3.807 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^{7/2}} \, dx\)

Optimal. Leaf size=316 \[ -\frac{2 a^4 \sqrt{a^2+2 a b x+b^2 x^2} (a B+5 A b)}{3 x^{3/2} (a+b x)}-\frac{10 a^3 b \sqrt{a^2+2 a b x+b^2 x^2} (a B+2 A b)}{\sqrt{x} (a+b x)}+\frac{20 a^2 b^2 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{10 a b^3 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (2 a B+A b)}{3 (a+b x)}+\frac{2 b^4 x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} (5 a B+A b)}{5 (a+b x)}-\frac{2 a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}+\frac{2 b^5 B x^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 (a+b x)} \]

[Out]

(-2*a^5*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^(5/2)*(a + b*x)) - (2*a^4*(5*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])/(3*x^(3/2)*(a + b*x)) - (10*a^3*b*(2*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(Sqrt[x]*(a + b*x)) + (2
0*a^2*b^2*(A*b + a*B)*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (10*a*b^3*(A*b + 2*a*B)*x^(3/2)*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (2*b^4*(A*b + 5*a*B)*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a +
b*x)) + (2*b^5*B*x^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(a + b*x))

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Rubi [A]  time = 0.119323, antiderivative size = 316, normalized size of antiderivative = 1., 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} \[ -\frac{2 a^4 \sqrt{a^2+2 a b x+b^2 x^2} (a B+5 A b)}{3 x^{3/2} (a+b x)}-\frac{10 a^3 b \sqrt{a^2+2 a b x+b^2 x^2} (a B+2 A b)}{\sqrt{x} (a+b x)}+\frac{20 a^2 b^2 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}+\frac{10 a b^3 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (2 a B+A b)}{3 (a+b x)}+\frac{2 b^4 x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} (5 a B+A b)}{5 (a+b x)}-\frac{2 a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}+\frac{2 b^5 B x^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^(7/2),x]

[Out]

(-2*a^5*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*x^(5/2)*(a + b*x)) - (2*a^4*(5*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])/(3*x^(3/2)*(a + b*x)) - (10*a^3*b*(2*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(Sqrt[x]*(a + b*x)) + (2
0*a^2*b^2*(A*b + a*B)*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(a + b*x) + (10*a*b^3*(A*b + 2*a*B)*x^(3/2)*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/(3*(a + b*x)) + (2*b^4*(A*b + 5*a*B)*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*(a +
b*x)) + (2*b^5*B*x^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(a + b*x))

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]

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])

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{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 )^5 (A+B x)}{x^{7/2}} \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (\frac{a^5 A b^5}{x^{7/2}}+\frac{a^4 b^5 (5 A b+a B)}{x^{5/2}}+\frac{5 a^3 b^6 (2 A b+a B)}{x^{3/2}}+\frac{10 a^2 b^7 (A b+a B)}{\sqrt{x}}+5 a b^8 (A b+2 a B) \sqrt{x}+b^9 (A b+5 a B) x^{3/2}+b^{10} B x^{5/2}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=-\frac{2 a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^{5/2} (a+b x)}-\frac{2 a^4 (5 A b+a B) \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}-\frac{10 a^3 b (2 A b+a B) \sqrt{a^2+2 a b x+b^2 x^2}}{\sqrt{x} (a+b x)}+\frac{20 a^2 b^2 (A b+a B) \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x}+\frac{10 a b^3 (A b+2 a B) x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac{2 b^4 (A b+5 a B) x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac{2 b^5 B x^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0473069, size = 122, normalized size = 0.39 \[ -\frac{2 \sqrt{(a+b x)^2} \left (1050 a^3 b^2 x^2 (A-B x)-350 a^2 b^3 x^3 (3 A+B x)+175 a^4 b x (A+3 B x)+7 a^5 (3 A+5 B x)-35 a b^4 x^4 (5 A+3 B x)-3 b^5 x^5 (7 A+5 B x)\right )}{105 x^{5/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^(7/2),x]

[Out]

(-2*Sqrt[(a + b*x)^2]*(1050*a^3*b^2*x^2*(A - B*x) - 350*a^2*b^3*x^3*(3*A + B*x) + 175*a^4*b*x*(A + 3*B*x) - 35
*a*b^4*x^4*(5*A + 3*B*x) + 7*a^5*(3*A + 5*B*x) - 3*b^5*x^5*(7*A + 5*B*x)))/(105*x^(5/2)*(a + b*x))

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Maple [A]  time = 0.006, size = 140, normalized size = 0.4 \begin{align*} -{\frac{-30\,B{b}^{5}{x}^{6}-42\,A{x}^{5}{b}^{5}-210\,B{x}^{5}a{b}^{4}-350\,A{x}^{4}a{b}^{4}-700\,B{x}^{4}{a}^{2}{b}^{3}-2100\,A{x}^{3}{a}^{2}{b}^{3}-2100\,B{x}^{3}{a}^{3}{b}^{2}+2100\,A{x}^{2}{a}^{3}{b}^{2}+1050\,B{x}^{2}{a}^{4}b+350\,A{a}^{4}bx+70\,B{a}^{5}x+42\,A{a}^{5}}{105\, \left ( bx+a \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}{x}^{-{\frac{5}{2}}}} \end{align*}

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)^(5/2)/x^(7/2),x)

[Out]

-2/105*(-15*B*b^5*x^6-21*A*b^5*x^5-105*B*a*b^4*x^5-175*A*a*b^4*x^4-350*B*a^2*b^3*x^4-1050*A*a^2*b^3*x^3-1050*B
*a^3*b^2*x^3+1050*A*a^3*b^2*x^2+525*B*a^4*b*x^2+175*A*a^4*b*x+35*B*a^5*x+21*A*a^5)*((b*x+a)^2)^(5/2)/x^(5/2)/(
b*x+a)^5

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Maxima [A]  time = 1.05934, size = 316, normalized size = 1. \begin{align*} \frac{2}{15} \,{\left ({\left (3 \, b^{5} x^{2} + 5 \, a b^{4} x\right )} \sqrt{x} + \frac{20 \,{\left (a b^{4} x^{2} + 3 \, a^{2} b^{3} x\right )}}{\sqrt{x}} + \frac{90 \,{\left (a^{2} b^{3} x^{2} - a^{3} b^{2} x\right )}}{x^{\frac{3}{2}}} - \frac{20 \,{\left (3 \, a^{3} b^{2} x^{2} + a^{4} b x\right )}}{x^{\frac{5}{2}}} - \frac{5 \, a^{4} b x^{2} + 3 \, a^{5} x}{x^{\frac{7}{2}}}\right )} A + \frac{2}{105} \,{\left (3 \,{\left (5 \, b^{5} x^{2} + 7 \, a b^{4} x\right )} x^{\frac{3}{2}} + 28 \,{\left (3 \, a b^{4} x^{2} + 5 \, a^{2} b^{3} x\right )} \sqrt{x} + \frac{210 \,{\left (a^{2} b^{3} x^{2} + 3 \, a^{3} b^{2} x\right )}}{\sqrt{x}} + \frac{420 \,{\left (a^{3} b^{2} x^{2} - a^{4} b x\right )}}{x^{\frac{3}{2}}} - \frac{35 \,{\left (3 \, a^{4} b x^{2} + a^{5} x\right )}}{x^{\frac{5}{2}}}\right )} B \end{align*}

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)^(5/2)/x^(7/2),x, algorithm="maxima")

[Out]

2/15*((3*b^5*x^2 + 5*a*b^4*x)*sqrt(x) + 20*(a*b^4*x^2 + 3*a^2*b^3*x)/sqrt(x) + 90*(a^2*b^3*x^2 - a^3*b^2*x)/x^
(3/2) - 20*(3*a^3*b^2*x^2 + a^4*b*x)/x^(5/2) - (5*a^4*b*x^2 + 3*a^5*x)/x^(7/2))*A + 2/105*(3*(5*b^5*x^2 + 7*a*
b^4*x)*x^(3/2) + 28*(3*a*b^4*x^2 + 5*a^2*b^3*x)*sqrt(x) + 210*(a^2*b^3*x^2 + 3*a^3*b^2*x)/sqrt(x) + 420*(a^3*b
^2*x^2 - a^4*b*x)/x^(3/2) - 35*(3*a^4*b*x^2 + a^5*x)/x^(5/2))*B

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Fricas [A]  time = 1.25527, size = 270, normalized size = 0.85 \begin{align*} \frac{2 \,{\left (15 \, B b^{5} x^{6} - 21 \, A a^{5} + 21 \,{\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 175 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 1050 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 525 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 35 \,{\left (B a^{5} + 5 \, A a^{4} b\right )} x\right )}}{105 \, x^{\frac{5}{2}}} \end{align*}

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)^(5/2)/x^(7/2),x, algorithm="fricas")

[Out]

2/105*(15*B*b^5*x^6 - 21*A*a^5 + 21*(5*B*a*b^4 + A*b^5)*x^5 + 175*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 1050*(B*a^3*b^
2 + A*a^2*b^3)*x^3 - 525*(B*a^4*b + 2*A*a^3*b^2)*x^2 - 35*(B*a^5 + 5*A*a^4*b)*x)/x^(5/2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}{x^{\frac{7}{2}}}\, dx \end{align*}

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)**(5/2)/x**(7/2),x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**(7/2), x)

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Giac [A]  time = 1.18535, size = 265, normalized size = 0.84 \begin{align*} \frac{2}{7} \, B b^{5} x^{\frac{7}{2}} \mathrm{sgn}\left (b x + a\right ) + 2 \, B a b^{4} x^{\frac{5}{2}} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{5} \, A b^{5} x^{\frac{5}{2}} \mathrm{sgn}\left (b x + a\right ) + \frac{20}{3} \, B a^{2} b^{3} x^{\frac{3}{2}} \mathrm{sgn}\left (b x + a\right ) + \frac{10}{3} \, A a b^{4} x^{\frac{3}{2}} \mathrm{sgn}\left (b x + a\right ) + 20 \, B a^{3} b^{2} \sqrt{x} \mathrm{sgn}\left (b x + a\right ) + 20 \, A a^{2} b^{3} \sqrt{x} \mathrm{sgn}\left (b x + a\right ) - \frac{2 \,{\left (75 \, B a^{4} b x^{2} \mathrm{sgn}\left (b x + a\right ) + 150 \, A a^{3} b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, B a^{5} x \mathrm{sgn}\left (b x + a\right ) + 25 \, A a^{4} b x \mathrm{sgn}\left (b x + a\right ) + 3 \, A a^{5} \mathrm{sgn}\left (b x + a\right )\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

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)^(5/2)/x^(7/2),x, algorithm="giac")

[Out]

2/7*B*b^5*x^(7/2)*sgn(b*x + a) + 2*B*a*b^4*x^(5/2)*sgn(b*x + a) + 2/5*A*b^5*x^(5/2)*sgn(b*x + a) + 20/3*B*a^2*
b^3*x^(3/2)*sgn(b*x + a) + 10/3*A*a*b^4*x^(3/2)*sgn(b*x + a) + 20*B*a^3*b^2*sqrt(x)*sgn(b*x + a) + 20*A*a^2*b^
3*sqrt(x)*sgn(b*x + a) - 2/15*(75*B*a^4*b*x^2*sgn(b*x + a) + 150*A*a^3*b^2*x^2*sgn(b*x + a) + 5*B*a^5*x*sgn(b*
x + a) + 25*A*a^4*b*x*sgn(b*x + a) + 3*A*a^5*sgn(b*x + a))/x^(5/2)