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

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

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Rubi [A]  time = 0.14, 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 = {770, 21, 44} \begin {gather*} -\frac {e^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\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 (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 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 770

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 = 0.08, size = 116, normalized size = 0.55 \begin {gather*} \frac {-(b d-a e) \left (11 a^2 e^2+a b e (15 e x-7 d)+b^2 \left (2 d^2-3 d e x+6 e^2 x^2\right )\right )+6 e^3 (a+b x)^3 \log (d+e x)-6 e^3 (a+b x)^3 \log (a+b x)}{6 \left ((a+b x)^2\right )^{3/2} (b d-a e)^4} \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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IntegrateAlgebraic [B]  time = 20.15, size = 7693, normalized size = 36.63 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [B]  time = 0.45, size = 425, normalized size = 2.02 \begin {gather*} -\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^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (a^{3} b^{4} d^{4} - 4 \, a^{4} b^{3} d^{3} e + 6 \, a^{5} b^{2} d^{2} e^{2} - 4 \, a^{6} b d e^{3} + a^{7} e^{4} + {\left (b^{7} d^{4} - 4 \, a b^{6} d^{3} e + 6 \, a^{2} b^{5} d^{2} e^{2} - 4 \, a^{3} b^{4} d e^{3} + a^{4} b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{6} d^{4} - 4 \, a^{2} b^{5} d^{3} e + 6 \, a^{3} b^{4} d^{2} e^{2} - 4 \, a^{4} b^{3} d e^{3} + a^{5} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{5} d^{4} - 4 \, a^{3} b^{4} d^{3} e + 6 \, a^{4} b^{3} d^{2} e^{2} - 4 \, a^{5} b^{2} d e^{3} + a^{6} b e^{4}\right )} x\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 - 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 + 6*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*log(b*x + a) - 6
*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*log(e*x + d))/(a^3*b^4*d^4 - 4*a^4*b^3*d^3*e + 6*a^
5*b^2*d^2*e^2 - 4*a^6*b*d*e^3 + a^7*e^4 + (b^7*d^4 - 4*a*b^6*d^3*e + 6*a^2*b^5*d^2*e^2 - 4*a^3*b^4*d*e^3 + a^4
*b^3*e^4)*x^3 + 3*(a*b^6*d^4 - 4*a^2*b^5*d^3*e + 6*a^3*b^4*d^2*e^2 - 4*a^4*b^3*d*e^3 + a^5*b^2*e^4)*x^2 + 3*(a
^2*b^5*d^4 - 4*a^3*b^4*d^3*e + 6*a^4*b^3*d^2*e^2 - 4*a^5*b^2*d*e^3 + a^6*b*e^4)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}}\,{d x} \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]

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

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maple [A]  time = 0.06, size = 251, normalized size = 1.20 \begin {gather*} -\frac {\left (6 b^{3} e^{3} x^{3} \ln \left (b x +a \right )-6 b^{3} e^{3} x^{3} \ln \left (e x +d \right )+18 a \,b^{2} e^{3} x^{2} \ln \left (b x +a \right )-18 a \,b^{2} e^{3} x^{2} \ln \left (e x +d \right )+18 a^{2} b \,e^{3} x \ln \left (b x +a \right )-18 a^{2} b \,e^{3} x \ln \left (e x +d \right )-6 a \,b^{2} e^{3} x^{2}+6 b^{3} d \,e^{2} x^{2}+6 a^{3} e^{3} \ln \left (b x +a \right )-6 a^{3} e^{3} \ln \left (e x +d \right )-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}}} \end {gather*}

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)

[Out]

-1/6*(6*b^3*e^3*x^3*ln(b*x+a)-6*b^3*e^3*x^3*ln(e*x+d)+18*a*b^2*e^3*x^2*ln(b*x+a)-18*a*b^2*e^3*x^2*ln(e*x+d)+18
*a^2*b*e^3*x*ln(b*x+a)-18*ln(e*x+d)*x*a^2*b*e^3-6*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+6*a^3*e^3*ln(b*x+a)-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*a*b)/e>0)', see `assume?`
for more details)Is ((2*a*b)/e    -(2*b^2*d)/e^2)    ^2    -(4*b^2       *((-(2*a*b*d)/e)        +(b^2*d^2)/e^
2+a^2))     /e^2 zero or nonzero?

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

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

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