3.21.39 \(\int \frac {a+b x}{(d+e x) (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [2039]

Optimal. Leaf size=210 \[ -\frac {e^2}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e}{2 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^3 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {784, 21, 46} \begin {gather*} -\frac {e^3 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {e^3 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {e^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac {e}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {1}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 784

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {a+b x}{(d+e x) \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 {a+b x}{\left (a b+b^2 x\right )^5 (d+e x)} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \frac {1}{(a+b x)^4 (d+e x)} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {b}{(b d-a e) (a+b x)^4}-\frac {b e}{(b d-a e)^2 (a+b x)^3}+\frac {b e^2}{(b d-a e)^3 (a+b x)^2}-\frac {b e^3}{(b d-a e)^4 (a+b x)}+\frac {e^4}{(b d-a e)^4 (d+e x)}\right ) \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {e^2}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e}{2 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^3 (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (a+b x) \log (d+e x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 116, normalized size = 0.55 \begin {gather*} \frac {-\left ((b d-a e) \left (11 a^2 e^2+a b e (-7 d+15 e x)+b^2 \left (2 d^2-3 d e x+6 e^2 x^2\right )\right )\right )-6 e^3 (a+b x)^3 \log (a+b x)+6 e^3 (a+b x)^3 \log (d+e x)}{6 (b d-a e)^4 \left ((a+b x)^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*d - a*e)*(11*a^2*e^2 + a*b*e*(-7*d + 15*e*x) + b^2*(2*d^2 - 3*d*e*x + 6*e^2*x^2))) - 6*e^3*(a + b*x)^3*L
og[a + b*x] + 6*e^3*(a + b*x)^3*Log[d + e*x])/(6*(b*d - a*e)^4*((a + b*x)^2)^(3/2))

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Maple [A]
time = 0.11, size = 251, normalized size = 1.20

method	result	size
default	\(-\frac {\left (6 \ln \left (b x +a \right ) b^{3} e^{3} x^{3}-6 \ln \left (e x +d \right ) b^{3} e^{3} x^{3}+18 \ln \left (b x +a \right ) a \,b^{2} e^{3} x^{2}-18 \ln \left (e x +d \right ) a \,b^{2} e^{3} x^{2}+18 \ln \left (b x +a \right ) a^{2} b \,e^{3} x -18 \ln \left (e x +d \right ) a^{2} b \,e^{3} x -6 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+6 \ln \left (b x +a \right ) a^{3} e^{3}-6 \ln \left (e x +d \right ) a^{3} e^{3}-15 a^{2} b \,e^{3} x +18 a \,b^{2} d \,e^{2} x -3 b^{3} d^{2} e x -11 a^{3} e^{3}+18 a^{2} b d \,e^{2}-9 a \,b^{2} d^{2} e +2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{6 \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(251\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {b^{2} e^{2} x^{2}}{a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}}+\frac {\left (5 a e -b d \right ) b e x}{2 a^{3} e^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}}+\frac {11 a^{2} e^{2}-7 a b d e +2 b^{2} d^{2}}{6 a^{3} e^{3}-18 a^{2} b d \,e^{2}+18 a \,b^{2} d^{2} e -6 b^{3} d^{3}}\right )}{\left (b x +a \right )^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{3} \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{3} \ln \left (b x +a \right )}{\left (b x +a \right ) \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\)	\(341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*ln(b*x+a)*b^3*e^3*x^3-6*ln(e*x+d)*b^3*e^3*x^3+18*ln(b*x+a)*a*b^2*e^3*x^2-18*ln(e*x+d)*a*b^2*e^3*x^2+18
*ln(b*x+a)*a^2*b*e^3*x-18*ln(e*x+d)*a^2*b*e^3*x-6*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+6*ln(b*x+a)*a^3*e^3-6*ln(e*x+d
)*a^3*e^3-15*a^2*b*e^3*x+18*a*b^2*d*e^2*x-3*b^3*d^2*e*x-11*a^3*e^3+18*a^2*b*d*e^2-9*a*b^2*d^2*e+2*b^3*d^3)*(b*
x+a)^2/(a*e-b*d)^4/((b*x+a)^2)^(5/2)

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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*x+a)/(e*x+d)/(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((2*%e^-1*a*b-2*%e^-2*b^2*d)^2>
0)', see `as

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (155) = 310\).
time = 2.46, size = 394, normalized size = 1.88 \begin {gather*} -\frac {2 \, b^{3} d^{3} + 6 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} e^{3} \log \left (b x + a\right ) - 6 \, {\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} e^{3} \log \left (x e + d\right ) - {\left (6 \, a b^{2} x^{2} + 15 \, a^{2} b x + 11 \, a^{3}\right )} e^{3} + 6 \, {\left (b^{3} d x^{2} + 3 \, a b^{2} d x + 3 \, a^{2} b d\right )} e^{2} - 3 \, {\left (b^{3} d^{2} x + 3 \, a b^{2} d^{2}\right )} e}{6 \, {\left (b^{7} d^{4} x^{3} + 3 \, a b^{6} d^{4} x^{2} + 3 \, a^{2} b^{5} d^{4} x + a^{3} b^{4} d^{4} + {\left (a^{4} b^{3} x^{3} + 3 \, a^{5} b^{2} x^{2} + 3 \, a^{6} b x + a^{7}\right )} e^{4} - 4 \, {\left (a^{3} b^{4} d x^{3} + 3 \, a^{4} b^{3} d x^{2} + 3 \, a^{5} b^{2} d x + a^{6} b d\right )} e^{3} + 6 \, {\left (a^{2} b^{5} d^{2} x^{3} + 3 \, a^{3} b^{4} d^{2} x^{2} + 3 \, a^{4} b^{3} d^{2} x + a^{5} b^{2} d^{2}\right )} e^{2} - 4 \, {\left (a b^{6} d^{3} x^{3} + 3 \, a^{2} b^{5} d^{3} x^{2} + 3 \, a^{3} b^{4} d^{3} x + a^{4} b^{3} d^{3}\right )} e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^3*d^3 + 6*(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*e^3*log(b*x + a) - 6*(b^3*x^3 + 3*a*b^2*x^2 + 3*
a^2*b*x + a^3)*e^3*log(x*e + d) - (6*a*b^2*x^2 + 15*a^2*b*x + 11*a^3)*e^3 + 6*(b^3*d*x^2 + 3*a*b^2*d*x + 3*a^2
*b*d)*e^2 - 3*(b^3*d^2*x + 3*a*b^2*d^2)*e)/(b^7*d^4*x^3 + 3*a*b^6*d^4*x^2 + 3*a^2*b^5*d^4*x + a^3*b^4*d^4 + (a
^4*b^3*x^3 + 3*a^5*b^2*x^2 + 3*a^6*b*x + a^7)*e^4 - 4*(a^3*b^4*d*x^3 + 3*a^4*b^3*d*x^2 + 3*a^5*b^2*d*x + a^6*b
*d)*e^3 + 6*(a^2*b^5*d^2*x^3 + 3*a^3*b^4*d^2*x^2 + 3*a^4*b^3*d^2*x + a^5*b^2*d^2)*e^2 - 4*(a*b^6*d^3*x^3 + 3*a
^2*b^5*d^3*x^2 + 3*a^3*b^4*d^3*x + a^4*b^3*d^3)*e)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 1.89, size = 302, normalized size = 1.44 \begin {gather*} -\frac {b e^{3} \log \left ({\left | b x + a \right |}\right )}{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 )} + \frac {e^{4} \log \left ({\left | x e + d \right |}\right )}{b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{4} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{5} \mathrm {sgn}\left (b x + a\right )} - \frac {2 \, b^{3} d^{3} - 9 \, a b^{2} d^{2} e + 18 \, a^{2} b d e^{2} - 11 \, a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + 5 \, a^{2} b e^{3}\right )} x}{6 \, {\left (b d - a e\right )}^{4} {\left (b x + a\right )}^{3} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*e^3*log(abs(b*x + a))/(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^4*log(abs(x*e + d))/(b^4*d^4*e*sgn(b*x + a) - 4*a*
b^3*d^3*e^2*sgn(b*x + a) + 6*a^2*b^2*d^2*e^3*sgn(b*x + a) - 4*a^3*b*d*e^4*sgn(b*x + a) + a^4*e^5*sgn(b*x + a))
- 1/6*(2*b^3*d^3 - 9*a*b^2*d^2*e + 18*a^2*b*d*e^2 - 11*a^3*e^3 + 6*(b^3*d*e^2 - a*b^2*e^3)*x^2 - 3*(b^3*d^2*e
- 6*a*b^2*d*e^2 + 5*a^2*b*e^3)*x)/((b*d - a*e)^4*(b*x + a)^3*sgn(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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