3.14.9 \(\int \frac {(d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=48 \[ -\frac {(d+e x)^4}{4 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {646, 37} \begin {gather*} -\frac {(d+e x)^4}{4 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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

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Mathematica [B]  time = 0.05, size = 106, normalized size = 2.21 \begin {gather*} \frac {-a^3 e^3-a^2 b e^2 (d+4 e x)-a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )-\left (b^3 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )}{4 b^4 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

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

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IntegrateAlgebraic [B]  time = 1.84, size = 598, normalized size = 12.46 \begin {gather*} \frac {-2 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (a^6 e^3-3 a^5 b d e^2-a^5 b e^3 x+3 a^4 b^2 d^2 e+3 a^4 b^2 d e^2 x+a^4 b^2 e^3 x^2-a^3 b^3 d^3-3 a^3 b^3 d^2 e x-3 a^3 b^3 d e^2 x^2-a^3 b^3 e^3 x^3+a^2 b^4 d^3 x+3 a^2 b^4 d^2 e x^2+3 a^2 b^4 d e^2 x^3+2 a^2 b^4 e^3 x^4-a b^5 d^3 x^2-3 a b^5 d^2 e x^3-2 a b^5 d e^2 x^4+2 a b^5 e^3 x^5+b^6 d^3 x^3+4 b^6 d^2 e x^4+6 b^6 d e^2 x^5+4 b^6 e^3 x^6\right )-2 \left (a^7 b e^3-3 a^6 b^2 d e^2+3 a^5 b^3 d^2 e-a^4 b^4 d^3-a^3 b^5 e^3 x^4-a^2 b^6 d e^2 x^4-4 a^2 b^6 e^3 x^5-a b^7 d^2 e x^4-4 a b^7 d e^2 x^5-6 a b^7 e^3 x^6-b^8 d^3 x^4-4 b^8 d^2 e x^5-6 b^8 d e^2 x^6-4 b^8 e^3 x^7\right )}{b^4 x^4 \sqrt {a^2+2 a b x+b^2 x^2} \left (-8 a^3 b^5-24 a^2 b^6 x-24 a b^7 x^2-8 b^8 x^3\right )+b^4 \sqrt {b^2} x^4 \left (8 a^4 b^4+32 a^3 b^5 x+48 a^2 b^6 x^2+32 a b^7 x^3+8 b^8 x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^3/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(-2*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-(a^3*b^3*d^3) + 3*a^4*b^2*d^2*e - 3*a^5*b*d*e^2 + a^6*e^3 + a^2*
b^4*d^3*x - 3*a^3*b^3*d^2*e*x + 3*a^4*b^2*d*e^2*x - a^5*b*e^3*x - a*b^5*d^3*x^2 + 3*a^2*b^4*d^2*e*x^2 - 3*a^3*
b^3*d*e^2*x^2 + a^4*b^2*e^3*x^2 + b^6*d^3*x^3 - 3*a*b^5*d^2*e*x^3 + 3*a^2*b^4*d*e^2*x^3 - a^3*b^3*e^3*x^3 + 4*
b^6*d^2*e*x^4 - 2*a*b^5*d*e^2*x^4 + 2*a^2*b^4*e^3*x^4 + 6*b^6*d*e^2*x^5 + 2*a*b^5*e^3*x^5 + 4*b^6*e^3*x^6) - 2
*(-(a^4*b^4*d^3) + 3*a^5*b^3*d^2*e - 3*a^6*b^2*d*e^2 + a^7*b*e^3 - b^8*d^3*x^4 - a*b^7*d^2*e*x^4 - a^2*b^6*d*e
^2*x^4 - a^3*b^5*e^3*x^4 - 4*b^8*d^2*e*x^5 - 4*a*b^7*d*e^2*x^5 - 4*a^2*b^6*e^3*x^5 - 6*b^8*d*e^2*x^6 - 6*a*b^7
*e^3*x^6 - 4*b^8*e^3*x^7))/(b^4*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^3*b^5 - 24*a^2*b^6*x - 24*a*b^7*x^2 -
8*b^8*x^3) + b^4*Sqrt[b^2]*x^4*(8*a^4*b^4 + 32*a^3*b^5*x + 48*a^2*b^6*x^2 + 32*a*b^7*x^3 + 8*b^8*x^4))

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fricas [B]  time = 0.41, size = 143, normalized size = 2.98 \begin {gather*} -\frac {4 \, b^{3} e^{3} x^{3} + b^{3} d^{3} + a b^{2} d^{2} e + a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + 4 \, {\left (b^{3} d^{2} e + a b^{2} d e^{2} + a^{2} b e^{3}\right )} x}{4 \, {\left (b^{8} x^{4} + 4 \, a b^{7} x^{3} + 6 \, a^{2} b^{6} x^{2} + 4 \, a^{3} b^{5} x + a^{4} b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/4*(4*b^3*e^3*x^3 + b^3*d^3 + a*b^2*d^2*e + a^2*b*d*e^2 + a^3*e^3 + 6*(b^3*d*e^2 + a*b^2*e^3)*x^2 + 4*(b^3*d
^2*e + a*b^2*d*e^2 + a^2*b*e^3)*x)/(b^8*x^4 + 4*a*b^7*x^3 + 6*a^2*b^6*x^2 + 4*a^3*b^5*x + a^4*b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.05, size = 119, normalized size = 2.48 \begin {gather*} -\frac {\left (b x +a \right ) \left (4 b^{3} e^{3} x^{3}+6 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+4 a^{2} b \,e^{3} x +4 a \,b^{2} d \,e^{2} x +4 b^{3} d^{2} e x +a^{3} e^{3}+a^{2} b d \,e^{2}+a \,b^{2} d^{2} e +b^{3} d^{3}\right )}{4 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

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

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maxima [B]  time = 1.14, size = 238, normalized size = 4.96 \begin {gather*} -\frac {e^{3} x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {d^{2} e}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, a^{2} e^{3}}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} - \frac {3 \, d e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {a e^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a d e^{2}}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {2 \, a^{2} e^{3}}{3 \, b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {d^{3}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {3 \, a d^{2} e}{4 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {3 \, a^{2} d e^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {a^{3} e^{3}}{4 \, b^{8} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

-e^3*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - d^2*e/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 2/3*a^2*e^3/((b
^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) - 3/2*d*e^2/(b^5*(x + a/b)^2) - 1/2*a*e^3/(b^6*(x + a/b)^2) + 2*a*d*e^2/(b^
6*(x + a/b)^3) + 2/3*a^2*e^3/(b^7*(x + a/b)^3) - 1/4*d^3/(b^5*(x + a/b)^4) + 3/4*a*d^2*e/(b^6*(x + a/b)^4) - 3
/4*a^2*d*e^2/(b^7*(x + a/b)^4) + 1/4*a^3*e^3/(b^8*(x + a/b)^4)

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mupad [B]  time = 0.75, size = 254, normalized size = 5.29 \begin {gather*} \frac {\left (\frac {2\,a\,e^3-3\,b\,d\,e^2}{2\,b^4}+\frac {a\,e^3}{2\,b^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^3}-\frac {\left (\frac {d^3}{4\,b}-\frac {a\,\left (\frac {3\,d^2\,e}{4\,b}+\frac {a\,\left (\frac {a\,e^3}{4\,b^2}-\frac {3\,d\,e^2}{4\,b}\right )}{b}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^5}-\frac {\left (\frac {a^2\,e^3-3\,a\,b\,d\,e^2+3\,b^2\,d^2\,e}{3\,b^4}+\frac {a\,\left (\frac {a\,e^3}{3\,b^3}+\frac {e^2\,\left (a\,e-3\,b\,d\right )}{3\,b^3}\right )}{b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{{\left (a+b\,x\right )}^4}-\frac {e^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{b^4\,{\left (a+b\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((d + e*x)**3/((a + b*x)**2)**(5/2), x)

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