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

Optimal. Leaf size=148 \[ \frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2}}{3 e^3 (a+b x)}-\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}{e^3 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x) \sqrt {d+e x}} \]

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Rubi [A]  time = 0.07, antiderivative size = 148, 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)^{3/2}}{3 e^3 (a+b x)}-\frac {4 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)}{e^3 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x) \sqrt {d+e 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)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)*Sqrt[d + e*x]) - (4*b*(b*d - a*e)*Sqrt[d + e*x
]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)) + (2*b^2*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*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}}{(d+e x)^{3/2}} \, 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 )}{(d+e x)^{3/2}} \, 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}{(d+e x)^{3/2}} \, 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 (d+e x)^{3/2}}-\frac {2 b (b d-a e)}{e^2 \sqrt {d+e x}}+\frac {b^2 \sqrt {d+e x}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {2 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) \sqrt {d+e x}}-\frac {4 b (b d-a e) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}+\frac {2 b^2 (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 = 0.04, size = 78, normalized size = 0.53 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (3 a^2 e^2-6 a b e (2 d+e x)+b^2 \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^2*e^2 - 6*a*b*e*(2*d + e*x) + b^2*(8*d^2 + 4*d*e*x - e^2*x^2)))/(3*e^3*(a + b*x)*Sq
rt[d + e*x])

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IntegrateAlgebraic [A]  time = 10.07, size = 99, normalized size = 0.67 \begin {gather*} \frac {2 \sqrt {\frac {(a e+b e x)^2}{e^2}} \left (-3 a^2 e^2+6 a b e (d+e x)+6 a b d e-3 b^2 d^2+b^2 (d+e x)^2-6 b^2 d (d+e x)\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^2*d^2 + 6*a*b*d*e - 3*a^2*e^2 - 6*b^2*d*(d + e*x) + 6*a*b*e*(d + e*x) + b^2
*(d + e*x)^2))/(3*e^2*Sqrt[d + e*x]*(a*e + b*e*x))

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fricas [A]  time = 0.41, size = 73, normalized size = 0.49 \begin {gather*} \frac {2 \, {\left (b^{2} e^{2} x^{2} - 8 \, b^{2} d^{2} + 12 \, a b d e - 3 \, a^{2} e^{2} - 2 \, {\left (2 \, b^{2} d e - 3 \, a b 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^2*e^2*x^2 - 8*b^2*d^2 + 12*a*b*d*e - 3*a^2*e^2 - 2*(2*b^2*d*e - 3*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 = 119, normalized size = 0.80 \begin {gather*} \frac {2}{3} \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{2} e^{6} \mathrm {sgn}\left (b x + a\right ) - 6 \, \sqrt {x e + d} b^{2} d e^{6} \mathrm {sgn}\left (b x + a\right ) + 6 \, \sqrt {x e + d} a b e^{7} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-9\right )} - \frac {2 \, {\left (b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} 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^2*e^6*sgn(b*x + a) - 6*sqrt(x*e + d)*b^2*d*e^6*sgn(b*x + a) + 6*sqrt(x*e + d)*a*b*e^7*s
gn(b*x + a))*e^(-9) - 2*(b^2*d^2*sgn(b*x + a) - 2*a*b*d*e*sgn(b*x + a) + a^2*e^2*sgn(b*x + a))*e^(-3)/sqrt(x*e
+ d)

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maple [A]  time = 0.05, size = 79, normalized size = 0.53 \begin {gather*} -\frac {2 \left (-b^{2} x^{2} e^{2}-6 a b \,e^{2} x +4 b^{2} d e x +3 a^{2} e^{2}-12 a b d e +8 b^{2} 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^2*e^2*x^2-6*a*b*e^2*x+4*b^2*d*e*x+3*a^2*e^2-12*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 = 75, normalized size = 0.51 \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 = 2.65, size = 90, normalized size = 0.61 \begin {gather*} \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (\frac {4\,x\,\left (3\,a\,e-2\,b\,d\right )}{3\,e^2}+\frac {2\,b\,x^2}{3\,e}-\frac {6\,a^2\,e^2-24\,a\,b\,d\,e+16\,b^2\,d^2}{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)*((4*x*(3*a*e - 2*b*d))/(3*e^2) + (2*b*x^2)/(3*e) - (6*a^2*e^2 + 16*b^2*d^2 - 24*a*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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