3.1596 \(\int \frac {(d+e x)^4}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=210 \[ \frac {6 e^2 (a+b x) (b d-a e)^2 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 e (b d-a e)^3}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^4}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 x (a+b x) (4 b d-3 a e)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac {e^3 x (a+b x) (4 b d-3 a e)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 e^2 (a+b x) (b d-a e)^2 \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 e (b d-a e)^3}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^4}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(-4*e*(b*d - a*e)^3)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*d - a*e)^4/(2*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (e^3*(4*b*d - 3*a*e)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*x^2*(a + b*x))/(2*b^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (6*e^2*(b*d - a*e)^2*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*
x^2])

Rule 43

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 646

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*} \int \frac {(d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^4}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {e^3 (4 b d-3 a e)}{b^7}+\frac {e^4 x}{b^6}+\frac {(b d-a e)^4}{b^7 (a+b x)^3}+\frac {4 e (b d-a e)^3}{b^7 (a+b x)^2}+\frac {6 e^2 (b d-a e)^2}{b^7 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 e (b d-a e)^3}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^4}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (4 b d-3 a e) x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {6 e^2 (b d-a e)^2 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 174, normalized size = 0.83 \[ \frac {7 a^4 e^4+2 a^3 b e^3 (e x-10 d)+a^2 b^2 e^2 \left (18 d^2-16 d e x-11 e^2 x^2\right )-4 a b^3 e \left (d^3-6 d^2 e x-4 d e^2 x^2+e^3 x^3\right )+12 e^2 (a+b x)^2 (b d-a e)^2 \log (a+b x)+b^4 \left (-d^4-8 d^3 e x+8 d e^3 x^3+e^4 x^4\right )}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(7*a^4*e^4 + 2*a^3*b*e^3*(-10*d + e*x) + a^2*b^2*e^2*(18*d^2 - 16*d*e*x - 11*e^2*x^2) - 4*a*b^3*e*(d^3 - 6*d^2
*e*x - 4*d*e^2*x^2 + e^3*x^3) + b^4*(-d^4 - 8*d^3*e*x + 8*d*e^3*x^3 + e^4*x^4) + 12*e^2*(b*d - a*e)^2*(a + b*x
)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])

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fricas [A]  time = 1.03, size = 292, normalized size = 1.39 \[ \frac {b^{4} e^{4} x^{4} - b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 20 \, a^{3} b d e^{3} + 7 \, a^{4} e^{4} + 4 \, {\left (2 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + {\left (16 \, a b^{3} d e^{3} - 11 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (4 \, b^{4} d^{3} e - 12 \, a b^{3} d^{2} e^{2} + 8 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x + 12 \, {\left (a^{2} b^{2} d^{2} e^{2} - 2 \, a^{3} b d e^{3} + a^{4} e^{4} + {\left (b^{4} d^{2} e^{2} - 2 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 2 \, {\left (a b^{3} d^{2} e^{2} - 2 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*(b^4*e^4*x^4 - b^4*d^4 - 4*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 20*a^3*b*d*e^3 + 7*a^4*e^4 + 4*(2*b^4*d*e^3
- a*b^3*e^4)*x^3 + (16*a*b^3*d*e^3 - 11*a^2*b^2*e^4)*x^2 - 2*(4*b^4*d^3*e - 12*a*b^3*d^2*e^2 + 8*a^2*b^2*d*e^3
- a^3*b*e^4)*x + 12*(a^2*b^2*d^2*e^2 - 2*a^3*b*d*e^3 + a^4*e^4 + (b^4*d^2*e^2 - 2*a*b^3*d*e^3 + a^2*b^2*e^4)*
x^2 + 2*(a*b^3*d^2*e^2 - 2*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.06, size = 341, normalized size = 1.62 \[ \frac {\left (b^{4} e^{4} x^{4}+12 a^{2} b^{2} e^{4} x^{2} \ln \left (b x +a \right )-24 a \,b^{3} d \,e^{3} x^{2} \ln \left (b x +a \right )-4 a \,b^{3} e^{4} x^{3}+12 b^{4} d^{2} e^{2} x^{2} \ln \left (b x +a \right )+8 b^{4} d \,e^{3} x^{3}+24 a^{3} b \,e^{4} x \ln \left (b x +a \right )-48 a^{2} b^{2} d \,e^{3} x \ln \left (b x +a \right )-11 a^{2} b^{2} e^{4} x^{2}+24 a \,b^{3} d^{2} e^{2} x \ln \left (b x +a \right )+16 a \,b^{3} d \,e^{3} x^{2}+12 a^{4} e^{4} \ln \left (b x +a \right )-24 a^{3} b d \,e^{3} \ln \left (b x +a \right )+2 a^{3} b \,e^{4} x +12 a^{2} b^{2} d^{2} e^{2} \ln \left (b x +a \right )-16 a^{2} b^{2} d \,e^{3} x +24 a \,b^{3} d^{2} e^{2} x -8 b^{4} d^{3} e x +7 a^{4} e^{4}-20 a^{3} b d \,e^{3}+18 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e -b^{4} d^{4}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2*(b^4*e^4*x^4+12*ln(b*x+a)*x^2*a^2*b^2*e^4-24*ln(b*x+a)*x^2*a*b^3*d*e^3+12*ln(b*x+a)*x^2*b^4*d^2*e^2-4*a*b^
3*e^4*x^3+8*b^4*d*e^3*x^3+24*ln(b*x+a)*x*a^3*b*e^4-48*ln(b*x+a)*x*a^2*b^2*d*e^3+24*ln(b*x+a)*x*a*b^3*d^2*e^2-1
1*a^2*b^2*e^4*x^2+16*a*b^3*d*e^3*x^2+12*a^4*e^4*ln(b*x+a)-24*a^3*b*d*e^3*ln(b*x+a)+12*a^2*b^2*d^2*e^2*ln(b*x+a
)+2*a^3*b*e^4*x-16*a^2*b^2*d*e^3*x+24*a*b^3*d^2*e^2*x-8*b^4*d^3*e*x+7*e^4*a^4-20*d*e^3*a^3*b+18*b^2*a^2*d^2*e^
2-4*a*b^3*d^3*e-b^4*d^4)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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maxima [B]  time = 1.13, size = 397, normalized size = 1.89 \[ \frac {e^{4} x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {4 \, d e^{3} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, a e^{4} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {6 \, d^{2} e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {12 \, a d e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {6 \, a^{2} e^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {4 \, d^{3} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {8 \, a^{2} d e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {5 \, a^{3} e^{4}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {12 \, a d^{2} e^{2} x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {24 \, a^{2} d e^{3} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {12 \, a^{3} e^{4} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {d^{4}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a d^{3} e}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, a^{2} d^{2} e^{2}}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {22 \, a^{3} d e^{3}}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, a^{4} e^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*e^4*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 4*d*e^3*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*a*e^4*
x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 6*d^2*e^2*log(x + a/b)/b^3 - 12*a*d*e^3*log(x + a/b)/b^4 + 6*a^2*e^4
*log(x + a/b)/b^5 - 4*d^3*e/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 8*a^2*d*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b
^4) - 5*a^3*e^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) + 12*a*d^2*e^2*x/(b^4*(x + a/b)^2) - 24*a^2*d*e^3*x/(b^5*(
x + a/b)^2) + 12*a^3*e^4*x/(b^6*(x + a/b)^2) - 1/2*d^4/(b^3*(x + a/b)^2) + 2*a*d^3*e/(b^4*(x + a/b)^2) + 9*a^2
*d^2*e^2/(b^5*(x + a/b)^2) - 22*a^3*d*e^3/(b^6*(x + a/b)^2) + 23/2*a^4*e^4/(b^7*(x + a/b)^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int((d + e*x)^4/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((d + e*x)**4/((a + b*x)**2)**(3/2), x)

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