3.2087 \(\int \frac{(a+b x) (d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=172 \[ \frac{3 e^3 \sqrt{d+e x}}{64 b^2 (a+b x) (b d-a e)^2}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (a+b x)^2 (b d-a e)}-\frac{3 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{5/2} (b d-a e)^{5/2}}-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4} \]

[Out]

-(e*Sqrt[d + e*x])/(8*b^2*(a + b*x)^3) - (e^2*Sqrt[d + e*x])/(32*b^2*(b*d - a*e)*(a + b*x)^2) + (3*e^3*Sqrt[d
+ e*x])/(64*b^2*(b*d - a*e)^2*(a + b*x)) - (d + e*x)^(3/2)/(4*b*(a + b*x)^4) - (3*e^4*ArcTanh[(Sqrt[b]*Sqrt[d
+ e*x])/Sqrt[b*d - a*e]])/(64*b^(5/2)*(b*d - a*e)^(5/2))

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Rubi [A]  time = 0.0908595, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {27, 47, 51, 63, 208} \[ \frac{3 e^3 \sqrt{d+e x}}{64 b^2 (a+b x) (b d-a e)^2}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (a+b x)^2 (b d-a e)}-\frac{3 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{5/2} (b d-a e)^{5/2}}-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e*Sqrt[d + e*x])/(8*b^2*(a + b*x)^3) - (e^2*Sqrt[d + e*x])/(32*b^2*(b*d - a*e)*(a + b*x)^2) + (3*e^3*Sqrt[d
+ e*x])/(64*b^2*(b*d - a*e)^2*(a + b*x)) - (d + e*x)^(3/2)/(4*b*(a + b*x)^4) - (3*e^4*ArcTanh[(Sqrt[b]*Sqrt[d
+ e*x])/Sqrt[b*d - a*e]])/(64*b^(5/2)*(b*d - a*e)^(5/2))

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 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 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 )^3} \, dx &=\int \frac{(d+e x)^{3/2}}{(a+b x)^5} \, dx\\ &=-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac{(3 e) \int \frac{\sqrt{d+e x}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^3}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac{e^2 \int \frac{1}{(a+b x)^3 \sqrt{d+e x}} \, dx}{16 b^2}\\ &=-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^3}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x)^2}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4}-\frac{\left (3 e^3\right ) \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{64 b^2 (b d-a e)}\\ &=-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^3}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x)^2}+\frac{3 e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 (a+b x)}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac{\left (3 e^4\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{128 b^2 (b d-a e)^2}\\ &=-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^3}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x)^2}+\frac{3 e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 (a+b x)}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4}+\frac{\left (3 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{64 b^2 (b d-a e)^2}\\ &=-\frac{e \sqrt{d+e x}}{8 b^2 (a+b x)^3}-\frac{e^2 \sqrt{d+e x}}{32 b^2 (b d-a e) (a+b x)^2}+\frac{3 e^3 \sqrt{d+e x}}{64 b^2 (b d-a e)^2 (a+b x)}-\frac{(d+e x)^{3/2}}{4 b (a+b x)^4}-\frac{3 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{5/2} (b d-a e)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0163519, size = 52, normalized size = 0.3 \[ \frac{2 e^4 (d+e x)^{5/2} \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};-\frac{b (d+e x)}{a e-b d}\right )}{5 (a e-b d)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^4*(d + e*x)^(5/2)*Hypergeometric2F1[5/2, 5, 7/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^5)

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Maple [A]  time = 0.013, size = 222, normalized size = 1.3 \begin{align*}{\frac{3\,{e}^{4}b}{64\, \left ( bex+ae \right ) ^{4} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) } \left ( ex+d \right ) ^{{\frac{7}{2}}}}+{\frac{11\,{e}^{4}}{64\, \left ( bex+ae \right ) ^{4} \left ( ae-bd \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{11\,{e}^{4}}{64\, \left ( bex+ae \right ) ^{4}b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{3\,{e}^{5}a}{64\, \left ( bex+ae \right ) ^{4}{b}^{2}}\sqrt{ex+d}}+{\frac{3\,{e}^{4}d}{64\, \left ( bex+ae \right ) ^{4}b}\sqrt{ex+d}}+{\frac{3\,{e}^{4}}{64\,{b}^{2} \left ({a}^{2}{e}^{2}-2\,abde+{b}^{2}{d}^{2} \right ) }\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

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)^3,x)

[Out]

3/64*e^4/(b*e*x+a*e)^4*b/(a^2*e^2-2*a*b*d*e+b^2*d^2)*(e*x+d)^(7/2)+11/64*e^4/(b*e*x+a*e)^4/(a*e-b*d)*(e*x+d)^(
5/2)-11/64*e^4/(b*e*x+a*e)^4/b*(e*x+d)^(3/2)-3/64*e^5/(b*e*x+a*e)^4/b^2*(e*x+d)^(1/2)*a+3/64*e^4/(b*e*x+a*e)^4
/b*(e*x+d)^(1/2)*d+3/64*e^4/b^2/(a^2*e^2-2*a*b*d*e+b^2*d^2)/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b
*d)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.13822, size = 2101, normalized size = 12.22 \begin{align*} \text{result too large to display} \end{align*}

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)^3,x, algorithm="fricas")

[Out]

[1/128*(3*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*lo
g((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(16*b^5*d^4 - 40*a*b^4*d^3*e + 26
*a^2*b^3*d^2*e^2 + a^3*b^2*d*e^3 - 3*a^4*b*e^4 - 3*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (2*b^5*d^2*e^2 - 13*a*b^4*d*e
^3 + 11*a^2*b^3*e^4)*x^2 + (24*b^5*d^3*e - 68*a*b^4*d^2*e^2 + 55*a^2*b^3*d*e^3 - 11*a^3*b^2*e^4)*x)*sqrt(e*x +
d))/(a^4*b^6*d^3 - 3*a^5*b^5*d^2*e + 3*a^6*b^4*d*e^2 - a^7*b^3*e^3 + (b^10*d^3 - 3*a*b^9*d^2*e + 3*a^2*b^8*d*
e^2 - a^3*b^7*e^3)*x^4 + 4*(a*b^9*d^3 - 3*a^2*b^8*d^2*e + 3*a^3*b^7*d*e^2 - a^4*b^6*e^3)*x^3 + 6*(a^2*b^8*d^3
- 3*a^3*b^7*d^2*e + 3*a^4*b^6*d*e^2 - a^5*b^5*e^3)*x^2 + 4*(a^3*b^7*d^3 - 3*a^4*b^6*d^2*e + 3*a^5*b^5*d*e^2 -
a^6*b^4*e^3)*x), 1/64*(3*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b
^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (16*b^5*d^4 - 40*a*b^4*d^3*e + 26*a^2
*b^3*d^2*e^2 + a^3*b^2*d*e^3 - 3*a^4*b*e^4 - 3*(b^5*d*e^3 - a*b^4*e^4)*x^3 + (2*b^5*d^2*e^2 - 13*a*b^4*d*e^3 +
11*a^2*b^3*e^4)*x^2 + (24*b^5*d^3*e - 68*a*b^4*d^2*e^2 + 55*a^2*b^3*d*e^3 - 11*a^3*b^2*e^4)*x)*sqrt(e*x + d))
/(a^4*b^6*d^3 - 3*a^5*b^5*d^2*e + 3*a^6*b^4*d*e^2 - a^7*b^3*e^3 + (b^10*d^3 - 3*a*b^9*d^2*e + 3*a^2*b^8*d*e^2
- a^3*b^7*e^3)*x^4 + 4*(a*b^9*d^3 - 3*a^2*b^8*d^2*e + 3*a^3*b^7*d*e^2 - a^4*b^6*e^3)*x^3 + 6*(a^2*b^8*d^3 - 3*
a^3*b^7*d^2*e + 3*a^4*b^6*d*e^2 - a^5*b^5*e^3)*x^2 + 4*(a^3*b^7*d^3 - 3*a^4*b^6*d^2*e + 3*a^5*b^5*d*e^2 - a^6*
b^4*e^3)*x)]

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

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)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.20988, size = 390, normalized size = 2.27 \begin{align*} \frac{3 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{4}}{64 \,{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} \sqrt{-b^{2} d + a b e}} + \frac{3 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{3} e^{4} - 11 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} d e^{4} - 11 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d^{2} e^{4} + 3 \, \sqrt{x e + d} b^{3} d^{3} e^{4} + 11 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{2} e^{5} + 22 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} d e^{5} - 9 \, \sqrt{x e + d} a b^{2} d^{2} e^{5} - 11 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b e^{6} + 9 \, \sqrt{x e + d} a^{2} b d e^{6} - 3 \, \sqrt{x e + d} a^{3} e^{7}}{64 \,{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{4}} \end{align*}

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)^3,x, algorithm="giac")

[Out]

3/64*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^4/((b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*sqrt(-b^2*d + a*b
*e)) + 1/64*(3*(x*e + d)^(7/2)*b^3*e^4 - 11*(x*e + d)^(5/2)*b^3*d*e^4 - 11*(x*e + d)^(3/2)*b^3*d^2*e^4 + 3*sqr
t(x*e + d)*b^3*d^3*e^4 + 11*(x*e + d)^(5/2)*a*b^2*e^5 + 22*(x*e + d)^(3/2)*a*b^2*d*e^5 - 9*sqrt(x*e + d)*a*b^2
*d^2*e^5 - 11*(x*e + d)^(3/2)*a^2*b*e^6 + 9*sqrt(x*e + d)*a^2*b*d*e^6 - 3*sqrt(x*e + d)*a^3*e^7)/((b^4*d^2 - 2
*a*b^3*d*e + a^2*b^2*e^2)*((x*e + d)*b - b*d + a*e)^4)