\(\int \frac {(d+e x)^4}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [1606]

Optimal result

Integrand size = 28, antiderivative size = 209 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 e^3 (b d-a e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 e (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e^2 (b d-a e)^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 209, 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} \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {4 e^3 (b d-a e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e^2 (b d-a e)^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 e (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

Int[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-4*e^3*(b*d - a*e))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*d - a*e)^4/(4*b^5*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) - (4*e*(b*d - a*e)^3)/(3*b^5*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*e^2*(b*d - a*e)^2)/(
b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*(a + b*x)*Log[a + b*x])/(b^5*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 (b^4 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^4}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {(b d-a e)^4}{b^9 (a+b x)^5}+\frac {4 e (b d-a e)^3}{b^9 (a+b x)^4}+\frac {6 e^2 (b d-a e)^2}{b^9 (a+b x)^3}+\frac {4 e^3 (b d-a e)}{b^9 (a+b x)^2}+\frac {e^4}{b^9 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {4 e^3 (b d-a e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^4}{4 b^5 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 e (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 e^2 (b d-a e)^2}{b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.66 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {-\left ((b d-a e) \left (25 a^3 e^3+a^2 b e^2 (13 d+88 e x)+a b^2 e \left (7 d^2+40 d e x+108 e^2 x^2\right )+b^3 \left (3 d^3+16 d^2 e x+36 d e^2 x^2+48 e^3 x^3\right )\right )\right )+12 e^4 (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt {(a+b x)^2}} \]

[In]

Integrate[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-((b*d - a*e)*(25*a^3*e^3 + a^2*b*e^2*(13*d + 88*e*x) + a*b^2*e*(7*d^2 + 40*d*e*x + 108*e^2*x^2) + b^3*(3*d^3
+ 16*d^2*e*x + 36*d*e^2*x^2 + 48*e^3*x^3))) + 12*e^4*(a + b*x)^4*Log[a + b*x])/(12*b^5*(a + b*x)^3*Sqrt[(a +
b*x)^2])

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.97

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {4 e^{3} \left (a e -b d \right ) x^{3}}{b^{2}}+\frac {3 e^{2} \left (3 a^{2} e^{2}-2 a b d e -b^{2} d^{2}\right ) x^{2}}{b^{3}}+\frac {2 e \left (11 a^{3} e^{3}-6 a^{2} b d \,e^{2}-3 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) x}{3 b^{4}}+\frac {25 e^{4} a^{4}-12 b \,e^{3} d \,a^{3}-6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e -3 b^{4} d^{4}}{12 b^{5}}\right )}{\left (b x +a \right )^{5}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{4} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(202\)
default	\(\frac {\left (12 \ln \left (b x +a \right ) b^{4} e^{4} x^{4}+48 \ln \left (b x +a \right ) x^{3} a \,b^{3} e^{4}+72 \ln \left (b x +a \right ) x^{2} a^{2} b^{2} e^{4}+48 x^{3} a \,b^{3} e^{4}-48 x^{3} b^{4} d \,e^{3}+48 \ln \left (b x +a \right ) x \,a^{3} b \,e^{4}+108 x^{2} a^{2} b^{2} e^{4}-72 x^{2} a \,b^{3} d \,e^{3}-36 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (b x +a \right ) a^{4} e^{4}+88 x \,a^{3} b \,e^{4}-48 x \,a^{2} b^{2} d \,e^{3}-24 x a \,b^{3} d^{2} e^{2}-16 x \,b^{4} d^{3} e +25 e^{4} a^{4}-12 b \,e^{3} d \,a^{3}-6 b^{2} e^{2} d^{2} a^{2}-4 a \,b^{3} d^{3} e -3 b^{4} d^{4}\right ) \left (b x +a \right )}{12 b^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(267\)

[In]

int((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

((b*x+a)^2)^(1/2)/(b*x+a)^5*(4*e^3*(a*e-b*d)/b^2*x^3+3*e^2*(3*a^2*e^2-2*a*b*d*e-b^2*d^2)/b^3*x^2+2/3*e*(11*a^3
*e^3-6*a^2*b*d*e^2-3*a*b^2*d^2*e-2*b^3*d^3)/b^4*x+1/12*(25*a^4*e^4-12*a^3*b*d*e^3-6*a^2*b^2*d^2*e^2-4*a*b^3*d^
3*e-3*b^4*d^4)/b^5)+((b*x+a)^2)^(1/2)/(b*x+a)*e^4/b^5*ln(b*x+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.28 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {3 \, b^{4} d^{4} + 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} + 12 \, a^{3} b d e^{3} - 25 \, a^{4} e^{4} + 48 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 36 \, {\left (b^{4} d^{2} e^{2} + 2 \, a b^{3} d e^{3} - 3 \, a^{2} b^{2} e^{4}\right )} x^{2} + 8 \, {\left (2 \, b^{4} d^{3} e + 3 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} - 11 \, a^{3} b e^{4}\right )} x - 12 \, {\left (b^{4} e^{4} x^{4} + 4 \, a b^{3} e^{4} x^{3} + 6 \, a^{2} b^{2} e^{4} x^{2} + 4 \, a^{3} b e^{4} x + a^{4} e^{4}\right )} \log \left (b x + a\right )}{12 \, {\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}\right )}} \]

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(3*b^4*d^4 + 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 12*a^3*b*d*e^3 - 25*a^4*e^4 + 48*(b^4*d*e^3 - a*b^3*e^4
)*x^3 + 36*(b^4*d^2*e^2 + 2*a*b^3*d*e^3 - 3*a^2*b^2*e^4)*x^2 + 8*(2*b^4*d^3*e + 3*a*b^3*d^2*e^2 + 6*a^2*b^2*d*
e^3 - 11*a^3*b*e^4)*x - 12*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*log(b
*x + a))/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5)

Sympy [F]

\[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((d + e*x)**4/((a + b*x)**2)**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 323 vs. \(2 (150) = 300\).

Time = 0.22 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.55 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, e^{4} {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{3} \, d e^{3} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, d^{3} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{2} \, d^{2} e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {d^{4}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \]

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/12*e^4*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^
3*b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5) - 1/3*d*e^3*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((
b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4
)) - 1/3*d^3*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/2*d^2*e^2*(6/(b^5*(x + a/
b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*(x + a/b)^4)) - 1/4*d^4/(b^5*(x + a/b)^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.96 \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^{4} \log \left ({\left | b x + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {48 \, {\left (b^{3} d e^{3} - a b^{2} e^{4}\right )} x^{3} + 36 \, {\left (b^{3} d^{2} e^{2} + 2 \, a b^{2} d e^{3} - 3 \, a^{2} b e^{4}\right )} x^{2} + 8 \, {\left (2 \, b^{3} d^{3} e + 3 \, a b^{2} d^{2} e^{2} + 6 \, a^{2} b d e^{3} - 11 \, a^{3} e^{4}\right )} x + \frac {3 \, b^{4} d^{4} + 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} + 12 \, a^{3} b d e^{3} - 25 \, a^{4} e^{4}}{b}}{12 \, {\left (b x + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x + a\right )} \]

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

e^4*log(abs(b*x + a))/(b^5*sgn(b*x + a)) - 1/12*(48*(b^3*d*e^3 - a*b^2*e^4)*x^3 + 36*(b^3*d^2*e^2 + 2*a*b^2*d*
e^3 - 3*a^2*b*e^4)*x^2 + 8*(2*b^3*d^3*e + 3*a*b^2*d^2*e^2 + 6*a^2*b*d*e^3 - 11*a^3*e^4)*x + (3*b^4*d^4 + 4*a*b
^3*d^3*e + 6*a^2*b^2*d^2*e^2 + 12*a^3*b*d*e^3 - 25*a^4*e^4)/b)/((b*x + a)^4*b^4*sgn(b*x + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

[In]

int((d + e*x)^4/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

int((d + e*x)^4/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)