\(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^{7/2}} \, dx\) [1698]

Optimal result

Integrand size = 30, antiderivative size = 316 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=\frac {2 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^6 (a+b x) (d+e x)^{5/2}}-\frac {10 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^{3/2}}+\frac {20 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) \sqrt {d+e x}}+\frac {20 b^3 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}-\frac {10 b^4 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x)}+\frac {2 b^5 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^6 (a+b x)} \]

[Out]

2/5*(-a*e+b*d)^5*((b*x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d)^(5/2)-10/3*b*(-a*e+b*d)^4*((b*x+a)^2)^(1/2)/e^6/(b*x+a)
/(e*x+d)^(3/2)-10/3*b^4*(-a*e+b*d)*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)+2/5*b^5*(e*x+d)^(5/2)*((b*x+a)^
2)^(1/2)/e^6/(b*x+a)+20*b^2*(-a*e+b*d)^3*((b*x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d)^(1/2)+20*b^3*(-a*e+b*d)^2*(e*x+
d)^(1/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {660, 45} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=\frac {20 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^6 (a+b x) \sqrt {d+e x}}-\frac {10 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{3 e^6 (a+b x) (d+e x)^{3/2}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{5 e^6 (a+b x) (d+e x)^{5/2}}+\frac {2 b^5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^6 (a+b x)}-\frac {10 b^4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^6 (a+b x)}+\frac {20 b^3 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}{e^6 (a+b x)} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^(7/2),x]

[Out]

(2*(b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^6*(a + b*x)*(d + e*x)^(5/2)) - (10*b*(b*d - a*e)^4*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(3*e^6*(a + b*x)*(d + e*x)^(3/2)) + (20*b^2*(b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/(e^6*(a + b*x)*Sqrt[d + e*x]) + (20*b^3*(b*d - a*e)^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^6*(a
+ b*x)) - (10*b^4*(b*d - a*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^6*(a + b*x)) + (2*b^5*(d +
e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^6*(a + b*x))

Rule 45

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] && NeQ[b*c - a*d, 0] && 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])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[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] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{(d+e x)^{7/2}} \, dx}{b^4 \left (a b+b^2 x\right )} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b^5 (b d-a e)^5}{e^5 (d+e x)^{7/2}}+\frac {5 b^6 (b d-a e)^4}{e^5 (d+e x)^{5/2}}-\frac {10 b^7 (b d-a e)^3}{e^5 (d+e x)^{3/2}}+\frac {10 b^8 (b d-a e)^2}{e^5 \sqrt {d+e x}}-\frac {5 b^9 (b d-a e) \sqrt {d+e x}}{e^5}+\frac {b^{10} (d+e x)^{3/2}}{e^5}\right ) \, dx}{b^4 \left (a b+b^2 x\right )} \\ & = \frac {2 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^6 (a+b x) (d+e x)^{5/2}}-\frac {10 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^{3/2}}+\frac {20 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) \sqrt {d+e x}}+\frac {20 b^3 (b d-a e)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}-\frac {10 b^4 (b d-a e) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x)}+\frac {2 b^5 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^6 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (3 a^5 e^5+5 a^4 b e^4 (2 d+5 e x)+10 a^3 b^2 e^3 \left (8 d^2+20 d e x+15 e^2 x^2\right )-30 a^2 b^3 e^2 \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )+5 a b^4 e \left (128 d^4+320 d^3 e x+240 d^2 e^2 x^2+40 d e^3 x^3-5 e^4 x^4\right )-b^5 \left (256 d^5+640 d^4 e x+480 d^3 e^2 x^2+80 d^2 e^3 x^3-10 d e^4 x^4+3 e^5 x^5\right )\right )}{15 e^6 (a+b x) (d+e x)^{5/2}} \]

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(d + e*x)^(7/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(3*a^5*e^5 + 5*a^4*b*e^4*(2*d + 5*e*x) + 10*a^3*b^2*e^3*(8*d^2 + 20*d*e*x + 15*e^2*x^2)
- 30*a^2*b^3*e^2*(16*d^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3) + 5*a*b^4*e*(128*d^4 + 320*d^3*e*x + 240*d^2
*e^2*x^2 + 40*d*e^3*x^3 - 5*e^4*x^4) - b^5*(256*d^5 + 640*d^4*e*x + 480*d^3*e^2*x^2 + 80*d^2*e^3*x^3 - 10*d*e^
4*x^4 + 3*e^5*x^5)))/(15*e^6*(a + b*x)*(d + e*x)^(5/2))

Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.68

method	result	size
risch	\(\frac {2 b^{3} \left (3 x^{2} b^{2} e^{2}+25 x a b \,e^{2}-19 b^{2} d e x +150 a^{2} e^{2}-275 a b d e +128 b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 e^{6} \left (b x +a \right )}-\frac {2 \left (150 x^{2} b^{2} e^{2}+25 x a b \,e^{2}+275 b^{2} d e x +3 a^{2} e^{2}+19 a b d e +128 b^{2} d^{2}\right ) \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 e^{6} \sqrt {e x +d}\, \left (x^{2} e^{2}+2 d e x +d^{2}\right ) \left (b x +a \right )}\)	\(215\)
gosper	\(-\frac {2 \left (-3 x^{5} e^{5} b^{5}-25 x^{4} a \,b^{4} e^{5}+10 x^{4} b^{5} d \,e^{4}-150 x^{3} a^{2} b^{3} e^{5}+200 x^{3} a \,b^{4} d \,e^{4}-80 x^{3} b^{5} d^{2} e^{3}+150 x^{2} a^{3} b^{2} e^{5}-900 x^{2} a^{2} b^{3} d \,e^{4}+1200 x^{2} a \,b^{4} d^{2} e^{3}-480 x^{2} b^{5} d^{3} e^{2}+25 a^{4} b \,e^{5} x +200 a^{3} b^{2} d \,e^{4} x -1200 x \,a^{2} b^{3} d^{2} e^{3}+1600 x a \,b^{4} d^{3} e^{2}-640 b^{5} d^{4} e x +3 a^{5} e^{5}+10 a^{4} b d \,e^{4}+80 a^{3} b^{2} d^{2} e^{3}-480 a^{2} b^{3} d^{3} e^{2}+640 a \,b^{4} d^{4} e -256 b^{5} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{6} \left (b x +a \right )^{5}}\)	\(289\)
default	\(-\frac {2 \left (-3 x^{5} e^{5} b^{5}-25 x^{4} a \,b^{4} e^{5}+10 x^{4} b^{5} d \,e^{4}-150 x^{3} a^{2} b^{3} e^{5}+200 x^{3} a \,b^{4} d \,e^{4}-80 x^{3} b^{5} d^{2} e^{3}+150 x^{2} a^{3} b^{2} e^{5}-900 x^{2} a^{2} b^{3} d \,e^{4}+1200 x^{2} a \,b^{4} d^{2} e^{3}-480 x^{2} b^{5} d^{3} e^{2}+25 a^{4} b \,e^{5} x +200 a^{3} b^{2} d \,e^{4} x -1200 x \,a^{2} b^{3} d^{2} e^{3}+1600 x a \,b^{4} d^{3} e^{2}-640 b^{5} d^{4} e x +3 a^{5} e^{5}+10 a^{4} b d \,e^{4}+80 a^{3} b^{2} d^{2} e^{3}-480 a^{2} b^{3} d^{3} e^{2}+640 a \,b^{4} d^{4} e -256 b^{5} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{6} \left (b x +a \right )^{5}}\)	\(289\)

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/15*b^3*(3*b^2*e^2*x^2+25*a*b*e^2*x-19*b^2*d*e*x+150*a^2*e^2-275*a*b*d*e+128*b^2*d^2)*(e*x+d)^(1/2)/e^6*((b*x
+a)^2)^(1/2)/(b*x+a)-2/15*(150*b^2*e^2*x^2+25*a*b*e^2*x+275*b^2*d*e*x+3*a^2*e^2+19*a*b*d*e+128*b^2*d^2)*(a^3*e
^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/e^6/(e*x+d)^(1/2)/(e^2*x^2+2*d*e*x+d^2)*((b*x+a)^2)^(1/2)/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (3 \, b^{5} e^{5} x^{5} + 256 \, b^{5} d^{5} - 640 \, a b^{4} d^{4} e + 480 \, a^{2} b^{3} d^{3} e^{2} - 80 \, a^{3} b^{2} d^{2} e^{3} - 10 \, a^{4} b d e^{4} - 3 \, a^{5} e^{5} - 5 \, {\left (2 \, b^{5} d e^{4} - 5 \, a b^{4} e^{5}\right )} x^{4} + 10 \, {\left (8 \, b^{5} d^{2} e^{3} - 20 \, a b^{4} d e^{4} + 15 \, a^{2} b^{3} e^{5}\right )} x^{3} + 30 \, {\left (16 \, b^{5} d^{3} e^{2} - 40 \, a b^{4} d^{2} e^{3} + 30 \, a^{2} b^{3} d e^{4} - 5 \, a^{3} b^{2} e^{5}\right )} x^{2} + 5 \, {\left (128 \, b^{5} d^{4} e - 320 \, a b^{4} d^{3} e^{2} + 240 \, a^{2} b^{3} d^{2} e^{3} - 40 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{9} x^{3} + 3 \, d e^{8} x^{2} + 3 \, d^{2} e^{7} x + d^{3} e^{6}\right )}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

2/15*(3*b^5*e^5*x^5 + 256*b^5*d^5 - 640*a*b^4*d^4*e + 480*a^2*b^3*d^3*e^2 - 80*a^3*b^2*d^2*e^3 - 10*a^4*b*d*e^
4 - 3*a^5*e^5 - 5*(2*b^5*d*e^4 - 5*a*b^4*e^5)*x^4 + 10*(8*b^5*d^2*e^3 - 20*a*b^4*d*e^4 + 15*a^2*b^3*e^5)*x^3 +
30*(16*b^5*d^3*e^2 - 40*a*b^4*d^2*e^3 + 30*a^2*b^3*d*e^4 - 5*a^3*b^2*e^5)*x^2 + 5*(128*b^5*d^4*e - 320*a*b^4*
d^3*e^2 + 240*a^2*b^3*d^2*e^3 - 40*a^3*b^2*d*e^4 - 5*a^4*b*e^5)*x)*sqrt(e*x + d)/(e^9*x^3 + 3*d*e^8*x^2 + 3*d^
2*e^7*x + d^3*e^6)

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{\frac {7}{2}}}\, dx \]

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2)/(e*x+d)**(7/2),x)

[Out]

Integral(((a + b*x)**2)**(5/2)/(d + e*x)**(7/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (3 \, b^{5} e^{5} x^{5} + 256 \, b^{5} d^{5} - 640 \, a b^{4} d^{4} e + 480 \, a^{2} b^{3} d^{3} e^{2} - 80 \, a^{3} b^{2} d^{2} e^{3} - 10 \, a^{4} b d e^{4} - 3 \, a^{5} e^{5} - 5 \, {\left (2 \, b^{5} d e^{4} - 5 \, a b^{4} e^{5}\right )} x^{4} + 10 \, {\left (8 \, b^{5} d^{2} e^{3} - 20 \, a b^{4} d e^{4} + 15 \, a^{2} b^{3} e^{5}\right )} x^{3} + 30 \, {\left (16 \, b^{5} d^{3} e^{2} - 40 \, a b^{4} d^{2} e^{3} + 30 \, a^{2} b^{3} d e^{4} - 5 \, a^{3} b^{2} e^{5}\right )} x^{2} + 5 \, {\left (128 \, b^{5} d^{4} e - 320 \, a b^{4} d^{3} e^{2} + 240 \, a^{2} b^{3} d^{2} e^{3} - 40 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x\right )}}{15 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )} \sqrt {e x + d}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

2/15*(3*b^5*e^5*x^5 + 256*b^5*d^5 - 640*a*b^4*d^4*e + 480*a^2*b^3*d^3*e^2 - 80*a^3*b^2*d^2*e^3 - 10*a^4*b*d*e^
4 - 3*a^5*e^5 - 5*(2*b^5*d*e^4 - 5*a*b^4*e^5)*x^4 + 10*(8*b^5*d^2*e^3 - 20*a*b^4*d*e^4 + 15*a^2*b^3*e^5)*x^3 +
30*(16*b^5*d^3*e^2 - 40*a*b^4*d^2*e^3 + 30*a^2*b^3*d*e^4 - 5*a^3*b^2*e^5)*x^2 + 5*(128*b^5*d^4*e - 320*a*b^4*
d^3*e^2 + 240*a^2*b^3*d^2*e^3 - 40*a^3*b^2*d*e^4 - 5*a^4*b*e^5)*x)/((e^8*x^2 + 2*d*e^7*x + d^2*e^6)*sqrt(e*x +
d))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (150 \, {\left (e x + d\right )}^{2} b^{5} d^{3} \mathrm {sgn}\left (b x + a\right ) - 25 \, {\left (e x + d\right )} b^{5} d^{4} \mathrm {sgn}\left (b x + a\right ) + 3 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 450 \, {\left (e x + d\right )}^{2} a b^{4} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 100 \, {\left (e x + d\right )} a b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 15 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 450 \, {\left (e x + d\right )}^{2} a^{2} b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 150 \, {\left (e x + d\right )} a^{2} b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 30 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 150 \, {\left (e x + d\right )}^{2} a^{3} b^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 100 \, {\left (e x + d\right )} a^{3} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) - 30 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 25 \, {\left (e x + d\right )} a^{4} b e^{4} \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - 3 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{6}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{5} e^{24} \mathrm {sgn}\left (b x + a\right ) - 25 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{5} d e^{24} \mathrm {sgn}\left (b x + a\right ) + 150 \, \sqrt {e x + d} b^{5} d^{2} e^{24} \mathrm {sgn}\left (b x + a\right ) + 25 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{4} e^{25} \mathrm {sgn}\left (b x + a\right ) - 300 \, \sqrt {e x + d} a b^{4} d e^{25} \mathrm {sgn}\left (b x + a\right ) + 150 \, \sqrt {e x + d} a^{2} b^{3} e^{26} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, e^{30}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

2/15*(150*(e*x + d)^2*b^5*d^3*sgn(b*x + a) - 25*(e*x + d)*b^5*d^4*sgn(b*x + a) + 3*b^5*d^5*sgn(b*x + a) - 450*
(e*x + d)^2*a*b^4*d^2*e*sgn(b*x + a) + 100*(e*x + d)*a*b^4*d^3*e*sgn(b*x + a) - 15*a*b^4*d^4*e*sgn(b*x + a) +
450*(e*x + d)^2*a^2*b^3*d*e^2*sgn(b*x + a) - 150*(e*x + d)*a^2*b^3*d^2*e^2*sgn(b*x + a) + 30*a^2*b^3*d^3*e^2*s
gn(b*x + a) - 150*(e*x + d)^2*a^3*b^2*e^3*sgn(b*x + a) + 100*(e*x + d)*a^3*b^2*d*e^3*sgn(b*x + a) - 30*a^3*b^2
*d^2*e^3*sgn(b*x + a) - 25*(e*x + d)*a^4*b*e^4*sgn(b*x + a) + 15*a^4*b*d*e^4*sgn(b*x + a) - 3*a^5*e^5*sgn(b*x
+ a))/((e*x + d)^(5/2)*e^6) + 2/15*(3*(e*x + d)^(5/2)*b^5*e^24*sgn(b*x + a) - 25*(e*x + d)^(3/2)*b^5*d*e^24*sg
n(b*x + a) + 150*sqrt(e*x + d)*b^5*d^2*e^24*sgn(b*x + a) + 25*(e*x + d)^(3/2)*a*b^4*e^25*sgn(b*x + a) - 300*sq
rt(e*x + d)*a*b^4*d*e^25*sgn(b*x + a) + 150*sqrt(e*x + d)*a^2*b^3*e^26*sgn(b*x + a))/e^30

Mupad [B] (verification not implemented)

Time = 10.54 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{7/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {6\,a^5\,e^5+20\,a^4\,b\,d\,e^4+160\,a^3\,b^2\,d^2\,e^3-960\,a^2\,b^3\,d^3\,e^2+1280\,a\,b^4\,d^4\,e-512\,b^5\,d^5}{15\,b\,e^8}-\frac {2\,b^4\,x^5}{5\,e^3}-\frac {2\,b^3\,x^4\,\left (5\,a\,e-2\,b\,d\right )}{3\,e^4}+\frac {x\,\left (50\,a^4\,b\,e^5+400\,a^3\,b^2\,d\,e^4-2400\,a^2\,b^3\,d^2\,e^3+3200\,a\,b^4\,d^3\,e^2-1280\,b^5\,d^4\,e\right )}{15\,b\,e^8}-\frac {4\,b^2\,x^3\,\left (15\,a^2\,e^2-20\,a\,b\,d\,e+8\,b^2\,d^2\right )}{3\,e^5}+\frac {4\,b\,x^2\,\left (5\,a^3\,e^3-30\,a^2\,b\,d\,e^2+40\,a\,b^2\,d^2\,e-16\,b^3\,d^3\right )}{e^6}\right )}{x^3\,\sqrt {d+e\,x}+\frac {a\,d^2\,\sqrt {d+e\,x}}{b\,e^2}+\frac {x^2\,\left (15\,a\,e^8+30\,b\,d\,e^7\right )\,\sqrt {d+e\,x}}{15\,b\,e^8}+\frac {d\,x\,\left (2\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}} \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2)/(d + e*x)^(7/2),x)

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((6*a^5*e^5 - 512*b^5*d^5 - 960*a^2*b^3*d^3*e^2 + 160*a^3*b^2*d^2*e^3 + 1280
*a*b^4*d^4*e + 20*a^4*b*d*e^4)/(15*b*e^8) - (2*b^4*x^5)/(5*e^3) - (2*b^3*x^4*(5*a*e - 2*b*d))/(3*e^4) + (x*(50
*a^4*b*e^5 - 1280*b^5*d^4*e + 3200*a*b^4*d^3*e^2 + 400*a^3*b^2*d*e^4 - 2400*a^2*b^3*d^2*e^3))/(15*b*e^8) - (4*
b^2*x^3*(15*a^2*e^2 + 8*b^2*d^2 - 20*a*b*d*e))/(3*e^5) + (4*b*x^2*(5*a^3*e^3 - 16*b^3*d^3 + 40*a*b^2*d^2*e - 3
0*a^2*b*d*e^2))/e^6))/(x^3*(d + e*x)^(1/2) + (a*d^2*(d + e*x)^(1/2))/(b*e^2) + (x^2*(15*a*e^8 + 30*b*d*e^7)*(d
+ e*x)^(1/2))/(15*b*e^8) + (d*x*(2*a*e + b*d)*(d + e*x)^(1/2))/(b*e^2))