3.16.76 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^2} \, dx\) [1576]

Optimal. Leaf size=292 \[ -\frac {10 b^2 (b d-a e)^3 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}+\frac {5 b^3 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}-\frac {5 b^4 (b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x)}+\frac {b^5 (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^6 (a+b x)}+\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^6 (a+b x)} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 292, 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 = {660, 45} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{e^6 (a+b x) (d+e x)}+\frac {5 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4 \log (d+e x)}{e^6 (a+b x)}-\frac {10 b^2 x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^5 (a+b x)}+\frac {b^5 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4}{4 e^6 (a+b x)}-\frac {5 b^4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}{3 e^6 (a+b x)}+\frac {5 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^2}{e^6 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^5}{(d+e x)^2} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {10 b^7 (b d-a e)^3}{e^5}-\frac {b^5 (b d-a e)^5}{e^5 (d+e x)^2}+\frac {5 b^6 (b d-a e)^4}{e^5 (d+e x)}+\frac {10 b^8 (b d-a e)^2 (d+e x)}{e^5}-\frac {5 b^9 (b d-a e) (d+e x)^2}{e^5}+\frac {b^{10} (d+e x)^3}{e^5}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac {10 b^2 (b d-a e)^3 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac {(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}+\frac {5 b^3 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}-\frac {5 b^4 (b d-a e) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x)}+\frac {b^5 (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^6 (a+b x)}+\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^6 (a+b x)}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 246, normalized size = 0.84 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (60 a^4 b d e^4-12 a^5 e^5+120 a^3 b^2 e^3 \left (-d^2+d e x+e^2 x^2\right )+60 a^2 b^3 e^2 \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+20 a b^4 e \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )+b^5 \left (12 d^5-48 d^4 e x-30 d^3 e^2 x^2+10 d^2 e^3 x^3-5 d e^4 x^4+3 e^5 x^5\right )+60 b (b d-a e)^4 (d+e x) \log (d+e x)\right )}{12 e^6 (a+b x) (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(60*a^4*b*d*e^4 - 12*a^5*e^5 + 120*a^3*b^2*e^3*(-d^2 + d*e*x + e^2*x^2) + 60*a^2*b^3*e^2*(2
*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^3) + 20*a*b^4*e*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4
*x^4) + b^5*(12*d^5 - 48*d^4*e*x - 30*d^3*e^2*x^2 + 10*d^2*e^3*x^3 - 5*d*e^4*x^4 + 3*e^5*x^5) + 60*b*(b*d - a*
e)^4*(d + e*x)*Log[d + e*x]))/(12*e^6*(a + b*x)*(d + e*x))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(455\) vs. \(2(222)=444\).
time = 0.64, size = 456, normalized size = 1.56

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{2} \left (\frac {1}{4} b^{3} x^{4} e^{3}+\frac {5}{3} a \,b^{2} e^{3} x^{3}-\frac {2}{3} b^{3} d \,e^{2} x^{3}+5 a^{2} b \,e^{3} x^{2}-5 a \,b^{2} d \,e^{2} x^{2}+\frac {3}{2} b^{3} d^{2} e \,x^{2}+10 e^{3} a^{3} x -20 a^{2} b d \,e^{2} x +15 a \,b^{2} d^{2} e x -4 b^{3} d^{3} x \right )}{\left (b x +a \right ) e^{5}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a^{5} e^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}{\left (b x +a \right ) e^{6} \left (e x +d \right )}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, b \left (e^{4} a^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{6}}\)	\(307\)
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (360 \ln \left (e x +d \right ) a^{2} b^{3} d^{2} e^{3} x -240 \ln \left (e x +d \right ) a \,b^{4} d^{3} e^{2} x -40 a \,b^{4} d \,e^{4} x^{3}+3 b^{5} e^{5} x^{5}+180 a \,b^{4} d^{3} e^{2} x +60 \ln \left (e x +d \right ) a^{4} b \,e^{5} x +12 b^{5} d^{5}+60 \ln \left (e x +d \right ) a^{4} b d \,e^{4}-240 \ln \left (e x +d \right ) a^{3} b^{2} d^{2} e^{3}+360 \ln \left (e x +d \right ) a^{2} b^{3} d^{3} e^{2}-240 \ln \left (e x +d \right ) a \,b^{4} d^{4} e +60 \ln \left (e x +d \right ) b^{5} d^{5}-180 a^{2} b^{3} d \,e^{4} x^{2}+120 a \,b^{4} d^{2} e^{3} x^{2}+120 a^{3} b^{2} d \,e^{4} x -240 a^{2} b^{3} d^{2} e^{3} x -12 a^{5} e^{5}+60 \ln \left (e x +d \right ) b^{5} d^{4} e x -240 \ln \left (e x +d \right ) a^{3} b^{2} d \,e^{4} x +20 a \,b^{4} e^{5} x^{4}-5 b^{5} d \,e^{4} x^{4}+60 a^{2} b^{3} e^{5} x^{3}+10 b^{5} d^{2} e^{3} x^{3}+120 a^{3} b^{2} e^{5} x^{2}-30 b^{5} d^{3} e^{2} x^{2}-48 b^{5} d^{4} e x +60 a^{4} b d \,e^{4}-120 a^{3} b^{2} d^{2} e^{3}+120 a^{2} b^{3} d^{3} e^{2}-60 a \,b^{4} d^{4} e \right )}{12 \left (b x +a \right )^{5} e^{6} \left (e x +d \right )}\)	\(456\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((b*x+a)^2)^(5/2)*(360*ln(e*x+d)*a^2*b^3*d^2*e^3*x-240*ln(e*x+d)*a*b^4*d^3*e^2*x-40*a*b^4*d*e^4*x^3+3*b^5
*e^5*x^5+180*a*b^4*d^3*e^2*x+60*ln(e*x+d)*a^4*b*e^5*x+12*b^5*d^5+60*ln(e*x+d)*a^4*b*d*e^4-240*ln(e*x+d)*a^3*b^
2*d^2*e^3+360*ln(e*x+d)*a^2*b^3*d^3*e^2-240*ln(e*x+d)*a*b^4*d^4*e+60*ln(e*x+d)*b^5*d^5-180*a^2*b^3*d*e^4*x^2+1
20*a*b^4*d^2*e^3*x^2+120*a^3*b^2*d*e^4*x-240*a^2*b^3*d^2*e^3*x-12*a^5*e^5+60*ln(e*x+d)*b^5*d^4*e*x-240*ln(e*x+
d)*a^3*b^2*d*e^4*x+20*a*b^4*e^5*x^4-5*b^5*d*e^4*x^4+60*a^2*b^3*e^5*x^3+10*b^5*d^2*e^3*x^3+120*a^3*b^2*e^5*x^2-
30*b^5*d^3*e^2*x^2-48*b^5*d^4*e*x+60*a^4*b*d*e^4-120*a^3*b^2*d^2*e^3+120*a^2*b^3*d^3*e^2-60*a*b^4*d^4*e)/(b*x+
a)^5/e^6/(e*x+d)

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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)^(5/2)/(e*x+d)^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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 3.04, size = 353, normalized size = 1.21 \begin {gather*} \frac {12 \, b^{5} d^{5} + {\left (3 \, b^{5} x^{5} + 20 \, a b^{4} x^{4} + 60 \, a^{2} b^{3} x^{3} + 120 \, a^{3} b^{2} x^{2} - 12 \, a^{5}\right )} e^{5} - 5 \, {\left (b^{5} d x^{4} + 8 \, a b^{4} d x^{3} + 36 \, a^{2} b^{3} d x^{2} - 24 \, a^{3} b^{2} d x - 12 \, a^{4} b d\right )} e^{4} + 10 \, {\left (b^{5} d^{2} x^{3} + 12 \, a b^{4} d^{2} x^{2} - 24 \, a^{2} b^{3} d^{2} x - 12 \, a^{3} b^{2} d^{2}\right )} e^{3} - 30 \, {\left (b^{5} d^{3} x^{2} - 6 \, a b^{4} d^{3} x - 4 \, a^{2} b^{3} d^{3}\right )} e^{2} - 12 \, {\left (4 \, b^{5} d^{4} x + 5 \, a b^{4} d^{4}\right )} e + 60 \, {\left (b^{5} d^{5} + a^{4} b x e^{5} - {\left (4 \, a^{3} b^{2} d x - a^{4} b d\right )} e^{4} + 2 \, {\left (3 \, a^{2} b^{3} d^{2} x - 2 \, a^{3} b^{2} d^{2}\right )} e^{3} - 2 \, {\left (2 \, a b^{4} d^{3} x - 3 \, a^{2} b^{3} d^{3}\right )} e^{2} + {\left (b^{5} d^{4} x - 4 \, a b^{4} d^{4}\right )} e\right )} \log \left (x e + d\right )}{12 \, {\left (x e^{7} + d e^{6}\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)^(5/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

1/12*(12*b^5*d^5 + (3*b^5*x^5 + 20*a*b^4*x^4 + 60*a^2*b^3*x^3 + 120*a^3*b^2*x^2 - 12*a^5)*e^5 - 5*(b^5*d*x^4 +
8*a*b^4*d*x^3 + 36*a^2*b^3*d*x^2 - 24*a^3*b^2*d*x - 12*a^4*b*d)*e^4 + 10*(b^5*d^2*x^3 + 12*a*b^4*d^2*x^2 - 24
*a^2*b^3*d^2*x - 12*a^3*b^2*d^2)*e^3 - 30*(b^5*d^3*x^2 - 6*a*b^4*d^3*x - 4*a^2*b^3*d^3)*e^2 - 12*(4*b^5*d^4*x
+ 5*a*b^4*d^4)*e + 60*(b^5*d^5 + a^4*b*x*e^5 - (4*a^3*b^2*d*x - a^4*b*d)*e^4 + 2*(3*a^2*b^3*d^2*x - 2*a^3*b^2*
d^2)*e^3 - 2*(2*a*b^4*d^3*x - 3*a^2*b^3*d^3)*e^2 + (b^5*d^4*x - 4*a*b^4*d^4)*e)*log(x*e + d))/(x*e^7 + d*e^6)

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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 {5}{2}}}{\left (d + e x\right )^{2}}\, 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)**(5/2)/(e*x+d)**2,x)

[Out]

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

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Giac [A]
time = 1.77, size = 382, normalized size = 1.31 \begin {gather*} 5 \, {\left (b^{5} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} b e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{12} \, {\left (3 \, b^{5} x^{4} e^{6} \mathrm {sgn}\left (b x + a\right ) - 8 \, b^{5} d x^{3} e^{5} \mathrm {sgn}\left (b x + a\right ) + 18 \, b^{5} d^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) - 48 \, b^{5} d^{3} x e^{3} \mathrm {sgn}\left (b x + a\right ) + 20 \, a b^{4} x^{3} e^{6} \mathrm {sgn}\left (b x + a\right ) - 60 \, a b^{4} d x^{2} e^{5} \mathrm {sgn}\left (b x + a\right ) + 180 \, a b^{4} d^{2} x e^{4} \mathrm {sgn}\left (b x + a\right ) + 60 \, a^{2} b^{3} x^{2} e^{6} \mathrm {sgn}\left (b x + a\right ) - 240 \, a^{2} b^{3} d x e^{5} \mathrm {sgn}\left (b x + a\right ) + 120 \, a^{3} b^{2} x e^{6} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-8\right )} + \frac {{\left (b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )}}{x e + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*(b^5*d^4*sgn(b*x + a) - 4*a*b^4*d^3*e*sgn(b*x + a) + 6*a^2*b^3*d^2*e^2*sgn(b*x + a) - 4*a^3*b^2*d*e^3*sgn(b*
x + a) + a^4*b*e^4*sgn(b*x + a))*e^(-6)*log(abs(x*e + d)) + 1/12*(3*b^5*x^4*e^6*sgn(b*x + a) - 8*b^5*d*x^3*e^5
*sgn(b*x + a) + 18*b^5*d^2*x^2*e^4*sgn(b*x + a) - 48*b^5*d^3*x*e^3*sgn(b*x + a) + 20*a*b^4*x^3*e^6*sgn(b*x + a
) - 60*a*b^4*d*x^2*e^5*sgn(b*x + a) + 180*a*b^4*d^2*x*e^4*sgn(b*x + a) + 60*a^2*b^3*x^2*e^6*sgn(b*x + a) - 240
*a^2*b^3*d*x*e^5*sgn(b*x + a) + 120*a^3*b^2*x*e^6*sgn(b*x + a))*e^(-8) + (b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e
*sgn(b*x + a) + 10*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3*b^2*d^2*e^3*sgn(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a)
- a^5*e^5*sgn(b*x + a))*e^(-6)/(x*e + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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