3.16.65 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^6} \, dx\) [1565]

Optimal. Leaf size=98 \[ \frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (b d-a e) (d+e x)^5}+\frac {b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{20 (b d-a e)^2 (d+e x)^4} \]

[Out]

1/5*(b*x+a)^3*((b*x+a)^2)^(1/2)/(-a*e+b*d)/(e*x+d)^5+1/20*b*(b*x+a)^3*((b*x+a)^2)^(1/2)/(-a*e+b*d)^2/(e*x+d)^4

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Rubi [A]
time = 0.03, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {660, 47, 37} \begin {gather*} \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{20 (d+e x)^4 (b d-a e)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{5 (d+e x)^5 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^6,x]

[Out]

((a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(b*d - a*e)*(d + e*x)^5) + (b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/(20*(b*d - a*e)^2*(d + e*x)^4)

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 47

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] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^6} \, dx &=\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)^6} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (b d-a e) (d+e x)^5}+\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)^5} \, dx}{5 b (b d-a e) \left (a b+b^2 x\right )}\\ &=\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (b d-a e) (d+e x)^5}+\frac {b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{20 (b d-a e)^2 (d+e x)^4}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 112, normalized size = 1.14 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (4 a^3 e^3+3 a^2 b e^2 (d+5 e x)+2 a b^2 e \left (d^2+5 d e x+10 e^2 x^2\right )+b^3 \left (d^3+5 d^2 e x+10 d e^2 x^2+10 e^3 x^3\right )\right )}{20 e^4 (a+b x) (d+e x)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^6,x]

[Out]

-1/20*(Sqrt[(a + b*x)^2]*(4*a^3*e^3 + 3*a^2*b*e^2*(d + 5*e*x) + 2*a*b^2*e*(d^2 + 5*d*e*x + 10*e^2*x^2) + b^3*(
d^3 + 5*d^2*e*x + 10*d*e^2*x^2 + 10*e^3*x^3)))/(e^4*(a + b*x)*(d + e*x)^5)

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Maple [A]
time = 0.64, size = 131, normalized size = 1.34

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b^{3} x^{3}}{2 e}-\frac {b^{2} \left (2 a e +b d \right ) x^{2}}{2 e^{2}}-\frac {b \left (3 a^{2} e^{2}+2 a b d e +b^{2} d^{2}\right ) x}{4 e^{3}}-\frac {4 e^{3} a^{3}+3 a^{2} b d \,e^{2}+2 a \,b^{2} d^{2} e +b^{3} d^{3}}{20 e^{4}}\right )}{\left (b x +a \right ) \left (e x +d \right )^{5}}\)	\(126\)
gosper	\(-\frac {\left (10 b^{3} e^{3} x^{3}+20 a \,b^{2} e^{3} x^{2}+10 b^{3} d \,e^{2} x^{2}+15 a^{2} b \,e^{3} x +10 a \,b^{2} d \,e^{2} x +5 b^{3} d^{2} e x +4 e^{3} a^{3}+3 a^{2} b d \,e^{2}+2 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 e^{4} \left (e x +d \right )^{5} \left (b x +a \right )^{3}}\)	\(131\)
default	\(-\frac {\left (10 b^{3} e^{3} x^{3}+20 a \,b^{2} e^{3} x^{2}+10 b^{3} d \,e^{2} x^{2}+15 a^{2} b \,e^{3} x +10 a \,b^{2} d \,e^{2} x +5 b^{3} d^{2} e x +4 e^{3} a^{3}+3 a^{2} b d \,e^{2}+2 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 e^{4} \left (e x +d \right )^{5} \left (b x +a \right )^{3}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x,method=_RETURNVERBOSE)

[Out]

-1/20/e^4*(10*b^3*e^3*x^3+20*a*b^2*e^3*x^2+10*b^3*d*e^2*x^2+15*a^2*b*e^3*x+10*a*b^2*d*e^2*x+5*b^3*d^2*e*x+4*a^
3*e^3+3*a^2*b*d*e^2+2*a*b^2*d^2*e+b^3*d^3)*((b*x+a)^2)^(3/2)/(e*x+d)^5/(b*x+a)^3

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 4.00, size = 144, normalized size = 1.47 \begin {gather*} -\frac {b^{3} d^{3} + {\left (10 \, b^{3} x^{3} + 20 \, a b^{2} x^{2} + 15 \, a^{2} b x + 4 \, a^{3}\right )} e^{3} + {\left (10 \, b^{3} d x^{2} + 10 \, a b^{2} d x + 3 \, a^{2} b d\right )} e^{2} + {\left (5 \, b^{3} d^{2} x + 2 \, a b^{2} d^{2}\right )} e}{20 \, {\left (x^{5} e^{9} + 5 \, d x^{4} e^{8} + 10 \, d^{2} x^{3} e^{7} + 10 \, d^{3} x^{2} e^{6} + 5 \, d^{4} x e^{5} + d^{5} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

-1/20*(b^3*d^3 + (10*b^3*x^3 + 20*a*b^2*x^2 + 15*a^2*b*x + 4*a^3)*e^3 + (10*b^3*d*x^2 + 10*a*b^2*d*x + 3*a^2*b
*d)*e^2 + (5*b^3*d^2*x + 2*a*b^2*d^2)*e)/(x^5*e^9 + 5*d*x^4*e^8 + 10*d^2*x^3*e^7 + 10*d^3*x^2*e^6 + 5*d^4*x*e^
5 + d^5*e^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**6,x)

[Out]

Integral(((a + b*x)**2)**(3/2)/(d + e*x)**6, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (76) = 152\).
time = 1.73, size = 169, normalized size = 1.72 \begin {gather*} -\frac {{\left (10 \, b^{3} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, b^{3} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) + 20 \, a b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, a b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 15 \, a^{2} b x e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )}}{20 \, {\left (x e + d\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

-1/20*(10*b^3*x^3*e^3*sgn(b*x + a) + 10*b^3*d*x^2*e^2*sgn(b*x + a) + 5*b^3*d^2*x*e*sgn(b*x + a) + b^3*d^3*sgn(
b*x + a) + 20*a*b^2*x^2*e^3*sgn(b*x + a) + 10*a*b^2*d*x*e^2*sgn(b*x + a) + 2*a*b^2*d^2*e*sgn(b*x + a) + 15*a^2
*b*x*e^3*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) + 4*a^3*e^3*sgn(b*x + a))*e^(-4)/(x*e + d)^5

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Mupad [B]
time = 0.63, size = 284, normalized size = 2.90 \begin {gather*} \frac {\left (\frac {2\,b^3\,d-3\,a\,b^2\,e}{3\,e^4}+\frac {b^3\,d}{3\,e^4}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}-\frac {\left (\frac {3\,a^2\,b\,e^2-3\,a\,b^2\,d\,e+b^3\,d^2}{4\,e^4}+\frac {d\,\left (\frac {b^3\,d}{4\,e^3}-\frac {b^2\,\left (3\,a\,e-b\,d\right )}{4\,e^3}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {\left (\frac {a^3}{5\,e}-\frac {d\,\left (\frac {3\,a^2\,b}{5\,e}-\frac {d\,\left (\frac {3\,a\,b^2}{5\,e}-\frac {b^3\,d}{5\,e^2}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,e^4\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(3/2)/(d + e*x)^6,x)

[Out]

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

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