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

Optimal. Leaf size=100 \[ -\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} \sqrt {b d-a e}}-\frac {3 e \sqrt {d+e x}}{4 b^2 (a+b x)}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {27, 47, 63, 208} \begin {gather*} -\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} \sqrt {b d-a e}}-\frac {3 e \sqrt {d+e x}}{4 b^2 (a+b x)}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2} \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]

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

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 {(d+e x)^{3/2}}{(a+b x)^3} \, dx\\ &=-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}+\frac {(3 e) \int \frac {\sqrt {d+e x}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac {3 e \sqrt {d+e x}}{4 b^2 (a+b x)}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}+\frac {\left (3 e^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{8 b^2}\\ &=-\frac {3 e \sqrt {d+e x}}{4 b^2 (a+b x)}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}+\frac {(3 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^2}\\ &=-\frac {3 e \sqrt {d+e x}}{4 b^2 (a+b x)}-\frac {(d+e x)^{3/2}}{2 b (a+b x)^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{5/2} \sqrt {b d-a e}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 90, normalized size = 0.90 \begin {gather*} \frac {3 e^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )}{4 b^{5/2} \sqrt {a e-b d}}-\frac {\sqrt {d+e x} (3 a e+2 b d+5 b e x)}{4 b^2 (a+b x)^2} \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]

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

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IntegrateAlgebraic [A]  time = 0.41, size = 116, normalized size = 1.16 \begin {gather*} -\frac {3 e^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{4 b^{5/2} \sqrt {a e-b d}}-\frac {e^2 \sqrt {d+e x} (3 a e+5 b (d+e x)-3 b d)}{4 b^2 (a e+b (d+e x)-b d)^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]

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

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fricas [B]  time = 0.43, size = 383, normalized size = 3.83 \begin {gather*} \left [\frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (2 \, b^{3} d^{2} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{4} d - a^{3} b^{3} e + {\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \, {\left (a b^{5} d - a^{2} b^{4} e\right )} x\right )}}, \frac {3 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (2 \, b^{3} d^{2} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \, {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{4} d - a^{3} b^{3} e + {\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \, {\left (a b^{5} d - a^{2} b^{4} e\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*(3*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*
b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(2*b^3*d^2 + a*b^2*d*e - 3*a^2*b*e^2 + 5*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x
+ d))/(a^2*b^4*d - a^3*b^3*e + (b^6*d - a*b^5*e)*x^2 + 2*(a*b^5*d - a^2*b^4*e)*x), 1/4*(3*(b^2*e^2*x^2 + 2*a*b
*e^2*x + a^2*e^2)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (2*b^3*d^2 +
a*b^2*d*e - 3*a^2*b*e^2 + 5*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4*d - a^3*b^3*e + (b^6*d - a*b^5*e
)*x^2 + 2*(a*b^5*d - a^2*b^4*e)*x)]

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giac [A]  time = 0.18, size = 112, normalized size = 1.12 \begin {gather*} \frac {3 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, \sqrt {-b^{2} d + a b e} b^{2}} - \frac {5 \, {\left (x e + d\right )}^{\frac {3}{2}} b e^{2} - 3 \, \sqrt {x e + d} b d e^{2} + 3 \, \sqrt {x e + d} a e^{3}}{4 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{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]

3/4*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/(sqrt(-b^2*d + a*b*e)*b^2) - 1/4*(5*(x*e + d)^(3/2)*b*e^2
- 3*sqrt(x*e + d)*b*d*e^2 + 3*sqrt(x*e + d)*a*e^3)/(((x*e + d)*b - b*d + a*e)^2*b^2)

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maple [A]  time = 0.07, size = 121, normalized size = 1.21 \begin {gather*} -\frac {3 \sqrt {e x +d}\, a \,e^{3}}{4 \left (b e x +a e \right )^{2} b^{2}}+\frac {3 \sqrt {e x +d}\, d \,e^{2}}{4 \left (b e x +a e \right )^{2} b}-\frac {5 \left (e x +d \right )^{\frac {3}{2}} e^{2}}{4 \left (b e x +a e \right )^{2} b}+\frac {3 e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{2}} \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]

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

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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 = 0.12, size = 135, normalized size = 1.35 \begin {gather*} \frac {3\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{4\,b^{5/2}\,\sqrt {a\,e-b\,d}}-\frac {\frac {5\,e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,b}+\frac {3\,e^2\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}{4\,b^2}}{b^2\,{\left (d+e\,x\right )}^2-\left (2\,b^2\,d-2\,a\,b\,e\right )\,\left (d+e\,x\right )+a^2\,e^2+b^2\,d^2-2\,a\,b\,d\,e} \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]

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

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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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