\(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{d+e x} \, dx\) [1560]

Optimal result

Integrand size = 28, antiderivative size = 166 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{d+e x} \, dx=\frac {b (b d-a e)^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {(b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2}+\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e}-\frac {(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^4 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 166, 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} \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{d+e x} \, dx=-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 \log (d+e x)}{e^4 (a+b x)}+\frac {b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x)}-\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^2}+\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e} \]

[In]

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

[Out]

(b*(b*d - a*e)^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)) - ((b*d - a*e)*(a + b*x)*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/(2*e^2) + ((a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e) - ((b*d - a*e)^3*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} \, 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^4 (b d-a e)^2}{e^3}-\frac {b^3 (b d-a e) \left (a b+b^2 x\right )}{e^2}+\frac {b^2 \left (a b+b^2 x\right )^2}{e}-\frac {b^3 (b d-a e)^3}{e^3 (d+e x)}\right ) \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {b (b d-a e)^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {(b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2}+\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e}-\frac {(b d-a e)^3 \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 = 1.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.55 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{d+e x} \, dx=\frac {\sqrt {(a+b x)^2} \left (b e x \left (18 a^2 e^2+9 a b e (-2 d+e x)+b^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 (b d-a e)^3 \log (d+e x)\right )}{6 e^4 (a+b x)} \]

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(b*e*x*(18*a^2*e^2 + 9*a*b*e*(-2*d + e*x) + b^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2)) - 6*(b*d - a
*e)^3*Log[d + e*x]))/(6*e^4*(a + b*x))

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b \left (\frac {1}{3} b^{2} e^{2} x^{3}+\frac {3}{2} x^{2} a b \,e^{2}-\frac {1}{2} b^{2} d e \,x^{2}+3 a^{2} e^{2} x -3 a b d e x +b^{2} d^{2} x \right )}{\left (b x +a \right ) e^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{4}}\)	\(141\)
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (2 e^{3} x^{3} b^{3}+9 x^{2} a \,b^{2} e^{3}-3 x^{2} b^{3} d \,e^{2}+6 \ln \left (e x +d \right ) a^{3} e^{3}-18 \ln \left (e x +d \right ) a^{2} b d \,e^{2}+18 \ln \left (e x +d \right ) a \,b^{2} d^{2} e -6 \ln \left (e x +d \right ) b^{3} d^{3}+18 a^{2} b \,e^{3} x -18 x a \,b^{2} d \,e^{2}+6 b^{3} d^{2} e x \right )}{6 \left (b x +a \right )^{3} e^{4}}\)	\(149\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.69 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{d+e x} \, dx=\frac {2 \, b^{3} e^{3} x^{3} - 3 \, {\left (b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} e - 3 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x - 6 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \log \left (e x + d\right )}{6 \, e^{4}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d),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 = 0.27 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{d+e x} \, dx=\frac {2 \, b^{3} e^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, b^{3} d e x^{2} \mathrm {sgn}\left (b x + a\right ) + 9 \, a b^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{3} d^{2} x \mathrm {sgn}\left (b x + a\right ) - 18 \, a b^{2} d e x \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b e^{2} x \mathrm {sgn}\left (b x + a\right )}{6 \, e^{3}} - \frac {{\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | e x + d \right |}\right )}{e^{4}} \]

[In]

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

[Out]

1/6*(2*b^3*e^2*x^3*sgn(b*x + a) - 3*b^3*d*e*x^2*sgn(b*x + a) + 9*a*b^2*e^2*x^2*sgn(b*x + a) + 6*b^3*d^2*x*sgn(
b*x + a) - 18*a*b^2*d*e*x*sgn(b*x + a) + 18*a^2*b*e^2*x*sgn(b*x + a))/e^3 - (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))*log(abs(e*x + d))/e^4

Mupad [F(-1)]

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

[In]

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

[Out]

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