3.1613 \(\int \frac {1}{(d+e x)^3 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=365 \[ \frac {5 b e^4 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^6}+\frac {e^4 (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^5}+\frac {15 b^2 e^4 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^7}-\frac {15 b^2 e^4 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^7}+\frac {10 b^2 e^3}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^6}-\frac {3 b^2 e^2}{(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {b^2 e}{(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^2}{4 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 365, 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, 44} \[ \frac {5 b e^4 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^6}+\frac {e^4 (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^5}+\frac {10 b^2 e^3}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^6}-\frac {3 b^2 e^2}{(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {15 b^2 e^4 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^7}-\frac {15 b^2 e^4 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^7}+\frac {b^2 e}{(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^2}{4 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(10*b^2*e^3)/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b^2/(4*(b*d - a*e)^3*(a + b*x)^3*Sqrt[a^2 + 2*a*b
*x + b^2*x^2]) + (b^2*e)/((b*d - a*e)^4*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b^2*e^2)/((b*d - a*e)^
5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (e^4*(a + b*x))/(2*(b*d - a*e)^5*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (5*b*e^4*(a + b*x))/((b*d - a*e)^6*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (15*b^2*e^4*(a + b*
x)*Log[a + b*x])/((b*d - a*e)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (15*b^2*e^4*(a + b*x)*Log[d + e*x])/((b*d - a
*e)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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 {1}{(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 {1}{\left (a b+b^2 x\right )^5 (d+e x)^3} \, 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 {1}{b^2 (b d-a e)^3 (a+b x)^5}-\frac {3 e}{b^2 (b d-a e)^4 (a+b x)^4}+\frac {6 e^2}{b^2 (b d-a e)^5 (a+b x)^3}-\frac {10 e^3}{b^2 (b d-a e)^6 (a+b x)^2}+\frac {15 e^4}{b^2 (b d-a e)^7 (a+b x)}-\frac {e^5}{b^5 (b d-a e)^5 (d+e x)^3}-\frac {5 e^5}{b^4 (b d-a e)^6 (d+e x)^2}-\frac {15 e^5}{b^3 (b d-a e)^7 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {10 b^2 e^3}{(b d-a e)^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2}{4 (b d-a e)^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 e}{(b d-a e)^4 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b^2 e^2}{(b d-a e)^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^4 (a+b x)}{2 (b d-a e)^5 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 b e^4 (a+b x)}{(b d-a e)^6 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {15 b^2 e^4 (a+b x) \log (a+b x)}{(b d-a e)^7 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 b^2 e^4 (a+b x) \log (d+e x)}{(b d-a e)^7 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 209, normalized size = 0.57 \[ \frac {-60 b^2 e^4 (a+b x)^3 \log (d+e x)+40 b^2 e^3 (a+b x)^2 (b d-a e)-12 b^2 e^2 (a+b x) (b d-a e)^2-\frac {b^2 (b d-a e)^4}{a+b x}+4 b^2 e (b d-a e)^3+60 b^2 e^4 (a+b x)^3 \log (a+b x)+\frac {20 b e^4 (a+b x)^3 (b d-a e)}{d+e x}+\frac {2 e^4 (a+b x)^3 (b d-a e)^2}{(d+e x)^2}}{4 \left ((a+b x)^2\right )^{3/2} (b d-a e)^7} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(4*b^2*e*(b*d - a*e)^3 - (b^2*(b*d - a*e)^4)/(a + b*x) - 12*b^2*e^2*(b*d - a*e)^2*(a + b*x) + 40*b^2*e^3*(b*d
- a*e)*(a + b*x)^2 + (2*e^4*(b*d - a*e)^2*(a + b*x)^3)/(d + e*x)^2 + (20*b*e^4*(b*d - a*e)*(a + b*x)^3)/(d + e
*x) + 60*b^2*e^4*(a + b*x)^3*Log[a + b*x] - 60*b^2*e^4*(a + b*x)^3*Log[d + e*x])/(4*(b*d - a*e)^7*((a + b*x)^2
)^(3/2))

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fricas [B]  time = 1.07, size = 1565, normalized size = 4.29 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/4*(b^6*d^6 - 8*a*b^5*d^5*e + 30*a^2*b^4*d^4*e^2 - 80*a^3*b^3*d^3*e^3 + 35*a^4*b^2*d^2*e^4 + 24*a^5*b*d*e^5
- 2*a^6*e^6 - 60*(b^6*d*e^5 - a*b^5*e^6)*x^5 - 30*(3*b^6*d^2*e^4 + 4*a*b^5*d*e^5 - 7*a^2*b^4*e^6)*x^4 - 20*(b^
6*d^3*e^3 + 15*a*b^5*d^2*e^4 - 3*a^2*b^4*d*e^5 - 13*a^3*b^3*e^6)*x^3 + 5*(b^6*d^4*e^2 - 16*a*b^5*d^3*e^3 - 66*
a^2*b^4*d^2*e^4 + 56*a^3*b^3*d*e^5 + 25*a^4*b^2*e^6)*x^2 - 2*(b^6*d^5*e - 10*a*b^5*d^4*e^2 + 60*a^2*b^4*d^3*e^
3 + 50*a^3*b^3*d^2*e^4 - 95*a^4*b^2*d*e^5 - 6*a^5*b*e^6)*x - 60*(b^6*e^6*x^6 + a^4*b^2*d^2*e^4 + 2*(b^6*d*e^5
+ 2*a*b^5*e^6)*x^5 + (b^6*d^2*e^4 + 8*a*b^5*d*e^5 + 6*a^2*b^4*e^6)*x^4 + 4*(a*b^5*d^2*e^4 + 3*a^2*b^4*d*e^5 +
a^3*b^3*e^6)*x^3 + (6*a^2*b^4*d^2*e^4 + 8*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^2 + 2*(2*a^3*b^3*d^2*e^4 + a^4*b^2*d*
e^5)*x)*log(b*x + a) + 60*(b^6*e^6*x^6 + a^4*b^2*d^2*e^4 + 2*(b^6*d*e^5 + 2*a*b^5*e^6)*x^5 + (b^6*d^2*e^4 + 8*
a*b^5*d*e^5 + 6*a^2*b^4*e^6)*x^4 + 4*(a*b^5*d^2*e^4 + 3*a^2*b^4*d*e^5 + a^3*b^3*e^6)*x^3 + (6*a^2*b^4*d^2*e^4
+ 8*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^2 + 2*(2*a^3*b^3*d^2*e^4 + a^4*b^2*d*e^5)*x)*log(e*x + d))/(a^4*b^7*d^9 - 7
*a^5*b^6*d^8*e + 21*a^6*b^5*d^7*e^2 - 35*a^7*b^4*d^6*e^3 + 35*a^8*b^3*d^5*e^4 - 21*a^9*b^2*d^4*e^5 + 7*a^10*b*
d^3*e^6 - a^11*d^2*e^7 + (b^11*d^7*e^2 - 7*a*b^10*d^6*e^3 + 21*a^2*b^9*d^5*e^4 - 35*a^3*b^8*d^4*e^5 + 35*a^4*b
^7*d^3*e^6 - 21*a^5*b^6*d^2*e^7 + 7*a^6*b^5*d*e^8 - a^7*b^4*e^9)*x^6 + 2*(b^11*d^8*e - 5*a*b^10*d^7*e^2 + 7*a^
2*b^9*d^6*e^3 + 7*a^3*b^8*d^5*e^4 - 35*a^4*b^7*d^4*e^5 + 49*a^5*b^6*d^3*e^6 - 35*a^6*b^5*d^2*e^7 + 13*a^7*b^4*
d*e^8 - 2*a^8*b^3*e^9)*x^5 + (b^11*d^9 + a*b^10*d^8*e - 29*a^2*b^9*d^7*e^2 + 91*a^3*b^8*d^6*e^3 - 119*a^4*b^7*
d^5*e^4 + 49*a^5*b^6*d^4*e^5 + 49*a^6*b^5*d^3*e^6 - 71*a^7*b^4*d^2*e^7 + 34*a^8*b^3*d*e^8 - 6*a^9*b^2*e^9)*x^4
+ 4*(a*b^10*d^9 - 4*a^2*b^9*d^8*e + a^3*b^8*d^7*e^2 + 21*a^4*b^7*d^6*e^3 - 49*a^5*b^6*d^5*e^4 + 49*a^6*b^5*d^
4*e^5 - 21*a^7*b^4*d^3*e^6 - a^8*b^3*d^2*e^7 + 4*a^9*b^2*d*e^8 - a^10*b*e^9)*x^3 + (6*a^2*b^9*d^9 - 34*a^3*b^8
*d^8*e + 71*a^4*b^7*d^7*e^2 - 49*a^5*b^6*d^6*e^3 - 49*a^6*b^5*d^5*e^4 + 119*a^7*b^4*d^4*e^5 - 91*a^8*b^3*d^3*e
^6 + 29*a^9*b^2*d^2*e^7 - a^10*b*d*e^8 - a^11*e^9)*x^2 + 2*(2*a^3*b^8*d^9 - 13*a^4*b^7*d^8*e + 35*a^5*b^6*d^7*
e^2 - 49*a^6*b^5*d^6*e^3 + 35*a^7*b^4*d^5*e^4 - 7*a^8*b^3*d^4*e^5 - 7*a^9*b^2*d^3*e^6 + 5*a^10*b*d^2*e^7 - a^1
1*d*e^8)*x)

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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(1/(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 = 0.07, size = 983, normalized size = 2.69 \[ -\frac {\left (60 b^{6} e^{6} x^{6} \ln \left (b x +a \right )-60 b^{6} e^{6} x^{6} \ln \left (e x +d \right )+240 a \,b^{5} e^{6} x^{5} \ln \left (b x +a \right )-240 a \,b^{5} e^{6} x^{5} \ln \left (e x +d \right )+120 b^{6} d \,e^{5} x^{5} \ln \left (b x +a \right )-120 b^{6} d \,e^{5} x^{5} \ln \left (e x +d \right )+360 a^{2} b^{4} e^{6} x^{4} \ln \left (b x +a \right )-360 a^{2} b^{4} e^{6} x^{4} \ln \left (e x +d \right )+480 a \,b^{5} d \,e^{5} x^{4} \ln \left (b x +a \right )-480 a \,b^{5} d \,e^{5} x^{4} \ln \left (e x +d \right )-60 a \,b^{5} e^{6} x^{5}+60 b^{6} d^{2} e^{4} x^{4} \ln \left (b x +a \right )-60 b^{6} d^{2} e^{4} x^{4} \ln \left (e x +d \right )+60 b^{6} d \,e^{5} x^{5}+240 a^{3} b^{3} e^{6} x^{3} \ln \left (b x +a \right )-240 a^{3} b^{3} e^{6} x^{3} \ln \left (e x +d \right )+720 a^{2} b^{4} d \,e^{5} x^{3} \ln \left (b x +a \right )-720 a^{2} b^{4} d \,e^{5} x^{3} \ln \left (e x +d \right )-210 a^{2} b^{4} e^{6} x^{4}+240 a \,b^{5} d^{2} e^{4} x^{3} \ln \left (b x +a \right )-240 a \,b^{5} d^{2} e^{4} x^{3} \ln \left (e x +d \right )+120 a \,b^{5} d \,e^{5} x^{4}+90 b^{6} d^{2} e^{4} x^{4}+60 a^{4} b^{2} e^{6} x^{2} \ln \left (b x +a \right )-60 a^{4} b^{2} e^{6} x^{2} \ln \left (e x +d \right )+480 a^{3} b^{3} d \,e^{5} x^{2} \ln \left (b x +a \right )-480 a^{3} b^{3} d \,e^{5} x^{2} \ln \left (e x +d \right )-260 a^{3} b^{3} e^{6} x^{3}+360 a^{2} b^{4} d^{2} e^{4} x^{2} \ln \left (b x +a \right )-360 a^{2} b^{4} d^{2} e^{4} x^{2} \ln \left (e x +d \right )-60 a^{2} b^{4} d \,e^{5} x^{3}+300 a \,b^{5} d^{2} e^{4} x^{3}+20 b^{6} d^{3} e^{3} x^{3}+120 a^{4} b^{2} d \,e^{5} x \ln \left (b x +a \right )-120 a^{4} b^{2} d \,e^{5} x \ln \left (e x +d \right )-125 a^{4} b^{2} e^{6} x^{2}+240 a^{3} b^{3} d^{2} e^{4} x \ln \left (b x +a \right )-240 a^{3} b^{3} d^{2} e^{4} x \ln \left (e x +d \right )-280 a^{3} b^{3} d \,e^{5} x^{2}+330 a^{2} b^{4} d^{2} e^{4} x^{2}+80 a \,b^{5} d^{3} e^{3} x^{2}-5 b^{6} d^{4} e^{2} x^{2}-12 a^{5} b \,e^{6} x +60 a^{4} b^{2} d^{2} e^{4} \ln \left (b x +a \right )-60 a^{4} b^{2} d^{2} e^{4} \ln \left (e x +d \right )-190 a^{4} b^{2} d \,e^{5} x +100 a^{3} b^{3} d^{2} e^{4} x +120 a^{2} b^{4} d^{3} e^{3} x -20 a \,b^{5} d^{4} e^{2} x +2 b^{6} d^{5} e x +2 a^{6} e^{6}-24 a^{5} b d \,e^{5}-35 a^{4} b^{2} d^{2} e^{4}+80 a^{3} b^{3} d^{3} e^{3}-30 a^{2} b^{4} d^{4} e^{2}+8 a \,b^{5} d^{5} e -b^{6} d^{6}\right ) \left (b x +a \right )}{4 \left (e x +d \right )^{2} \left (a e -b d \right )^{7} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/4*(480*a^3*b^3*d*e^5*x^2*ln(b*x+a)+360*a^2*b^4*d^2*e^4*x^2*ln(b*x+a)+720*a^2*b^4*d*e^5*x^3*ln(b*x+a)+240*a*
b^5*d^2*e^4*x^3*ln(b*x+a)+240*a^3*b^3*d^2*e^4*x*ln(b*x+a)+480*a*b^5*d*e^5*x^4*ln(b*x+a)-24*a^5*b*d*e^5-35*a^4*
b^2*d^2*e^4+120*a^4*b^2*d*e^5*x*ln(b*x+a)+8*a*b^5*d^5*e+2*a^6*e^6-b^6*d^6-210*a^2*b^4*e^6*x^4-260*a^3*b^3*e^6*
x^3-60*a*b^5*e^6*x^5+60*b^6*d*e^5*x^5+2*b^6*d^5*e*x-125*a^4*b^2*e^6*x^2-5*b^6*d^4*e^2*x^2-12*a^5*b*e^6*x+20*b^
6*d^3*e^3*x^3+80*a^3*b^3*d^3*e^3-30*a^2*b^4*d^4*e^2+90*x^4*b^6*d^2*e^4+60*ln(b*x+a)*x^6*b^6*e^6-60*ln(e*x+d)*x
^6*b^6*e^6-240*ln(e*x+d)*x^3*a*b^5*d^2*e^4-480*ln(e*x+d)*x^2*a^3*b^3*d*e^5-360*ln(e*x+d)*x^2*a^2*b^4*d^2*e^4-1
20*ln(e*x+d)*x*a^4*b^2*d*e^5-240*ln(e*x+d)*x*a^3*b^3*d^2*e^4-480*ln(e*x+d)*x^4*a*b^5*d*e^5-720*ln(e*x+d)*x^3*a
^2*b^4*d*e^5+240*ln(b*x+a)*x^5*a*b^5*e^6+120*ln(b*x+a)*x^5*b^6*d*e^5-240*ln(e*x+d)*x^5*a*b^5*e^6-120*ln(e*x+d)
*x^5*b^6*d*e^5-360*ln(e*x+d)*x^4*a^2*b^4*e^6-60*ln(e*x+d)*x^4*b^6*d^2*e^4-240*ln(e*x+d)*x^3*a^3*b^3*e^6-60*ln(
e*x+d)*x^2*a^4*b^2*e^6-60*ln(e*x+d)*a^4*b^2*d^2*e^4+120*a*b^5*d*e^5*x^4-60*a^2*b^4*d*e^5*x^3+360*a^2*b^4*e^6*x
^4*ln(b*x+a)+60*b^6*d^2*e^4*x^4*ln(b*x+a)+240*a^3*b^3*e^6*x^3*ln(b*x+a)+60*a^4*b^2*e^6*x^2*ln(b*x+a)+60*a^4*b^
2*d^2*e^4*ln(b*x+a)-190*a^4*b^2*d*e^5*x-280*a^3*b^3*d*e^5*x^2+100*a^3*b^3*d^2*e^4*x+120*a^2*b^4*d^3*e^3*x-20*a
*b^5*d^4*e^2*x+300*a*b^5*d^2*e^4*x^3+330*a^2*b^4*d^2*e^4*x^2+80*a*b^5*d^3*e^3*x^2)*(b*x+a)/(e*x+d)^2/(a*e-b*d)
^7/((b*x+a)^2)^(5/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (d+e\,x\right )}^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)

[Out]

int(1/((d + e*x)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d + e x\right )^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral(1/((d + e*x)**3*((a + b*x)**2)**(5/2)), x)

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