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

Optimal. Leaf size=214 \[ \frac {2 a^2 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{a+b x}+\frac {2 b^2 x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{5 (a+b x)}+\frac {2 a b x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac {2 b^3 B x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}-\frac {2 a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 76} \begin {gather*} \frac {2 b^2 x^{5/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{5 (a+b x)}+\frac {2 a b x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac {2 a^2 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{a+b x}-\frac {2 a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 b^3 B x^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(Sqrt[x]*(a + b*x)) + (2*a^2*(3*A*b + a*B)*Sqrt[x]*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/(a + b*x) + (2*a*b*(A*b + a*B)*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (2*b^2*(A*b + 3
*a*B)*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a + b*x)) + (2*b^3*B*x^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
7*(a + b*x))

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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

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Mathematica [A]  time = 0.03, size = 85, normalized size = 0.40 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (-35 a^3 (A-B x)+35 a^2 b x (3 A+B x)+7 a b^2 x^2 (5 A+3 B x)+b^3 x^3 (7 A+5 B x)\right )}{35 \sqrt {x} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(-35*a^3*(A - B*x) + 35*a^2*b*x*(3*A + B*x) + 7*a*b^2*x^2*(5*A + 3*B*x) + b^3*x^3*(7*A +
5*B*x)))/(35*Sqrt[x]*(a + b*x))

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IntegrateAlgebraic [A]  time = 6.45, size = 97, normalized size = 0.45 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (-35 a^3 A+35 a^3 B x+105 a^2 A b x+35 a^2 b B x^2+35 a A b^2 x^2+21 a b^2 B x^3+7 A b^3 x^3+5 b^3 B x^4\right )}{35 \sqrt {x} (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(-35*a^3*A + 105*a^2*A*b*x + 35*a^3*B*x + 35*a*A*b^2*x^2 + 35*a^2*b*B*x^2 + 7*A*b^3*x^3 +
21*a*b^2*B*x^3 + 5*b^3*B*x^4))/(35*Sqrt[x]*(a + b*x))

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fricas [A]  time = 0.43, size = 73, normalized size = 0.34 \begin {gather*} \frac {2 \, {\left (5 \, B b^{3} x^{4} - 35 \, A a^{3} + 7 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 35 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 35 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{35 \, \sqrt {x}} \end {gather*}

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)/x^(3/2),x, algorithm="fricas")

[Out]

2/35*(5*B*b^3*x^4 - 35*A*a^3 + 7*(3*B*a*b^2 + A*b^3)*x^3 + 35*(B*a^2*b + A*a*b^2)*x^2 + 35*(B*a^3 + 3*A*a^2*b)
*x)/sqrt(x)

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giac [A]  time = 0.17, size = 125, normalized size = 0.58 \begin {gather*} \frac {2}{7} \, B b^{3} x^{\frac {7}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{5} \, B a b^{2} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, A b^{3} x^{\frac {5}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{2} b x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, A a b^{2} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a^{2} b \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{\sqrt {x}} \end {gather*}

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)/x^(3/2),x, algorithm="giac")

[Out]

2/7*B*b^3*x^(7/2)*sgn(b*x + a) + 6/5*B*a*b^2*x^(5/2)*sgn(b*x + a) + 2/5*A*b^3*x^(5/2)*sgn(b*x + a) + 2*B*a^2*b
*x^(3/2)*sgn(b*x + a) + 2*A*a*b^2*x^(3/2)*sgn(b*x + a) + 2*B*a^3*sqrt(x)*sgn(b*x + a) + 6*A*a^2*b*sqrt(x)*sgn(
b*x + a) - 2*A*a^3*sgn(b*x + a)/sqrt(x)

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maple [A]  time = 0.05, size = 92, normalized size = 0.43 \begin {gather*} -\frac {2 \left (-5 B \,b^{3} x^{4}-7 A \,b^{3} x^{3}-21 B a \,b^{2} x^{3}-35 A a \,b^{2} x^{2}-35 B \,a^{2} b \,x^{2}-105 A \,a^{2} b x -35 B \,a^{3} x +35 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \left (b x +a \right )^{3} \sqrt {x}} \end {gather*}

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)/x^(3/2),x)

[Out]

-2/35*(-5*B*b^3*x^4-7*A*b^3*x^3-21*B*a*b^2*x^3-35*A*a*b^2*x^2-35*B*a^2*b*x^2-105*A*a^2*b*x-35*B*a^3*x+35*A*a^3
)*((b*x+a)^2)^(3/2)/x^(1/2)/(b*x+a)^3

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maxima [A]  time = 0.56, size = 133, normalized size = 0.62 \begin {gather*} \frac {2}{15} \, {\left ({\left (3 \, b^{3} x^{2} + 5 \, a b^{2} x\right )} \sqrt {x} + \frac {10 \, {\left (a b^{2} x^{2} + 3 \, a^{2} b x\right )}}{\sqrt {x}} + \frac {15 \, {\left (a^{2} b x^{2} - a^{3} x\right )}}{x^{\frac {3}{2}}}\right )} A + \frac {2}{105} \, {\left (3 \, {\left (5 \, b^{3} x^{2} + 7 \, a b^{2} x\right )} x^{\frac {3}{2}} + 14 \, {\left (3 \, a b^{2} x^{2} + 5 \, a^{2} b x\right )} \sqrt {x} + \frac {35 \, {\left (a^{2} b x^{2} + 3 \, a^{3} x\right )}}{\sqrt {x}}\right )} B \end {gather*}

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)/x^(3/2),x, algorithm="maxima")

[Out]

2/15*((3*b^3*x^2 + 5*a*b^2*x)*sqrt(x) + 10*(a*b^2*x^2 + 3*a^2*b*x)/sqrt(x) + 15*(a^2*b*x^2 - a^3*x)/x^(3/2))*A
+ 2/105*(3*(5*b^3*x^2 + 7*a*b^2*x)*x^(3/2) + 14*(3*a*b^2*x^2 + 5*a^2*b*x)*sqrt(x) + 35*(a^2*b*x^2 + 3*a^3*x)/
sqrt(x))*B

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mupad [B]  time = 1.64, size = 107, normalized size = 0.50 \begin {gather*} \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {x\,\left (70\,B\,a^3+210\,A\,b\,a^2\right )}{35\,b}-\frac {2\,A\,a^3}{b}+\frac {2\,B\,b^2\,x^4}{7}+\frac {x^3\,\left (14\,A\,b^3+42\,B\,a\,b^2\right )}{35\,b}+2\,a\,x^2\,\left (A\,b+B\,a\right )\right )}{x^{3/2}+\frac {a\,\sqrt {x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((x*(70*B*a^3 + 210*A*a^2*b))/(35*b) - (2*A*a^3)/b + (2*B*b^2*x^4)/7 + (x^3*(
14*A*b^3 + 42*B*a*b^2))/(35*b) + 2*a*x^2*(A*b + B*a)))/(x^(3/2) + (a*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 ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{x^{\frac {3}{2}}}\, dx \end {gather*}

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)/x**(3/2),x)

[Out]

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

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