\(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^2} \, dx\) [1561]

Optimal result

Integrand size = 28, antiderivative size = 183 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=-\frac {b^2 (2 b d-3 a e) x \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}+\frac {b^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x)}+\frac {(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x) (d+e x)}+\frac {3 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^4 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 45} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^4 (a+b x) (d+e x)}+\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^4 (a+b x)}-\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2} (2 b d-3 a e)}{e^3 (a+b x)}+\frac {b^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2 (a+b x)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (6 a^2 b d e^2-2 a^3 e^3+6 a b^2 e \left (-d^2+d e x+e^2 x^2\right )+b^3 \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+6 b (b d-a e)^2 (d+e x) \log (d+e x)\right )}{2 e^4 (a+b x) (d+e x)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.93 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \left (\frac {1}{2} b e \,x^{2}+3 a e x -2 b d x \right )}{\left (b x +a \right ) e^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{\left (b x +a \right ) e^{4} \left (e x +d \right )}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, b \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{4}}\)	\(156\)
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (e^{3} x^{3} b^{3}+6 \ln \left (e x +d \right ) a^{2} b \,e^{3} x -12 \ln \left (e x +d \right ) a \,b^{2} d \,e^{2} x +6 \ln \left (e x +d \right ) b^{3} d^{2} e x +6 x^{2} a \,b^{2} e^{3}-3 x^{2} b^{3} d \,e^{2}+6 \ln \left (e x +d \right ) a^{2} b d \,e^{2}-12 \ln \left (e x +d \right ) a \,b^{2} d^{2} e +6 \ln \left (e x +d \right ) b^{3} d^{3}+6 x a \,b^{2} d \,e^{2}-4 b^{3} d^{2} e x -2 a^{3} e^{3}+6 a^{2} b d \,e^{2}-6 a \,b^{2} d^{2} e +2 b^{3} d^{3}\right )}{2 \left (b x +a \right )^{3} e^{4} \left (e x +d \right )}\)	\(216\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\frac {b^{3} e^{3} x^{3} + 2 \, b^{3} d^{3} - 6 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} - 2 \, a^{3} e^{3} - 3 \, {\left (b^{3} d e^{2} - 2 \, a b^{2} e^{3}\right )} x^{2} - 2 \, {\left (2 \, b^{3} d^{2} e - 3 \, a b^{2} d e^{2}\right )} x + 6 \, {\left (b^{3} d^{3} - 2 \, a b^{2} d^{2} e + a^{2} b d e^{2} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x + d e^{4}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^2} \, dx=\frac {3 \, {\left (b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} + \frac {b^{3} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, b^{3} d e x \mathrm {sgn}\left (b x + a\right ) + 6 \, a b^{2} e^{2} x \mathrm {sgn}\left (b x + a\right )}{2 \, e^{4}} + \frac {b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )}{{\left (e x + d\right )} e^{4}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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