3.16.79 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{5/2}}{(d+e x)^5} \, dx\) [1579]

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

[Out]

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

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Rubi [A]
time = 0.10, 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 {5 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^6 (a+b x) (d+e x)^2}-\frac {5 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}{3 e^6 (a+b x) (d+e x)^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}{4 e^6 (a+b x) (d+e x)^4}+\frac {b^5 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac {5 b^4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^6 (a+b x)}-\frac {10 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^6 (a+b x) (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^5*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])/(4*e^6*(
a + b*x)*(d + e*x)^4) - (5*b*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^6*(a + b*x)*(d + e*x)^3) + (5*b
^2*(b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^6*(a + b*x)*(d + e*x)^2) - (10*b^3*(b*d - a*e)^2*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(e^6*(a + b*x)*(d + e*x)) - (5*b^4*(b*d - a*e)*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)^5} \, 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)^5} \, 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 {b^{10}}{e^5}-\frac {b^5 (b d-a e)^5}{e^5 (d+e x)^5}+\frac {5 b^6 (b d-a e)^4}{e^5 (d+e x)^4}-\frac {10 b^7 (b d-a e)^3}{e^5 (d+e x)^3}+\frac {10 b^8 (b d-a e)^2}{e^5 (d+e x)^2}-\frac {5 b^9 (b d-a e)}{e^5 (d+e x)}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac {b^5 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}}{4 e^6 (a+b x) (d+e x)^4}-\frac {5 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^3}+\frac {5 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)^2}-\frac {10 b^3 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)}-\frac {5 b^4 (b d-a e) \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 = 0.08, size = 243, normalized size = 0.83 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (3 a^5 e^5+5 a^4 b e^4 (d+4 e x)+10 a^3 b^2 e^3 \left (d^2+4 d e x+6 e^2 x^2\right )+30 a^2 b^3 e^2 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )-5 a b^4 d e \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )+b^5 \left (77 d^5+248 d^4 e x+252 d^3 e^2 x^2+48 d^2 e^3 x^3-48 d e^4 x^4-12 e^5 x^5\right )+60 b^4 (b d-a e) (d+e x)^4 \log (d+e x)\right )}{12 e^6 (a+b x) (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(Sqrt[(a + b*x)^2]*(3*a^5*e^5 + 5*a^4*b*e^4*(d + 4*e*x) + 10*a^3*b^2*e^3*(d^2 + 4*d*e*x + 6*e^2*x^2) + 3
0*a^2*b^3*e^2*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) - 5*a*b^4*d*e*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 +
48*e^3*x^3) + b^5*(77*d^5 + 248*d^4*e*x + 252*d^3*e^2*x^2 + 48*d^2*e^3*x^3 - 48*d*e^4*x^4 - 12*e^5*x^5) + 60*
b^4*(b*d - a*e)*(d + e*x)^4*Log[d + e*x]))/(e^6*(a + b*x)*(d + e*x)^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(457\) vs. \(2(222)=444\).
time = 0.64, size = 458, normalized size = 1.57

method	result	size
risch	\(\frac {b^{5} x \sqrt {\left (b x +a \right )^{2}}}{e^{5} \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\left (-10 e^{4} a^{2} b^{3}+20 a \,b^{4} d \,e^{3}-10 b^{5} d^{2} e^{2}\right ) x^{3}-5 b^{2} e \left (e^{3} a^{3}+3 a^{2} b d \,e^{2}-9 a \,b^{2} d^{2} e +5 b^{3} d^{3}\right ) x^{2}-\frac {5 b \left (e^{4} a^{4}+2 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-22 a \,b^{3} d^{3} e +13 b^{4} d^{4}\right ) x}{3}-\frac {3 a^{5} e^{5}+5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}+30 a^{2} b^{3} d^{3} e^{2}-125 a \,b^{4} d^{4} e +77 b^{5} d^{5}}{12 e}\right )}{\left (b x +a \right ) e^{5} \left (e x +d \right )^{4}}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, b^{4} \left (a e -b d \right ) \ln \left (e x +d \right )}{\left (b x +a \right ) e^{6}}\)	\(298\)
default	\(\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (240 \ln \left (e x +d \right ) a \,b^{4} d^{3} e^{2} x +240 a \,b^{4} d \,e^{4} x^{3}+12 b^{5} e^{5} x^{5}+440 a \,b^{4} d^{3} e^{2} x +240 \ln \left (e x +d \right ) a \,b^{4} d \,e^{4} x^{3}-360 \ln \left (e x +d \right ) b^{5} d^{3} e^{2} x^{2}-77 b^{5} d^{5}+60 \ln \left (e x +d \right ) a \,b^{4} d^{4} e -60 \ln \left (e x +d \right ) b^{5} d^{5}-180 a^{2} b^{3} d \,e^{4} x^{2}+540 a \,b^{4} d^{2} e^{3} x^{2}-40 a^{3} b^{2} d \,e^{4} x -120 a^{2} b^{3} d^{2} e^{3} x -3 a^{5} e^{5}-240 \ln \left (e x +d \right ) b^{5} d^{4} e x +360 \ln \left (e x +d \right ) a \,b^{4} d^{2} e^{3} x^{2}+60 \ln \left (e x +d \right ) a \,b^{4} e^{5} x^{4}-60 \ln \left (e x +d \right ) b^{5} d \,e^{4} x^{4}+48 b^{5} d \,e^{4} x^{4}-120 a^{2} b^{3} e^{5} x^{3}-48 b^{5} d^{2} e^{3} x^{3}-60 a^{3} b^{2} e^{5} x^{2}-252 b^{5} d^{3} e^{2} x^{2}-20 a^{4} b \,e^{5} x -248 b^{5} d^{4} e x -240 \ln \left (e x +d \right ) b^{5} d^{2} e^{3} x^{3}-5 a^{4} b d \,e^{4}-10 a^{3} b^{2} d^{2} e^{3}-30 a^{2} b^{3} d^{3} e^{2}+125 a \,b^{4} d^{4} e \right )}{12 \left (b x +a \right )^{5} e^{6} \left (e x +d \right )^{4}}\)	\(458\)

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)^5,x,method=_RETURNVERBOSE)

[Out]

1/12*((b*x+a)^2)^(5/2)*(240*ln(e*x+d)*a*b^4*d^3*e^2*x+240*a*b^4*d*e^4*x^3+12*b^5*e^5*x^5+440*a*b^4*d^3*e^2*x+2
40*ln(e*x+d)*a*b^4*d*e^4*x^3-360*ln(e*x+d)*b^5*d^3*e^2*x^2-77*b^5*d^5+60*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+540*a*b^4*d^2*e^3*x^2-40*a^3*b^2*d*e^4*x-120*a^2*b^3*d^2*e^3*x-3*a^5*e^5-240*ln(e*
x+d)*b^5*d^4*e*x+360*ln(e*x+d)*a*b^4*d^2*e^3*x^2+60*ln(e*x+d)*a*b^4*e^5*x^4-60*ln(e*x+d)*b^5*d*e^4*x^4+48*b^5*
d*e^4*x^4-120*a^2*b^3*e^5*x^3-48*b^5*d^2*e^3*x^3-60*a^3*b^2*e^5*x^2-252*b^5*d^3*e^2*x^2-20*a^4*b*e^5*x-248*b^5
*d^4*e*x-240*ln(e*x+d)*b^5*d^2*e^3*x^3-5*a^4*b*d*e^4-10*a^3*b^2*d^2*e^3-30*a^2*b^3*d^3*e^2+125*a*b^4*d^4*e)/(b
*x+a)^5/e^6/(e*x+d)^4

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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)^5,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 = 3.09, size = 388, normalized size = 1.33 \begin {gather*} -\frac {77 \, b^{5} d^{5} - {\left (12 \, b^{5} x^{5} - 120 \, a^{2} b^{3} x^{3} - 60 \, a^{3} b^{2} x^{2} - 20 \, a^{4} b x - 3 \, a^{5}\right )} e^{5} - {\left (48 \, b^{5} d x^{4} + 240 \, a b^{4} d x^{3} - 180 \, a^{2} b^{3} d x^{2} - 40 \, a^{3} b^{2} d x - 5 \, a^{4} b d\right )} e^{4} + 2 \, {\left (24 \, b^{5} d^{2} x^{3} - 270 \, a b^{4} d^{2} x^{2} + 60 \, a^{2} b^{3} d^{2} x + 5 \, a^{3} b^{2} d^{2}\right )} e^{3} + 2 \, {\left (126 \, b^{5} d^{3} x^{2} - 220 \, a b^{4} d^{3} x + 15 \, a^{2} b^{3} d^{3}\right )} e^{2} + {\left (248 \, b^{5} d^{4} x - 125 \, a b^{4} d^{4}\right )} e + 60 \, {\left (b^{5} d^{5} - a b^{4} x^{4} e^{5} + {\left (b^{5} d x^{4} - 4 \, a b^{4} d x^{3}\right )} e^{4} + 2 \, {\left (2 \, b^{5} d^{2} x^{3} - 3 \, a b^{4} d^{2} x^{2}\right )} e^{3} + 2 \, {\left (3 \, b^{5} d^{3} x^{2} - 2 \, a b^{4} d^{3} x\right )} e^{2} + {\left (4 \, b^{5} d^{4} x - a b^{4} d^{4}\right )} e\right )} \log \left (x e + d\right )}{12 \, {\left (x^{4} e^{10} + 4 \, d x^{3} e^{9} + 6 \, d^{2} x^{2} e^{8} + 4 \, d^{3} x e^{7} + d^{4} 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)^5,x, algorithm="fricas")

[Out]

-1/12*(77*b^5*d^5 - (12*b^5*x^5 - 120*a^2*b^3*x^3 - 60*a^3*b^2*x^2 - 20*a^4*b*x - 3*a^5)*e^5 - (48*b^5*d*x^4 +
240*a*b^4*d*x^3 - 180*a^2*b^3*d*x^2 - 40*a^3*b^2*d*x - 5*a^4*b*d)*e^4 + 2*(24*b^5*d^2*x^3 - 270*a*b^4*d^2*x^2
+ 60*a^2*b^3*d^2*x + 5*a^3*b^2*d^2)*e^3 + 2*(126*b^5*d^3*x^2 - 220*a*b^4*d^3*x + 15*a^2*b^3*d^3)*e^2 + (248*b
^5*d^4*x - 125*a*b^4*d^4)*e + 60*(b^5*d^5 - a*b^4*x^4*e^5 + (b^5*d*x^4 - 4*a*b^4*d*x^3)*e^4 + 2*(2*b^5*d^2*x^3
- 3*a*b^4*d^2*x^2)*e^3 + 2*(3*b^5*d^3*x^2 - 2*a*b^4*d^3*x)*e^2 + (4*b^5*d^4*x - a*b^4*d^4)*e)*log(x*e + d))/(
x^4*e^10 + 4*d*x^3*e^9 + 6*d^2*x^2*e^8 + 4*d^3*x*e^7 + d^4*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 )^{5}}\, 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)**5,x)

[Out]

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

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Giac [A]
time = 1.64, size = 370, normalized size = 1.27 \begin {gather*} b^{5} x e^{\left (-5\right )} \mathrm {sgn}\left (b x + a\right ) - 5 \, {\left (b^{5} d \mathrm {sgn}\left (b x + a\right ) - a b^{4} e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (77 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 125 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 30 \, 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 ) + 3 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right ) + 120 \, {\left (b^{5} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{4} d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{2} b^{3} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} + 60 \, {\left (5 \, b^{5} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 9 \, a b^{4} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{3} d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{3} b^{2} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 20 \, {\left (13 \, b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 22 \, a b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{4} b e^{5} \mathrm {sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-6\right )}}{12 \, {\left (x e + d\right )}^{4}} \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)^5,x, algorithm="giac")

[Out]

b^5*x*e^(-5)*sgn(b*x + a) - 5*(b^5*d*sgn(b*x + a) - a*b^4*e*sgn(b*x + a))*e^(-6)*log(abs(x*e + d)) - 1/12*(77*
b^5*d^5*sgn(b*x + a) - 125*a*b^4*d^4*e*sgn(b*x + a) + 30*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) + 3*a^5*e^5*sgn(b*x + a) + 120*(b^5*d^2*e^3*sgn(b*x + a) - 2*a*b^4*d*e^
4*sgn(b*x + a) + a^2*b^3*e^5*sgn(b*x + a))*x^3 + 60*(5*b^5*d^3*e^2*sgn(b*x + a) - 9*a*b^4*d^2*e^3*sgn(b*x + a)
+ 3*a^2*b^3*d*e^4*sgn(b*x + a) + a^3*b^2*e^5*sgn(b*x + a))*x^2 + 20*(13*b^5*d^4*e*sgn(b*x + a) - 22*a*b^4*d^3
*e^2*sgn(b*x + a) + 6*a^2*b^3*d^2*e^3*sgn(b*x + a) + 2*a^3*b^2*d*e^4*sgn(b*x + a) + a^4*b*e^5*sgn(b*x + a))*x)
*e^(-6)/(x*e + d)^4

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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 )}^5} \,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)^5,x)

[Out]

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

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