3.19.71 \(\int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {d+e x}} \, dx\)

Optimal. Leaf size=150 \[ \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^3 (a+b x)}-\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}{e^3 (a+b x)} \]

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Rubi [A]  time = 0.07, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {770, 21, 43} \begin {gather*} \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^3 (a+b x)}-\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)}{3 e^3 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}{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])/Sqrt[d + e*x],x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.05, size = 78, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (15 a^2 e^2+10 a b e (e x-2 d)+b^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*Sqrt[d + e*x]*(15*a^2*e^2 + 10*a*b*e*(-2*d + e*x) + b^2*(8*d^2 - 4*d*e*x + 3*e^2*x^2)))/(
15*e^3*(a + b*x))

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IntegrateAlgebraic [A]  time = 14.94, size = 100, normalized size = 0.67 \begin {gather*} \frac {2 \sqrt {d+e x} \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (15 a^2 e^2+10 a b e (d+e x)-30 a b d e+15 b^2 d^2+3 b^2 (d+e x)^2-10 b^2 d (d+e x)\right )}{15 e^2 (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])/Sqrt[d + e*x],x]

[Out]

(2*Sqrt[d + e*x]*Sqrt[(a*e + b*e*x)^2/e^2]*(15*b^2*d^2 - 30*a*b*d*e + 15*a^2*e^2 - 10*b^2*d*(d + e*x) + 10*a*b
*e*(d + e*x) + 3*b^2*(d + e*x)^2))/(15*e^2*(a*e + b*e*x))

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fricas [A]  time = 0.42, size = 64, normalized size = 0.43 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} - 20 \, a b d e + 15 \, a^{2} e^{2} - 2 \, {\left (2 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, 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)^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*b^2*e^2*x^2 + 8*b^2*d^2 - 20*a*b*d*e + 15*a^2*e^2 - 2*(2*b^2*d*e - 5*a*b*e^2)*x)*sqrt(e*x + d)/e^3

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giac [A]  time = 0.19, size = 103, normalized size = 0.69 \begin {gather*} \frac {2}{15} \, {\left (10 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a b e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} b^{2} e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + 15 \, \sqrt {x e + d} a^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\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)^(1/2),x, algorithm="giac")

[Out]

2/15*(10*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a*b*e^(-1)*sgn(b*x + a) + (3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/
2)*d + 15*sqrt(x*e + d)*d^2)*b^2*e^(-2)*sgn(b*x + a) + 15*sqrt(x*e + d)*a^2*sgn(b*x + a))*e^(-1)

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maple [A]  time = 0.05, size = 79, normalized size = 0.53 \begin {gather*} \frac {2 \sqrt {e x +d}\, \left (3 b^{2} x^{2} e^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {\left (b x +a \right )^{2}}}{15 \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)^(1/2),x)

[Out]

2/15*(e*x+d)^(1/2)*(3*b^2*e^2*x^2+10*a*b*e^2*x-4*b^2*d*e*x+15*a^2*e^2-20*a*b*d*e+8*b^2*d^2)*((b*x+a)^2)^(1/2)/
e^3/(b*x+a)

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maxima [A]  time = 0.66, size = 119, normalized size = 0.79 \begin {gather*} \frac {2 \, {\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e - {\left (b d e - 3 \, a e^{2}\right )} x\right )} a}{3 \, \sqrt {e x + d} e^{2}} + \frac {2 \, {\left (3 \, b e^{3} x^{3} + 8 \, b d^{3} - 10 \, a d^{2} e - {\left (b d e^{2} - 5 \, a e^{3}\right )} x^{2} + {\left (4 \, b d^{2} e - 5 \, a d e^{2}\right )} x\right )} b}{15 \, \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)^(1/2),x, algorithm="maxima")

[Out]

2/3*(b*e^2*x^2 - 2*b*d^2 + 3*a*d*e - (b*d*e - 3*a*e^2)*x)*a/(sqrt(e*x + d)*e^2) + 2/15*(3*b*e^3*x^3 + 8*b*d^3
- 10*a*d^2*e - (b*d*e^2 - 5*a*e^3)*x^2 + (4*b*d^2*e - 5*a*d*e^2)*x)*b/(sqrt(e*x + d)*e^3)

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mupad [B]  time = 2.41, size = 127, normalized size = 0.85 \begin {gather*} \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {2\,b\,x^3}{5}+\frac {2\,x^2\,\left (10\,a\,e-b\,d\right )}{15\,e}+\frac {30\,a^2\,d\,e^2-40\,a\,b\,d^2\,e+16\,b^2\,d^3}{15\,b\,e^3}+\frac {x\,\left (30\,a^2\,e^3-20\,a\,b\,d\,e^2+8\,b^2\,d^2\,e\right )}{15\,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)^(1/2),x)

[Out]

(((a + b*x)^2)^(1/2)*((2*b*x^3)/5 + (2*x^2*(10*a*e - b*d))/(15*e) + (16*b^2*d^3 + 30*a^2*d*e^2 - 40*a*b*d^2*e)
/(15*b*e^3) + (x*(30*a^2*e^3 + 8*b^2*d^2*e - 20*a*b*d*e^2))/(15*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}}}{\sqrt {d + e x}}\, 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)**(1/2),x)

[Out]

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

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