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

Optimal. Leaf size=223 \[ -\frac {b^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {2 b e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}-\frac {e (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^2}-\frac {3 b^2 e (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {3 b^2 e (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

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Rubi [A]  time = 0.15, antiderivative size = 223, 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 {b^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {2 b e (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}-\frac {e (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^2}-\frac {3 b^2 e (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {3 b^2 e (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2/((b*d - a*e)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (e*(a + b*x))/(2*(b*d - a*e)^2*(d + e*x)^2*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]) - (2*b*e*(a + b*x))/((b*d - a*e)^3*(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*b^2*e*(a +
b*x)*Log[a + b*x])/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*b^2*e*(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)^3 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {a+b x}{\left (a b+b^2 x\right )^3 (d+e x)^3} \, 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)^2 (d+e x)^3} \, 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^3}{(b d-a e)^3 (a+b x)^2}-\frac {3 b^3 e}{(b d-a e)^4 (a+b x)}+\frac {e^2}{(b d-a e)^2 (d+e x)^3}+\frac {2 b e^2}{(b d-a e)^3 (d+e x)^2}+\frac {3 b^2 e^2}{(b d-a e)^4 (d+e x)}\right ) \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {b^2}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (a+b x)}{2 (b d-a e)^2 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b e (a+b x)}{(b d-a e)^3 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 b^2 e (a+b x) \log (a+b x)}{(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2 e (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 = 0.10, size = 135, normalized size = 0.61 \begin {gather*} \frac {-(b d-a e) \left (-a^2 e^2+a b e (5 d+3 e x)+b^2 \left (2 d^2+9 d e x+6 e^2 x^2\right )\right )-6 b^2 e (a+b x) (d+e x)^2 \log (a+b x)+6 b^2 e (a+b x) (d+e x)^2 \log (d+e x)}{2 \sqrt {(a+b x)^2} (d+e x)^2 (b d-a e)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((b*d - a*e)*(-(a^2*e^2) + a*b*e*(5*d + 3*e*x) + b^2*(2*d^2 + 9*d*e*x + 6*e^2*x^2))) - 6*b^2*e*(a + b*x)*(d
+ e*x)^2*Log[a + b*x] + 6*b^2*e*(a + b*x)*(d + e*x)^2*Log[d + e*x])/(2*(b*d - a*e)^4*Sqrt[(a + b*x)^2]*(d + e*
x)^2)

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IntegrateAlgebraic [B]  time = 53.00, size = 5578, normalized size = 25.01 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [B]  time = 0.45, size = 495, normalized size = 2.22 \begin {gather*} -\frac {2 \, b^{3} d^{3} + 3 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} + a^{3} e^{3} + 6 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 3 \, {\left (3 \, b^{3} d^{2} e - 2 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x + 6 \, {\left (b^{3} e^{3} x^{3} + a b^{2} d^{2} e + {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2}\right )} x\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} e^{3} x^{3} + a b^{2} d^{2} e + {\left (2 \, b^{3} d e^{2} + a b^{2} e^{3}\right )} x^{2} + {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (a b^{4} d^{6} - 4 \, a^{2} b^{3} d^{5} e + 6 \, a^{3} b^{2} d^{4} e^{2} - 4 \, a^{4} b d^{3} e^{3} + a^{5} d^{2} e^{4} + {\left (b^{5} d^{4} e^{2} - 4 \, a b^{4} d^{3} e^{3} + 6 \, a^{2} b^{3} d^{2} e^{4} - 4 \, a^{3} b^{2} d e^{5} + a^{4} b e^{6}\right )} x^{3} + {\left (2 \, b^{5} d^{5} e - 7 \, a b^{4} d^{4} e^{2} + 8 \, a^{2} b^{3} d^{3} e^{3} - 2 \, a^{3} b^{2} d^{2} e^{4} - 2 \, a^{4} b d e^{5} + a^{5} e^{6}\right )} x^{2} + {\left (b^{5} d^{6} - 2 \, a b^{4} d^{5} e - 2 \, a^{2} b^{3} d^{4} e^{2} + 8 \, a^{3} b^{2} d^{3} e^{3} - 7 \, a^{4} b d^{2} e^{4} + 2 \, a^{5} d e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^3*d^3 + 3*a*b^2*d^2*e - 6*a^2*b*d*e^2 + a^3*e^3 + 6*(b^3*d*e^2 - a*b^2*e^3)*x^2 + 3*(3*b^3*d^2*e - 2
*a*b^2*d*e^2 - a^2*b*e^3)*x + 6*(b^3*e^3*x^3 + a*b^2*d^2*e + (2*b^3*d*e^2 + a*b^2*e^3)*x^2 + (b^3*d^2*e + 2*a*
b^2*d*e^2)*x)*log(b*x + a) - 6*(b^3*e^3*x^3 + a*b^2*d^2*e + (2*b^3*d*e^2 + a*b^2*e^3)*x^2 + (b^3*d^2*e + 2*a*b
^2*d*e^2)*x)*log(e*x + d))/(a*b^4*d^6 - 4*a^2*b^3*d^5*e + 6*a^3*b^2*d^4*e^2 - 4*a^4*b*d^3*e^3 + a^5*d^2*e^4 +
(b^5*d^4*e^2 - 4*a*b^4*d^3*e^3 + 6*a^2*b^3*d^2*e^4 - 4*a^3*b^2*d*e^5 + a^4*b*e^6)*x^3 + (2*b^5*d^5*e - 7*a*b^4
*d^4*e^2 + 8*a^2*b^3*d^3*e^3 - 2*a^3*b^2*d^2*e^4 - 2*a^4*b*d*e^5 + a^5*e^6)*x^2 + (b^5*d^6 - 2*a*b^4*d^5*e - 2
*a^2*b^3*d^4*e^2 + 8*a^3*b^2*d^3*e^3 - 7*a^4*b*d^2*e^4 + 2*a^5*d*e^5)*x)

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giac [B]  time = 2.85, size = 461, normalized size = 2.07 \begin {gather*} -\frac {3 \, a b^{2} e \log \left ({\left | b + \frac {a}{x} \right |}\right )}{a b^{4} d^{4} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) - 4 \, a^{2} b^{3} d^{3} e \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) + 6 \, a^{3} b^{2} d^{2} e^{2} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) - 4 \, a^{4} b d e^{3} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) + a^{5} e^{4} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right )} + \frac {3 \, b^{2} d e \log \left ({\left | \frac {d}{x} + e \right |}\right )}{b^{4} d^{5} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) - 4 \, a b^{3} d^{4} e \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) + 6 \, a^{2} b^{2} d^{3} e^{2} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) - 4 \, a^{3} b d^{2} e^{3} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) + a^{4} d e^{4} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right )} + \frac {2 \, b^{4} d^{3} e^{2} + 3 \, a b^{3} d^{2} e^{3} - 6 \, a^{2} b^{2} d e^{4} + a^{3} b e^{5} + \frac {4 \, b^{4} d^{4} e + 2 \, a b^{3} d^{3} e^{2} - 3 \, a^{2} b^{2} d^{2} e^{3} - 4 \, a^{3} b d e^{4} + a^{4} e^{5}}{x} + \frac {2 \, {\left (b^{4} d^{5} - a b^{3} d^{4} e + 3 \, a^{2} b^{2} d^{3} e^{2} - 4 \, a^{3} b d^{2} e^{3} + a^{4} d e^{4}\right )}}{x^{2}}}{2 \, {\left (b d - a e\right )}^{4} a {\left (b + \frac {a}{x}\right )} d^{2} {\left (\frac {d}{x} + e\right )}^{2} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 331, normalized size = 1.48 \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 )+6 a \,b^{2} e^{3} x^{2} \ln \left (b x +a \right )-6 a \,b^{2} e^{3} x^{2} \ln \left (e x +d \right )+12 b^{3} d \,e^{2} x^{2} \ln \left (b x +a \right )-12 b^{3} d \,e^{2} x^{2} \ln \left (e x +d \right )+12 a \,b^{2} d \,e^{2} x \ln \left (b x +a \right )-12 a \,b^{2} d \,e^{2} x \ln \left (e x +d \right )-6 a \,b^{2} e^{3} x^{2}+6 b^{3} d^{2} e x \ln \left (b x +a \right )-6 b^{3} d^{2} e x \ln \left (e x +d \right )+6 b^{3} d \,e^{2} x^{2}-3 a^{2} b \,e^{3} x +6 a \,b^{2} d^{2} e \ln \left (b x +a \right )-6 a \,b^{2} d^{2} e \ln \left (e x +d \right )-6 a \,b^{2} d \,e^{2} x +9 b^{3} d^{2} e x +a^{3} e^{3}-6 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{2 \left (e x +d \right )^{2} \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*ln(b*x+a)*x^3*b^3*e^3-6*ln(e*x+d)*x^3*b^3*e^3+6*ln(b*x+a)*x^2*a*b^2*e^3+12*ln(b*x+a)*x^2*b^3*d*e^2-6*l
n(e*x+d)*x^2*a*b^2*e^3-12*ln(e*x+d)*x^2*b^3*d*e^2+12*a*b^2*d*e^2*x*ln(b*x+a)+6*b^3*d^2*e*x*ln(b*x+a)-12*ln(e*x
+d)*x*a*b^2*d*e^2-6*ln(e*x+d)*x*b^3*d^2*e-6*a*b^2*e^3*x^2+6*b^3*d*e^2*x^2+6*a*b^2*d^2*e*ln(b*x+a)-6*ln(e*x+d)*
a*b^2*d^2*e-3*a^2*b*e^3*x-6*a*b^2*d*e^2*x+9*b^3*d^2*e*x+a^3*e^3-6*a^2*b*d*e^2+3*a*b^2*d^2*e+2*b^3*d^3)*(b*x+a)
^2/(e*x+d)^2/(a*e-b*d)^4/((b*x+a)^2)^(3/2)

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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)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),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 \begin {gather*} \int \frac {a+b\,x}{{\left (d+e\,x\right )}^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a + b*x)/((d + e*x)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), 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 )^{3} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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