3.1870 \(\int \frac{A+B x}{\sqrt{d+e x} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=215 \[ -\frac{\sqrt{d+e x} (A b-a B)}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{\sqrt{d+e x} (-a B e-3 A b e+4 b B d)}{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}+\frac{e (a+b x) (-a B e-3 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.204577, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {770, 78, 51, 63, 208} \[ -\frac{\sqrt{d+e x} (A b-a B)}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{\sqrt{d+e x} (-a B e-3 A b e+4 b B d)}{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}+\frac{e (a+b x) (-a B e-3 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 0.280961, size = 170, normalized size = 0.79 \[ \frac{(a+b x) \sqrt{d+e x} \left (\frac{(a+b x) (-a B e-3 A b e+4 b B d) \left (\sqrt{b} \sqrt{d+e x} (a e-b d)+e (a+b x) \sqrt{a e-b d} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )\right )}{2 \sqrt{b} \sqrt{d+e x} (b d-a e)^2}+a B-A b\right )}{2 b \left ((a+b x)^2\right )^{3/2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.022, size = 556, normalized size = 2.6 \begin{align*}{\frac{bx+a}{4\,be \left ( ae-bd \right ) ^{2}} \left ( 3\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{b}^{3}{e}^{3}+B\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){x}^{2}a{b}^{2}{e}^{3}-4\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{b}^{3}d{e}^{2}+6\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{2}{e}^{3}+2\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}b{e}^{3}-8\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{2}d{e}^{2}+3\,A\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{2}e+3\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}b{e}^{3}+B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{{\frac{3}{2}}}abe-4\,B\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{b}^{2}d+B\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){a}^{3}{e}^{3}-4\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}bd{e}^{2}+5\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}ab{e}^{2}-5\,A\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}de-B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}{e}^{2}-3\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}abde+4\,B\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{2}{d}^{2} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x)

[Out]

1/4*(b*x+a)/e*(3*A*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^2*b^3*e^3+B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)
*b)^(1/2))*x^2*a*b^2*e^3-4*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x^2*b^3*d*e^2+6*A*arctan((e*x+d)^(1/2
)*b/((a*e-b*d)*b)^(1/2))*x*a*b^2*e^3+2*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a^2*b*e^3-8*B*arctan((e
*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a*b^2*d*e^2+3*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b^2*e+3*A*arctan((e*x+d
)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^2*b*e^3+B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*a*b*e-4*B*((a*e-b*d)*b)^(1/2)*(e*
x+d)^(3/2)*b^2*d+B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^3*e^3-4*B*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b
)^(1/2))*a^2*b*d*e^2+5*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*e^2-5*A*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d
*e-B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2-3*B*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b*d*e+4*B*((a*e-b*d)*b)
^(1/2)*(e*x+d)^(1/2)*b^2*d^2)/((a*e-b*d)*b)^(1/2)/b/(a*e-b*d)^2/((b*x+a)^2)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x + A}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} \sqrt{e x + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*x + A)/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*sqrt(e*x + d)), x)

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Fricas [B]  time = 1.69729, size = 1646, normalized size = 7.66 \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)/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[-1/8*((4*B*a^2*b*d*e - (B*a^3 + 3*A*a^2*b)*e^2 + (4*B*b^3*d*e - (B*a*b^2 + 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d
*e - (B*a^2*b + 3*A*a*b^2)*e^2)*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*(2*(B*a*b^3 + A*b^4)*d^2 - (B*a^2*b^2 + 7*A*a*b^3)*d*e - (B*a^3*b - 5*A*a^2*b^2)*e^2 +
(4*B*b^4*d^2 - (5*B*a*b^3 + 3*A*b^4)*d*e + (B*a^2*b^2 + 3*A*a*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^3 - 3*a^
3*b^4*d^2*e + 3*a^4*b^3*d*e^2 - a^5*b^2*e^3 + (b^7*d^3 - 3*a*b^6*d^2*e + 3*a^2*b^5*d*e^2 - a^3*b^4*e^3)*x^2 +
2*(a*b^6*d^3 - 3*a^2*b^5*d^2*e + 3*a^3*b^4*d*e^2 - a^4*b^3*e^3)*x), -1/4*((4*B*a^2*b*d*e - (B*a^3 + 3*A*a^2*b)
*e^2 + (4*B*b^3*d*e - (B*a*b^2 + 3*A*b^3)*e^2)*x^2 + 2*(4*B*a*b^2*d*e - (B*a^2*b + 3*A*a*b^2)*e^2)*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)) + (2*(B*a*b^3 + A*b^4)*d^2 - (B*a^2*b^2
+ 7*A*a*b^3)*d*e - (B*a^3*b - 5*A*a^2*b^2)*e^2 + (4*B*b^4*d^2 - (5*B*a*b^3 + 3*A*b^4)*d*e + (B*a^2*b^2 + 3*A*a
*b^3)*e^2)*x)*sqrt(e*x + d))/(a^2*b^5*d^3 - 3*a^3*b^4*d^2*e + 3*a^4*b^3*d*e^2 - a^5*b^2*e^3 + (b^7*d^3 - 3*a*b
^6*d^2*e + 3*a^2*b^5*d*e^2 - a^3*b^4*e^3)*x^2 + 2*(a*b^6*d^3 - 3*a^2*b^5*d^2*e + 3*a^3*b^4*d*e^2 - a^4*b^3*e^3
)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{\sqrt{d + e x} \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.26288, size = 543, normalized size = 2.53 \begin{align*} -\frac{{\left (4 \, B b d e^{2} - B a e^{3} - 3 \, A b e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{4 \,{\left (b^{3} d^{2} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b^{2} d e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} b e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{-b^{2} d + a b e}} - \frac{4 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{2} d e^{2} - 4 \, \sqrt{x e + d} B b^{2} d^{2} e^{2} -{\left (x e + d\right )}^{\frac{3}{2}} B a b e^{3} - 3 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{2} e^{3} + 3 \, \sqrt{x e + d} B a b d e^{3} + 5 \, \sqrt{x e + d} A b^{2} d e^{3} + \sqrt{x e + d} B a^{2} e^{4} - 5 \, \sqrt{x e + d} A a b e^{4}}{4 \,{\left (b^{3} d^{2} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 2 \, a b^{2} d e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + a^{2} b e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

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