3.1819 \(\int \frac{(A+B x) (d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=209 \[ -\frac{e^2 (-5 a B e-A b e+6 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{7/2} (b d-a e)^{3/2}}-\frac{(d+e x)^{3/2} (-5 a B e-A b e+6 b B d)}{12 b^2 (a+b x)^2 (b d-a e)}-\frac{e \sqrt{d+e x} (-5 a B e-A b e+6 b B d)}{8 b^3 (a+b x) (b d-a e)}-\frac{(d+e x)^{5/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

[Out]

-(e*(6*b*B*d - A*b*e - 5*a*B*e)*Sqrt[d + e*x])/(8*b^3*(b*d - a*e)*(a + b*x)) - ((6*b*B*d - A*b*e - 5*a*B*e)*(d
+ e*x)^(3/2))/(12*b^2*(b*d - a*e)*(a + b*x)^2) - ((A*b - a*B)*(d + e*x)^(5/2))/(3*b*(b*d - a*e)*(a + b*x)^3)
- (e^2*(6*b*B*d - A*b*e - 5*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(7/2)*(b*d - a*e)^(3
/2))

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Rubi [A]  time = 0.180661, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {27, 78, 47, 63, 208} \[ -\frac{e^2 (-5 a B e-A b e+6 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{7/2} (b d-a e)^{3/2}}-\frac{(d+e x)^{3/2} (-5 a B e-A b e+6 b B d)}{12 b^2 (a+b x)^2 (b d-a e)}-\frac{e \sqrt{d+e x} (-5 a B e-A b e+6 b B d)}{8 b^3 (a+b x) (b d-a e)}-\frac{(d+e x)^{5/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

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]

-(e*(6*b*B*d - A*b*e - 5*a*B*e)*Sqrt[d + e*x])/(8*b^3*(b*d - a*e)*(a + b*x)) - ((6*b*B*d - A*b*e - 5*a*B*e)*(d
+ e*x)^(3/2))/(12*b^2*(b*d - a*e)*(a + b*x)^2) - ((A*b - a*B)*(d + e*x)^(5/2))/(3*b*(b*d - a*e)*(a + b*x)^3)
- (e^2*(6*b*B*d - A*b*e - 5*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(7/2)*(b*d - a*e)^(3
/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 78

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

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{(A+B x) (d+e x)^{3/2}}{(a+b x)^4} \, dx\\ &=-\frac{(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(6 b B d-A b e-5 a B e) \int \frac{(d+e x)^{3/2}}{(a+b x)^3} \, dx}{6 b (b d-a e)}\\ &=-\frac{(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(e (6 b B d-A b e-5 a B e)) \int \frac{\sqrt{d+e x}}{(a+b x)^2} \, dx}{8 b^2 (b d-a e)}\\ &=-\frac{e (6 b B d-A b e-5 a B e) \sqrt{d+e x}}{8 b^3 (b d-a e) (a+b x)}-\frac{(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac{\left (e^2 (6 b B d-A b e-5 a B e)\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{16 b^3 (b d-a e)}\\ &=-\frac{e (6 b B d-A b e-5 a B e) \sqrt{d+e x}}{8 b^3 (b d-a e) (a+b x)}-\frac{(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac{(e (6 b B d-A b e-5 a 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 )}{8 b^3 (b d-a e)}\\ &=-\frac{e (6 b B d-A b e-5 a B e) \sqrt{d+e x}}{8 b^3 (b d-a e) (a+b x)}-\frac{(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac{(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}-\frac{e^2 (6 b B d-A b e-5 a B e) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{8 b^{7/2} (b d-a e)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.487631, size = 177, normalized size = 0.85 \[ \frac{\frac{(a+b x) (-5 a B e-A b e+6 b B d) \left (3 \sqrt{b} e^2 (a+b x)^2 \sqrt{d+e x} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )-b (d+e x) \sqrt{a e-b d} (3 a e+2 b d+5 b e x)\right )}{\sqrt{a e-b d}}-8 b^3 (d+e x)^3 (A b-a B)}{24 b^4 (a+b x)^3 \sqrt{d+e x} (b d-a e)} \]

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]

(-8*b^3*(A*b - a*B)*(d + e*x)^3 + ((6*b*B*d - A*b*e - 5*a*B*e)*(a + b*x)*(-(b*Sqrt[-(b*d) + a*e]*(d + e*x)*(2*
b*d + 3*a*e + 5*b*e*x)) + 3*Sqrt[b]*e^2*(a + b*x)^2*Sqrt[d + e*x]*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) +
a*e]]))/Sqrt[-(b*d) + a*e])/(24*b^4*(b*d - a*e)*(a + b*x)^3*Sqrt[d + e*x])

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Maple [B]  time = 0.02, size = 487, normalized size = 2.3 \begin{align*}{\frac{{e}^{3}A}{8\, \left ( bex+ae \right ) ^{3} \left ( ae-bd \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{11\,{e}^{3}aB}{8\, \left ( bex+ae \right ) ^{3} \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{e}^{2}Bd}{4\, \left ( bex+ae \right ) ^{3} \left ( ae-bd \right ) } \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{{e}^{3}A}{3\, \left ( bex+ae \right ) ^{3}b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{e}^{3}aB}{3\, \left ( bex+ae \right ) ^{3}{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+2\,{\frac{{e}^{2} \left ( ex+d \right ) ^{3/2}Bd}{ \left ( bex+ae \right ) ^{3}b}}-{\frac{{e}^{4}Aa}{8\, \left ( bex+ae \right ) ^{3}{b}^{2}}\sqrt{ex+d}}+{\frac{{e}^{3}Ad}{8\, \left ( bex+ae \right ) ^{3}b}\sqrt{ex+d}}-{\frac{5\,B{a}^{2}{e}^{4}}{8\, \left ( bex+ae \right ) ^{3}{b}^{3}}\sqrt{ex+d}}+{\frac{11\,{e}^{3}Bda}{8\, \left ( bex+ae \right ) ^{3}{b}^{2}}\sqrt{ex+d}}-{\frac{3\,{e}^{2}B{d}^{2}}{4\, \left ( bex+ae \right ) ^{3}b}\sqrt{ex+d}}+{\frac{{e}^{3}A}{ \left ( 8\,ae-8\,bd \right ){b}^{2}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{5\,{e}^{3}aB}{ \left ( 8\,ae-8\,bd \right ){b}^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}-{\frac{3\,{e}^{2}Bd}{ \left ( 4\,ae-4\,bd \right ){b}^{2}}\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)^2,x)

[Out]

1/8*e^3/(b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(5/2)*A-11/8*e^3/(b*e*x+a*e)^3/(a*e-b*d)/b*(e*x+d)^(5/2)*a*B+5/4*e^2/(
b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(5/2)*B*d-1/3*e^3/(b*e*x+a*e)^3/b*(e*x+d)^(3/2)*A-5/3*e^3/(b*e*x+a*e)^3/b^2*(e*
x+d)^(3/2)*a*B+2*e^2/(b*e*x+a*e)^3/b*(e*x+d)^(3/2)*B*d-1/8*e^4/(b*e*x+a*e)^3/b^2*(e*x+d)^(1/2)*A*a+1/8*e^3/(b*
e*x+a*e)^3/b*(e*x+d)^(1/2)*A*d-5/8*e^4/(b*e*x+a*e)^3/b^3*(e*x+d)^(1/2)*a^2*B+11/8*e^3/(b*e*x+a*e)^3/b^2*(e*x+d
)^(1/2)*B*d*a-3/4*e^2/(b*e*x+a*e)^3/b*(e*x+d)^(1/2)*B*d^2+1/8*e^3/(a*e-b*d)/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*
x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*A+5/8*e^3/(a*e-b*d)/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d
)*b)^(1/2))*a*B-3/4*e^2/(a*e-b*d)/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*B*d

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.58115, size = 2277, normalized size = 10.89 \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)^2,x, algorithm="fricas")

[Out]

[1/48*(3*(6*B*a^3*b*d*e^2 - (5*B*a^4 + A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (5*B*a*b^3 + A*b^4)*e^3)*x^3 + 3*(6*B*a
*b^3*d*e^2 - (5*B*a^2*b^2 + A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (5*B*a^3*b + A*a^2*b^2)*e^3)*x)*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*(4*(B*a*b^4 + 2*A*
b^5)*d^3 + 2*(2*B*a^2*b^3 - 5*A*a*b^4)*d^2*e - (23*B*a^3*b^2 + A*a^2*b^3)*d*e^2 + 3*(5*B*a^4*b + A*a^3*b^2)*e^
3 + 3*(10*B*b^5*d^2*e - (21*B*a*b^4 - A*b^5)*d*e^2 + (11*B*a^2*b^3 - A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (5*B
*a*b^4 + 7*A*b^5)*d^2*e - (31*B*a^2*b^3 + 11*A*a*b^4)*d*e^2 + 4*(5*B*a^3*b^2 + A*a^2*b^3)*e^3)*x)*sqrt(e*x + d
))/(a^3*b^6*d^2 - 2*a^4*b^5*d*e + a^5*b^4*e^2 + (b^9*d^2 - 2*a*b^8*d*e + a^2*b^7*e^2)*x^3 + 3*(a*b^8*d^2 - 2*a
^2*b^7*d*e + a^3*b^6*e^2)*x^2 + 3*(a^2*b^7*d^2 - 2*a^3*b^6*d*e + a^4*b^5*e^2)*x), 1/24*(3*(6*B*a^3*b*d*e^2 - (
5*B*a^4 + A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (5*B*a*b^3 + A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (5*B*a^2*b^2 + A
*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (5*B*a^3*b + A*a^2*b^2)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^
2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (4*(B*a*b^4 + 2*A*b^5)*d^3 + 2*(2*B*a^2*b^3 - 5*A*a*b^4)*d^2*e - (
23*B*a^3*b^2 + A*a^2*b^3)*d*e^2 + 3*(5*B*a^4*b + A*a^3*b^2)*e^3 + 3*(10*B*b^5*d^2*e - (21*B*a*b^4 - A*b^5)*d*e
^2 + (11*B*a^2*b^3 - A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (5*B*a*b^4 + 7*A*b^5)*d^2*e - (31*B*a^2*b^3 + 11*A*a
*b^4)*d*e^2 + 4*(5*B*a^3*b^2 + A*a^2*b^3)*e^3)*x)*sqrt(e*x + d))/(a^3*b^6*d^2 - 2*a^4*b^5*d*e + a^5*b^4*e^2 +
(b^9*d^2 - 2*a*b^8*d*e + a^2*b^7*e^2)*x^3 + 3*(a*b^8*d^2 - 2*a^2*b^7*d*e + a^3*b^6*e^2)*x^2 + 3*(a^2*b^7*d^2 -
2*a^3*b^6*d*e + a^4*b^5*e^2)*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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.1809, size = 514, normalized size = 2.46 \begin{align*} \frac{{\left (6 \, B b d e^{2} - 5 \, B a e^{3} - A b e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{8 \,{\left (b^{4} d - a b^{3} e\right )} \sqrt{-b^{2} d + a b e}} - \frac{30 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{3} d e^{2} - 48 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{3} d^{2} e^{2} + 18 \, \sqrt{x e + d} B b^{3} d^{3} e^{2} - 33 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{2} e^{3} + 3 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{3} e^{3} + 88 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{2} d e^{3} + 8 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{3} d e^{3} - 51 \, \sqrt{x e + d} B a b^{2} d^{2} e^{3} - 3 \, \sqrt{x e + d} A b^{3} d^{2} e^{3} - 40 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b e^{4} - 8 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{2} e^{4} + 48 \, \sqrt{x e + d} B a^{2} b d e^{4} + 6 \, \sqrt{x e + d} A a b^{2} d e^{4} - 15 \, \sqrt{x e + d} B a^{3} e^{5} - 3 \, \sqrt{x e + d} A a^{2} b e^{5}}{24 \,{\left (b^{4} d - a b^{3} e\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \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)^2,x, algorithm="giac")

[Out]

1/8*(6*B*b*d*e^2 - 5*B*a*e^3 - A*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d - a*b^3*e)*sqrt(-
b^2*d + a*b*e)) - 1/24*(30*(x*e + d)^(5/2)*B*b^3*d*e^2 - 48*(x*e + d)^(3/2)*B*b^3*d^2*e^2 + 18*sqrt(x*e + d)*B
*b^3*d^3*e^2 - 33*(x*e + d)^(5/2)*B*a*b^2*e^3 + 3*(x*e + d)^(5/2)*A*b^3*e^3 + 88*(x*e + d)^(3/2)*B*a*b^2*d*e^3
+ 8*(x*e + d)^(3/2)*A*b^3*d*e^3 - 51*sqrt(x*e + d)*B*a*b^2*d^2*e^3 - 3*sqrt(x*e + d)*A*b^3*d^2*e^3 - 40*(x*e
+ d)^(3/2)*B*a^2*b*e^4 - 8*(x*e + d)^(3/2)*A*a*b^2*e^4 + 48*sqrt(x*e + d)*B*a^2*b*d*e^4 + 6*sqrt(x*e + d)*A*a*
b^2*d*e^4 - 15*sqrt(x*e + d)*B*a^3*e^5 - 3*sqrt(x*e + d)*A*a^2*b*e^5)/((b^4*d - a*b^3*e)*((x*e + d)*b - b*d +
a*e)^3)