3.8.25 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=216 \[ -\frac {6 a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{\sqrt {x} (a+b x)}+\frac {2 b^2 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{a+b x}-\frac {2 a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{3 x^{3/2} (a+b x)}+\frac {2 b^3 B x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}-\frac {2 a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 216, 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 a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{3 x^{3/2} (a+b x)}-\frac {6 a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{\sqrt {x} (a+b x)}+\frac {2 b^2 \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{a+b x}-\frac {2 a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}+\frac {2 b^3 B x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (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^(7/2),x]

[Out]

(-2*a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^(5/2)*(a + b*x)) - (2*a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])/(3*x^(3/2)*(a + b*x)) - (6*a*b*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(Sqrt[x]*(a + b*x)) + (2*b^2*
(A*b + 3*a*B)*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (2*b^3*B*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/(3*(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^{7/2}} \, 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^{7/2}} \, 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^{7/2}}+\frac {a^2 b^3 (3 A b+a B)}{x^{5/2}}+\frac {3 a b^4 (A b+a B)}{x^{3/2}}+\frac {b^5 (A b+3 a B)}{\sqrt {x}}+b^6 B \sqrt {x}\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}}{5 x^{5/2} (a+b x)}-\frac {2 a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac {6 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{\sqrt {x} (a+b x)}+\frac {2 b^2 (A b+3 a B) \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}+\frac {2 b^3 B x^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 83, normalized size = 0.38 \begin {gather*} -\frac {2 \sqrt {(a+b x)^2} \left (a^3 (3 A+5 B x)+15 a^2 b x (A+3 B x)+45 a b^2 x^2 (A-B x)-5 b^3 x^3 (3 A+B x)\right )}{15 x^{5/2} (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^(7/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(45*a*b^2*x^2*(A - B*x) - 5*b^3*x^3*(3*A + B*x) + 15*a^2*b*x*(A + 3*B*x) + a^3*(3*A + 5*
B*x)))/(15*x^(5/2)*(a + b*x))

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IntegrateAlgebraic [A]  time = 14.38, size = 97, normalized size = 0.45 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (-3 a^3 A-5 a^3 B x-15 a^2 A b x-45 a^2 b B x^2-45 a A b^2 x^2+45 a b^2 B x^3+15 A b^3 x^3+5 b^3 B x^4\right )}{15 x^{5/2} (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^(7/2),x]

[Out]

(2*Sqrt[(a + b*x)^2]*(-3*a^3*A - 15*a^2*A*b*x - 5*a^3*B*x - 45*a*A*b^2*x^2 - 45*a^2*b*B*x^2 + 15*A*b^3*x^3 + 4
5*a*b^2*B*x^3 + 5*b^3*B*x^4))/(15*x^(5/2)*(a + b*x))

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fricas [A]  time = 0.41, size = 73, normalized size = 0.34 \begin {gather*} \frac {2 \, {\left (5 \, B b^{3} x^{4} - 3 \, A a^{3} + 15 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} - 45 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} - 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \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^(7/2),x, algorithm="fricas")

[Out]

2/15*(5*B*b^3*x^4 - 3*A*a^3 + 15*(3*B*a*b^2 + A*b^3)*x^3 - 45*(B*a^2*b + A*a*b^2)*x^2 - 5*(B*a^3 + 3*A*a^2*b)*
x)/x^(5/2)

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giac [A]  time = 0.17, size = 124, normalized size = 0.57 \begin {gather*} \frac {2}{3} \, B b^{3} x^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right ) + 6 \, B a b^{2} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) + 2 \, A b^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right ) - \frac {2 \, {\left (45 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 45 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{3} \mathrm {sgn}\left (b x + a\right )\right )}}{15 \, x^{\frac {5}{2}}} \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^(7/2),x, algorithm="giac")

[Out]

2/3*B*b^3*x^(3/2)*sgn(b*x + a) + 6*B*a*b^2*sqrt(x)*sgn(b*x + a) + 2*A*b^3*sqrt(x)*sgn(b*x + a) - 2/15*(45*B*a^
2*b*x^2*sgn(b*x + a) + 45*A*a*b^2*x^2*sgn(b*x + a) + 5*B*a^3*x*sgn(b*x + a) + 15*A*a^2*b*x*sgn(b*x + a) + 3*A*
a^3*sgn(b*x + a))/x^(5/2)

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maple [A]  time = 0.07, size = 92, normalized size = 0.43 \begin {gather*} -\frac {2 \left (-5 B \,b^{3} x^{4}-15 A \,b^{3} x^{3}-45 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 B \,a^{2} b \,x^{2}+15 A \,a^{2} b x +5 B \,a^{3} x +3 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (b x +a \right )^{3} x^{\frac {5}{2}}} \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^(7/2),x)

[Out]

-2/15*(-5*B*b^3*x^4-15*A*b^3*x^3-45*B*a*b^2*x^3+45*A*a*b^2*x^2+45*B*a^2*b*x^2+15*A*a^2*b*x+5*B*a^3*x+3*A*a^3)*
((b*x+a)^2)^(3/2)/x^(5/2)/(b*x+a)^3

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maxima [A]  time = 0.65, size = 131, normalized size = 0.61 \begin {gather*} \frac {2}{3} \, B {\left (\frac {b^{3} x^{2} + 3 \, a b^{2} x}{\sqrt {x}} + \frac {6 \, {\left (a b^{2} x^{2} - a^{2} b x\right )}}{x^{\frac {3}{2}}} - \frac {3 \, a^{2} b x^{2} + a^{3} x}{x^{\frac {5}{2}}}\right )} + \frac {2}{15} \, A {\left (\frac {15 \, {\left (b^{3} x^{2} - a b^{2} x\right )}}{x^{\frac {3}{2}}} - \frac {10 \, {\left (3 \, a b^{2} x^{2} + a^{2} b x\right )}}{x^{\frac {5}{2}}} - \frac {5 \, a^{2} b x^{2} + 3 \, a^{3} x}{x^{\frac {7}{2}}}\right )} \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^(7/2),x, algorithm="maxima")

[Out]

2/3*B*((b^3*x^2 + 3*a*b^2*x)/sqrt(x) + 6*(a*b^2*x^2 - a^2*b*x)/x^(3/2) - (3*a^2*b*x^2 + a^3*x)/x^(5/2)) + 2/15
*A*(15*(b^3*x^2 - a*b^2*x)/x^(3/2) - 10*(3*a*b^2*x^2 + a^2*b*x)/x^(5/2) - (5*a^2*b*x^2 + 3*a^3*x)/x^(7/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{x^{7/2}} \,d x \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^(7/2),x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/x^(7/2), x)

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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 {7}{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**(7/2),x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**(7/2), x)

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