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

Optimal. Leaf size=160 \[ -\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (-a B e-A b e+2 b B d)}{e^3 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{e^3 (a+b x) \sqrt {d+e x}}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^3 (a+b x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {770, 77} \begin {gather*} -\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (-a B e-A b e+2 b B d)}{e^3 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{e^3 (a+b x) \sqrt {d+e x}}+\frac {2 b B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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]

Rubi steps

\begin {align*} \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^{3/2}} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right ) (A+B x)}{(d+e x)^{3/2}} \, dx}{a b+b^2 x}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b (b d-a e) (-B d+A e)}{e^2 (d+e x)^{3/2}}+\frac {b (-2 b B d+A b e+a B e)}{e^2 \sqrt {d+e x}}+\frac {b^2 B \sqrt {d+e x}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 (b d-a e) (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}-\frac {2 (2 b B d-A b e-a B e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}+\frac {2 b B (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 85, normalized size = 0.53 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (3 a e (-A e+2 B d+B e x)+3 A b e (2 d+e x)+b B \left (-8 d^2-4 d e x+e^2 x^2\right )\right )}{3 e^3 (a+b x) \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 20.63, size = 111, normalized size = 0.69 \begin {gather*} \frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (-3 a A e^2+3 a B e (d+e x)+3 a B d e+3 A b e (d+e x)+3 A b d e-3 b B d^2-6 b B d (d+e x)+b B (d+e x)^2\right )}{3 e^2 \sqrt {d+e x} (a e+b e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 79, normalized size = 0.49 \begin {gather*} \frac {2 \, {\left (B b e^{2} x^{2} - 8 \, B b d^{2} - 3 \, A a e^{2} + 6 \, {\left (B a + A b\right )} d e - {\left (4 \, B b d e - 3 \, {\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{4} x + d e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 148, normalized size = 0.92 \begin {gather*} \frac {2}{3} \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} B b e^{6} \mathrm {sgn}\left (b x + a\right ) - 6 \, \sqrt {x e + d} B b d e^{6} \mathrm {sgn}\left (b x + a\right ) + 3 \, \sqrt {x e + d} B a e^{7} \mathrm {sgn}\left (b x + a\right ) + 3 \, \sqrt {x e + d} A b e^{7} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-9\right )} - \frac {2 \, {\left (B b d^{2} \mathrm {sgn}\left (b x + a\right ) - B a d e \mathrm {sgn}\left (b x + a\right ) - A b d e \mathrm {sgn}\left (b x + a\right ) + A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )}}{\sqrt {x e + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((x*e + d)^(3/2)*B*b*e^6*sgn(b*x + a) - 6*sqrt(x*e + d)*B*b*d*e^6*sgn(b*x + a) + 3*sqrt(x*e + d)*B*a*e^7*s
gn(b*x + a) + 3*sqrt(x*e + d)*A*b*e^7*sgn(b*x + a))*e^(-9) - 2*(B*b*d^2*sgn(b*x + a) - B*a*d*e*sgn(b*x + a) -
A*b*d*e*sgn(b*x + a) + A*a*e^2*sgn(b*x + a))*e^(-3)/sqrt(x*e + d)

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maple [A]  time = 0.04, size = 89, normalized size = 0.56 \begin {gather*} -\frac {2 \left (-B b \,x^{2} e^{2}-3 A b \,e^{2} x -3 B a \,e^{2} x +4 B b d e x +3 A a \,e^{2}-6 A b d e -6 B a d e +8 B b \,d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{3 \sqrt {e x +d}\, \left (b x +a \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.77, size = 75, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (b e x + 2 \, b d - a e\right )} A}{\sqrt {e x + d} e^{2}} + \frac {2 \, {\left (b e^{2} x^{2} - 8 \, b d^{2} + 6 \, a d e - {\left (4 \, b d e - 3 \, a e^{2}\right )} x\right )} B}{3 \, \sqrt {e x + d} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.41, size = 109, normalized size = 0.68 \begin {gather*} \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,B\,x^2}{3\,e}-\frac {6\,A\,a\,e^2+16\,B\,b\,d^2-12\,A\,b\,d\,e-12\,B\,a\,d\,e}{3\,b\,e^3}+\frac {x\,\left (6\,A\,b\,e^2+6\,B\,a\,e^2-8\,B\,b\,d\,e\right )}{3\,b\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a + b*x)^2)^(1/2)*((2*B*x^2)/(3*e) - (6*A*a*e^2 + 16*B*b*d^2 - 12*A*b*d*e - 12*B*a*d*e)/(3*b*e^3) + (x*(6*A
*b*e^2 + 6*B*a*e^2 - 8*B*b*d*e))/(3*b*e^3)))/(x*(d + e*x)^(1/2) + (a*(d + e*x)^(1/2))/b)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \sqrt {\left (a + b x\right )^{2}}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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