\(\int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [1588]

Optimal result

Integrand size = 28, antiderivative size = 173 \[ \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {e (b d-a e)^2 x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e) (a+b x) (d+e x)^2}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^3}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e)^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 173, 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} \[ \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) (d+e x)^2 (b d-a e)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^3}{3 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (b d-a e)^3 \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e x (a+b x) (b d-a e)^2}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.52 \[ \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \left (b e x \left (6 a^2 e^2-3 a b e (6 d+e x)+b^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+6 (b d-a e)^3 \log (a+b x)\right )}{6 b^4 \sqrt {(a+b x)^2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.35 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, e \left (\frac {1}{3} b^{2} e^{2} x^{3}-\frac {1}{2} x^{2} a b \,e^{2}+\frac {3}{2} b^{2} d e \,x^{2}+a^{2} e^{2} x -3 a b d e x +3 b^{2} d^{2} x \right )}{\left (b x +a \right ) b^{3}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\)	\(142\)
default	\(-\frac {\left (b x +a \right ) \left (-2 e^{3} x^{3} b^{3}+3 x^{2} a \,b^{2} e^{3}-9 x^{2} b^{3} d \,e^{2}+6 \ln \left (b x +a \right ) a^{3} e^{3}-18 \ln \left (b x +a \right ) a^{2} b d \,e^{2}+18 \ln \left (b x +a \right ) a \,b^{2} d^{2} e -6 \ln \left (b x +a \right ) b^{3} d^{3}-6 a^{2} b \,e^{3} x +18 x a \,b^{2} d \,e^{2}-18 b^{3} d^{2} e x \right )}{6 \sqrt {\left (b x +a \right )^{2}}\, b^{4}}\)	\(147\)

[In]

int((e*x+d)^3/((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.67 \[ \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, b^{3} e^{3} x^{3} + 3 \, {\left (3 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (3 \, b^{3} d^{2} e - 3 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \log \left (b x + a\right )}{6 \, b^{4}} \]

[In]

integrate((e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (126) = 252\).

Time = 1.22 (sec) , antiderivative size = 462, normalized size of antiderivative = 2.67 \[ \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {e^{3} x^{2}}{3 b^{2}} + \frac {x \left (- \frac {5 a e^{3}}{3 b} + 3 d e^{2}\right )}{2 b^{2}} + \frac {- \frac {2 a^{2} e^{3}}{3 b^{2}} - \frac {3 a \left (- \frac {5 a e^{3}}{3 b} + 3 d e^{2}\right )}{2 b} + 3 d^{2} e}{b^{2}}\right ) + \frac {\left (\frac {a}{b} + x\right ) \left (- \frac {a^{2} \left (- \frac {5 a e^{3}}{3 b} + 3 d e^{2}\right )}{2 b^{2}} - \frac {a \left (- \frac {2 a^{2} e^{3}}{3 b^{2}} - \frac {3 a \left (- \frac {5 a e^{3}}{3 b} + 3 d e^{2}\right )}{2 b} + 3 d^{2} e\right )}{b} + d^{3}\right ) \log {\left (\frac {a}{b} + x \right )}}{\sqrt {b^{2} \left (\frac {a}{b} + x\right )^{2}}} & \text {for}\: b^{2} \neq 0 \\\frac {2 d^{3} \sqrt {a^{2} + 2 a b x} + \frac {3 d^{2} e \left (- a^{2} \sqrt {a^{2} + 2 a b x} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3}\right )}{a b} + \frac {3 d e^{2} \left (a^{4} \sqrt {a^{2} + 2 a b x} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5}\right )}{2 a^{2} b^{2}} + \frac {e^{3} \left (- a^{6} \sqrt {a^{2} + 2 a b x} + a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}} - \frac {3 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}\right )}{4 a^{3} b^{3}}}{2 a b} & \text {for}\: a b \neq 0 \\\frac {d^{3} x + \frac {3 d^{2} e x^{2}}{2} + d e^{2} x^{3} + \frac {e^{3} x^{4}}{4}}{\sqrt {a^{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x+d)**3/((b*x+a)**2)**(1/2),x)

[Out]

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(e**3*x**2/(3*b**2) + x*(-5*a*e**3/(3*b) + 3*d*e**2)/(2*b**2) + (-
2*a**2*e**3/(3*b**2) - 3*a*(-5*a*e**3/(3*b) + 3*d*e**2)/(2*b) + 3*d**2*e)/b**2) + (a/b + x)*(-a**2*(-5*a*e**3/
(3*b) + 3*d*e**2)/(2*b**2) - a*(-2*a**2*e**3/(3*b**2) - 3*a*(-5*a*e**3/(3*b) + 3*d*e**2)/(2*b) + 3*d**2*e)/b +
d**3)*log(a/b + x)/sqrt(b**2*(a/b + x)**2), Ne(b**2, 0)), ((2*d**3*sqrt(a**2 + 2*a*b*x) + 3*d**2*e*(-a**2*sqr
t(a**2 + 2*a*b*x) + (a**2 + 2*a*b*x)**(3/2)/3)/(a*b) + 3*d*e**2*(a**4*sqrt(a**2 + 2*a*b*x) - 2*a**2*(a**2 + 2*
a*b*x)**(3/2)/3 + (a**2 + 2*a*b*x)**(5/2)/5)/(2*a**2*b**2) + e**3*(-a**6*sqrt(a**2 + 2*a*b*x) + a**4*(a**2 + 2
*a*b*x)**(3/2) - 3*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/(4*a**3*b**3))/(2*a*b), Ne(a*b,
0)), ((d**3*x + 3*d**2*e*x**2/2 + d*e**2*x**3 + e**3*x**4/4)/sqrt(a**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, d e^{2} x^{2}}{2 \, b} - \frac {5 \, a e^{3} x^{2}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} e^{3} x^{2}}{3 \, b^{2}} - \frac {3 \, a d e^{2} x}{b^{2}} + \frac {5 \, a^{2} e^{3} x}{3 \, b^{3}} + \frac {d^{3} \log \left (x + \frac {a}{b}\right )}{b} - \frac {3 \, a d^{2} e \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {3 \, a^{2} d e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {a^{3} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{2} e}{b^{2}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e^{3}}{3 \, b^{4}} \]

[In]

integrate((e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^3}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 \, b^{2} e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 9 \, b^{2} d e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, b^{2} d^{2} e x \mathrm {sgn}\left (b x + a\right ) - 18 \, a b d e^{2} x \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} e^{3} x \mathrm {sgn}\left (b x + a\right )}{6 \, b^{3}} + \frac {{\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \]

[In]

integrate((e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((d + e*x)^3/((a + b*x)^2)^(1/2),x)

[Out]

int((d + e*x)^3/((a + b*x)^2)^(1/2), x)