} b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {279 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{3} e^{4} - 511 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{3} d e^{4} + 385 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{4} - 105 \, \sqrt {e x + d} b^{3} d^{3} e^{4} + 511 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{2} e^{5} - 770 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} d e^{5} + 315 \, \sqrt {e x + d} a b^{2} d^{2} e^{5} + 385 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b e^{6} - 315 \, \sqrt {e x + d} a^{2} b d e^{6} + 105 \, \sqrt {e x + d} a^{3} e^{7}}{192 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \]
35/64*e^4*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b* e)*b^4*sgn(b*x + a)) - 1/192*(279*(e*x + d)^(7/2)*b^3*e^4 - 511*(e*x + d)^ (5/2)*b^3*d*e^4 + 385*(e*x + d)^(3/2)*b^3*d^2*e^4 - 105*sqrt(e*x + d)*b^3* d^3*e^4 + 511*(e*x + d)^(5/2)*a*b^2*e^5 - 770*(e*x + d)^(3/2)*a*b^2*d*e^5 + 315*sqrt(e*x + d)*a*b^2*d^2*e^5 + 385*(e*x + d)^(3/2)*a^2*b*e^6 - 315*sq rt(e*x + d)*a^2*b*d*e^6 + 105*sqrt(e*x + d)*a^3*e^7)/(((e*x + d)*b - b*d + a*e)^4*b^4*sgn(b*x + a))
Timed out. \[ \int \frac {(d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{7/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]