\(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\) [1688]

Optimal result

Integrand size = 30, antiderivative size = 204 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x) (d+e x)^{3/2}}-\frac {6 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}}-\frac {6 b^2 (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {2 b^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 204, 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} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}{e^4 (a+b x)}-\frac {6 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) \sqrt {d+e x}}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{3 e^4 (a+b x) (d+e x)^{3/2}}+\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^4 (a+b x)} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

(2*(b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^4*(a + b*x)*(d + e*x)^(3/2)) - (6*b*(b*d - a*e)^2*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/(e^4*(a + b*x)*Sqrt[d + e*x]) - (6*b^2*(b*d - a*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(e^4*(a + b*x)) + (2*b^3*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^4*(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 )^3}{(d+e x)^{5/2}} \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b^3 (b d-a e)^3}{e^3 (d+e x)^{5/2}}+\frac {3 b^4 (b d-a e)^2}{e^3 (d+e x)^{3/2}}-\frac {3 b^5 (b d-a e)}{e^3 \sqrt {d+e x}}+\frac {b^6 \sqrt {d+e x}}{e^3}\right ) \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x) (d+e x)^{3/2}}-\frac {6 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) \sqrt {d+e x}}-\frac {6 b^2 (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {2 b^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^4 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.58 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (a^3 e^3+3 a^2 b e^2 (2 d+3 e x)-3 a b^2 e \left (8 d^2+12 d e x+3 e^2 x^2\right )+b^3 \left (16 d^3+24 d^2 e x+6 d e^2 x^2-e^3 x^3\right )\right )}{3 e^4 (a+b x) (d+e x)^{3/2}} \]

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^(5/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(a^3*e^3 + 3*a^2*b*e^2*(2*d + 3*e*x) - 3*a*b^2*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + b^3*(1
6*d^3 + 24*d^2*e*x + 6*d*e^2*x^2 - e^3*x^3)))/(3*e^4*(a + b*x)*(d + e*x)^(3/2))

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.53

method	result	size
risch	\(\frac {2 b^{2} \left (b e x +9 a e -8 b d \right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{3 e^{4} \left (b x +a \right )}-\frac {2 \left (9 b e x +a e +8 b d \right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 e^{4} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )}\)	\(108\)
gosper	\(-\frac {2 \left (-e^{3} x^{3} b^{3}-9 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+9 a^{2} b \,e^{3} x -36 x a \,b^{2} d \,e^{2}+24 b^{3} d^{2} e x +a^{3} e^{3}+6 a^{2} b d \,e^{2}-24 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4} \left (b x +a \right )^{3}}\)	\(131\)
default	\(-\frac {2 \left (-e^{3} x^{3} b^{3}-9 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+9 a^{2} b \,e^{3} x -36 x a \,b^{2} d \,e^{2}+24 b^{3} d^{2} e x +a^{3} e^{3}+6 a^{2} b d \,e^{2}-24 a \,b^{2} d^{2} e +16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{3 \left (e x +d \right )^{\frac {3}{2}} e^{4} \left (b x +a \right )^{3}}\)	\(131\)

[In]

int((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3*b^2*(b*e*x+9*a*e-8*b*d)*(e*x+d)^(1/2)/e^4*((b*x+a)^2)^(1/2)/(b*x+a)-2/3*(9*b*e*x+a*e+8*b*d)*(a^2*e^2-2*a*b
*d*e+b^2*d^2)/e^4/(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 -
3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)*sqrt(e*x + d)/(e^6*x^2 + 2*d*e^5*x + d^2*e^4)

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(5/2),x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.61 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )}}{3 \, {\left (e^{5} x + d e^{4}\right )} \sqrt {e x + d}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 -
3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)/((e^5*x + d*e^4)*sqrt(e*x + d))

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {2 \, {\left (9 \, {\left (e x + d\right )} b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 18 \, {\left (e x + d\right )} a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 9 \, {\left (e x + d\right )} a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, {\left (e x + d\right )}^{\frac {3}{2}} e^{4}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{3} e^{8} \mathrm {sgn}\left (b x + a\right ) - 9 \, \sqrt {e x + d} b^{3} d e^{8} \mathrm {sgn}\left (b x + a\right ) + 9 \, \sqrt {e x + d} a b^{2} e^{9} \mathrm {sgn}\left (b x + a\right )\right )}}{3 \, e^{12}} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

-2/3*(9*(e*x + d)*b^3*d^2*sgn(b*x + a) - b^3*d^3*sgn(b*x + a) - 18*(e*x + d)*a*b^2*d*e*sgn(b*x + a) + 3*a*b^2*
d^2*e*sgn(b*x + a) + 9*(e*x + d)*a^2*b*e^2*sgn(b*x + a) - 3*a^2*b*d*e^2*sgn(b*x + a) + a^3*e^3*sgn(b*x + a))/(
(e*x + d)^(3/2)*e^4) + 2/3*((e*x + d)^(3/2)*b^3*e^8*sgn(b*x + a) - 9*sqrt(e*x + d)*b^3*d*e^8*sgn(b*x + a) + 9*
sqrt(e*x + d)*a*b^2*e^9*sgn(b*x + a))/e^12

Mupad [B] (verification not implemented)

Time = 10.20 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,a^3\,e^3+12\,a^2\,b\,d\,e^2-48\,a\,b^2\,d^2\,e+32\,b^3\,d^3}{3\,b\,e^5}+\frac {2\,x\,\left (3\,a^2\,e^2-12\,a\,b\,d\,e+8\,b^2\,d^2\right )}{e^4}-\frac {2\,b^2\,x^3}{3\,e^2}-\frac {2\,b\,x^2\,\left (3\,a\,e-2\,b\,d\right )}{e^3}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (3\,a\,e^5+3\,b\,d\,e^4\right )\,\sqrt {d+e\,x}}{3\,b\,e^5}} \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/(d + e*x)^(5/2),x)

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*a^3*e^3 + 32*b^3*d^3 - 48*a*b^2*d^2*e + 12*a^2*b*d*e^2)/(3*b*e^5) + (2*x
*(3*a^2*e^2 + 8*b^2*d^2 - 12*a*b*d*e))/e^4 - (2*b^2*x^3)/(3*e^2) - (2*b*x^2*(3*a*e - 2*b*d))/e^3))/(x^2*(d + e
*x)^(1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(3*a*e^5 + 3*b*d*e^4)*(d + e*x)^(1/2))/(3*b*e^5))