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

Optimal. Leaf size=140 \[ \frac {15 e^2}{4 \sqrt {d+e x} (b d-a e)^3}-\frac {15 \sqrt {b} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{7/2}}+\frac {5 e}{4 (a+b x) \sqrt {d+e x} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)} \]

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Rubi [A]  time = 0.06, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \begin {gather*} \frac {15 e^2}{4 \sqrt {d+e x} (b d-a e)^3}-\frac {15 \sqrt {b} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{7/2}}+\frac {5 e}{4 (a+b x) \sqrt {d+e x} (b d-a e)^2}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(15*e^2)/(4*(b*d - a*e)^3*Sqrt[d + e*x]) - 1/(2*(b*d - a*e)*(a + b*x)^2*Sqrt[d + e*x]) + (5*e)/(4*(b*d - a*e)^
2*(a + b*x)*Sqrt[d + e*x]) - (15*Sqrt[b]*e^2*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^
(7/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {a+b x}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a+b x)^3 (d+e x)^{3/2}} \, dx\\ &=-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}-\frac {(5 e) \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}} \, dx}{4 (b d-a e)}\\ &=-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {\left (15 e^2\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^2}\\ &=\frac {15 e^2}{4 (b d-a e)^3 \sqrt {d+e x}}-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {\left (15 b e^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 (b d-a e)^3}\\ &=\frac {15 e^2}{4 (b d-a e)^3 \sqrt {d+e x}}-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}+\frac {(15 b e) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 (b d-a e)^3}\\ &=\frac {15 e^2}{4 (b d-a e)^3 \sqrt {d+e x}}-\frac {1}{2 (b d-a e) (a+b x)^2 \sqrt {d+e x}}+\frac {5 e}{4 (b d-a e)^2 (a+b x) \sqrt {d+e x}}-\frac {15 \sqrt {b} e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.36 \begin {gather*} -\frac {2 e^2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{\sqrt {d+e x} (a e-b d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^2*Hypergeometric2F1[-1/2, 3, 1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e)^3*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.54, size = 163, normalized size = 1.16 \begin {gather*} \frac {e^2 \left (8 a^2 e^2+25 a b e (d+e x)-16 a b d e+8 b^2 d^2+15 b^2 (d+e x)^2-25 b^2 d (d+e x)\right )}{4 \sqrt {d+e x} (b d-a e)^3 (-a e-b (d+e x)+b d)^2}+\frac {15 \sqrt {b} e^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 (a e-b d)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(8*b^2*d^2 - 16*a*b*d*e + 8*a^2*e^2 - 25*b^2*d*(d + e*x) + 25*a*b*e*(d + e*x) + 15*b^2*(d + e*x)^2))/(4*(
b*d - a*e)^3*Sqrt[d + e*x]*(b*d - a*e - b*(d + e*x))^2) + (15*Sqrt[b]*e^2*ArcTan[(Sqrt[b]*Sqrt[-(b*d) + a*e]*S
qrt[d + e*x])/(b*d - a*e)])/(4*(-(b*d) + a*e)^(7/2))

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fricas [B]  time = 0.45, size = 782, normalized size = 5.59 \begin {gather*} \left [-\frac {15 \, {\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} + {\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} + {\left (2 \, a b d e^{2} + a^{2} e^{3}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \, {\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} + {\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} + {\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} + {\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}, -\frac {15 \, {\left (b^{2} e^{3} x^{3} + a^{2} d e^{2} + {\left (b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{2} + {\left (2 \, a b d e^{2} + a^{2} e^{3}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (15 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} + 9 \, a b d e + 8 \, a^{2} e^{2} + 5 \, {\left (b^{2} d e + 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{3} d^{4} - 3 \, a^{3} b^{2} d^{3} e + 3 \, a^{4} b d^{2} e^{2} - a^{5} d e^{3} + {\left (b^{5} d^{3} e - 3 \, a b^{4} d^{2} e^{2} + 3 \, a^{2} b^{3} d e^{3} - a^{3} b^{2} e^{4}\right )} x^{3} + {\left (b^{5} d^{4} - a b^{4} d^{3} e - 3 \, a^{2} b^{3} d^{2} e^{2} + 5 \, a^{3} b^{2} d e^{3} - 2 \, a^{4} b e^{4}\right )} x^{2} + {\left (2 \, a b^{4} d^{4} - 5 \, a^{2} b^{3} d^{3} e + 3 \, a^{3} b^{2} d^{2} e^{2} + a^{4} b d e^{3} - a^{5} e^{4}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(15*(b^2*e^3*x^3 + a^2*d*e^2 + (b^2*d*e^2 + 2*a*b*e^3)*x^2 + (2*a*b*d*e^2 + a^2*e^3)*x)*sqrt(b/(b*d - a*
e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(15*b^2*e^2*x^2
- 2*b^2*d^2 + 9*a*b*d*e + 8*a^2*e^2 + 5*(b^2*d*e + 5*a*b*e^2)*x)*sqrt(e*x + d))/(a^2*b^3*d^4 - 3*a^3*b^2*d^3*
e + 3*a^4*b*d^2*e^2 - a^5*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^3*b^2*e^4)*x^3 + (b^5*d^4
- a*b^4*d^3*e - 3*a^2*b^3*d^2*e^2 + 5*a^3*b^2*d*e^3 - 2*a^4*b*e^4)*x^2 + (2*a*b^4*d^4 - 5*a^2*b^3*d^3*e + 3*a
^3*b^2*d^2*e^2 + a^4*b*d*e^3 - a^5*e^4)*x), -1/4*(15*(b^2*e^3*x^3 + a^2*d*e^2 + (b^2*d*e^2 + 2*a*b*e^3)*x^2 +
(2*a*b*d*e^2 + a^2*e^3)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x
+ b*d)) - (15*b^2*e^2*x^2 - 2*b^2*d^2 + 9*a*b*d*e + 8*a^2*e^2 + 5*(b^2*d*e + 5*a*b*e^2)*x)*sqrt(e*x + d))/(a^2
*b^3*d^4 - 3*a^3*b^2*d^3*e + 3*a^4*b*d^2*e^2 - a^5*d*e^3 + (b^5*d^3*e - 3*a*b^4*d^2*e^2 + 3*a^2*b^3*d*e^3 - a^
3*b^2*e^4)*x^3 + (b^5*d^4 - a*b^4*d^3*e - 3*a^2*b^3*d^2*e^2 + 5*a^3*b^2*d*e^3 - 2*a^4*b*e^4)*x^2 + (2*a*b^4*d^
4 - 5*a^2*b^3*d^3*e + 3*a^3*b^2*d^2*e^2 + a^4*b*d*e^3 - a^5*e^4)*x)]

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giac [B]  time = 0.18, size = 235, normalized size = 1.68 \begin {gather*} \frac {15 \, b \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, e^{2}}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {x e + d}} + \frac {7 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} e^{2} - 9 \, \sqrt {x e + d} b^{2} d e^{2} + 9 \, \sqrt {x e + d} a b e^{3}}{4 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*s
qrt(-b^2*d + a*b*e)) + 2*e^2/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(x*e + d)) + 1/4*(7*(x*e
+ d)^(3/2)*b^2*e^2 - 9*sqrt(x*e + d)*b^2*d*e^2 + 9*sqrt(x*e + d)*a*b*e^3)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b
*d*e^2 - a^3*e^3)*((x*e + d)*b - b*d + a*e)^2)

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maple [A]  time = 0.11, size = 179, normalized size = 1.28 \begin {gather*} -\frac {9 \sqrt {e x +d}\, a b \,e^{3}}{4 \left (a e -b d \right )^{3} \left (b e x +a e \right )^{2}}+\frac {9 \sqrt {e x +d}\, b^{2} d \,e^{2}}{4 \left (a e -b d \right )^{3} \left (b e x +a e \right )^{2}}-\frac {7 \left (e x +d \right )^{\frac {3}{2}} b^{2} e^{2}}{4 \left (a e -b d \right )^{3} \left (b e x +a e \right )^{2}}-\frac {15 b \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}-\frac {2 e^{2}}{\left (a e -b d \right )^{3} \sqrt {e x +d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/4*e^2/(a*e-b*d)^3*b^2/(b*e*x+a*e)^2*(e*x+d)^(3/2)-9/4*e^3/(a*e-b*d)^3*b/(b*e*x+a*e)^2*(e*x+d)^(1/2)*a+9/4*e
^2/(a*e-b*d)^3*b^2/(b*e*x+a*e)^2*(e*x+d)^(1/2)*d-15/4*e^2/(a*e-b*d)^3*b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/
2)/((a*e-b*d)*b)^(1/2)*b)-2*e^2/(a*e-b*d)^3/(e*x+d)^(1/2)

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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)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^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 positive or negative?

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mupad [B]  time = 2.32, size = 205, normalized size = 1.46 \begin {gather*} -\frac {\frac {2\,e^2}{a\,e-b\,d}+\frac {15\,b^2\,e^2\,{\left (d+e\,x\right )}^2}{4\,{\left (a\,e-b\,d\right )}^3}+\frac {25\,b\,e^2\,\left (d+e\,x\right )}{4\,{\left (a\,e-b\,d\right )}^2}}{b^2\,{\left (d+e\,x\right )}^{5/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{3/2}+\sqrt {d+e\,x}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}-\frac {15\,\sqrt {b}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}}\right )}{4\,{\left (a\,e-b\,d\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*e^2)/(a*e - b*d) + (15*b^2*e^2*(d + e*x)^2)/(4*(a*e - b*d)^3) + (25*b*e^2*(d + e*x))/(4*(a*e - b*d)^2))/
(b^2*(d + e*x)^(5/2) - (2*b^2*d - 2*a*b*e)*(d + e*x)^(3/2) + (d + e*x)^(1/2)*(a^2*e^2 + b^2*d^2 - 2*a*b*d*e))
- (15*b^(1/2)*e^2*atan((b^(1/2)*(d + e*x)^(1/2)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/(a*e - b*
d)^(7/2)))/(4*(a*e - b*d)^(7/2))

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sympy [F(-1)]  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)**(3/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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