3.693 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=297 \[ -\frac{a^4 \sqrt{a^2+2 a b x+b^2 x^2} (a B+5 A b)}{2 x^2 (a+b x)}-\frac{5 a^3 b \sqrt{a^2+2 a b x+b^2 x^2} (a B+2 A b)}{x (a+b x)}+\frac{5 a b^3 x \sqrt{a^2+2 a b x+b^2 x^2} (2 a B+A b)}{a+b x}+\frac{b^4 x^2 \sqrt{a^2+2 a b x+b^2 x^2} (5 a B+A b)}{2 (a+b x)}+\frac{10 a^2 b^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac{a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}+\frac{b^5 B x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.122422, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {770, 76} \[ -\frac{a^4 \sqrt{a^2+2 a b x+b^2 x^2} (a B+5 A b)}{2 x^2 (a+b x)}-\frac{5 a^3 b \sqrt{a^2+2 a b x+b^2 x^2} (a B+2 A b)}{x (a+b x)}+\frac{5 a b^3 x \sqrt{a^2+2 a b x+b^2 x^2} (2 a B+A b)}{a+b x}+\frac{b^4 x^2 \sqrt{a^2+2 a b x+b^2 x^2} (5 a B+A b)}{2 (a+b x)}+\frac{10 a^2 b^2 \log (x) \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{a+b x}-\frac{a^5 A \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}+\frac{b^5 B x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^4,x]

[Out]

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

Mathematica [A]  time = 0.0553524, size = 127, normalized size = 0.43 \[ \frac{\sqrt{(a+b x)^2} \left (60 a^2 b^2 x^3 \log (x) (a B+A b)-60 a^3 A b^2 x^2-15 a^4 b x (A+2 B x)+a^5 (-(2 A+3 B x))+60 a^2 b^3 B x^4+15 a b^4 x^4 (2 A+B x)+b^5 x^5 (3 A+2 B x)\right )}{6 x^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^4,x]

[Out]

(Sqrt[(a + b*x)^2]*(-60*a^3*A*b^2*x^2 + 60*a^2*b^3*B*x^4 + 15*a*b^4*x^4*(2*A + B*x) - 15*a^4*b*x*(A + 2*B*x) +
b^5*x^5*(3*A + 2*B*x) - a^5*(2*A + 3*B*x) + 60*a^2*b^2*(A*b + a*B)*x^3*Log[x]))/(6*x^3*(a + b*x))

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Maple [A]  time = 0.013, size = 144, normalized size = 0.5 \begin{align*}{\frac{2\,B{b}^{5}{x}^{6}+3\,A{x}^{5}{b}^{5}+15\,B{x}^{5}a{b}^{4}+60\,A\ln \left ( x \right ){x}^{3}{a}^{2}{b}^{3}+30\,A{x}^{4}a{b}^{4}+60\,B\ln \left ( x \right ){x}^{3}{a}^{3}{b}^{2}+60\,B{x}^{4}{a}^{2}{b}^{3}-60\,A{x}^{2}{a}^{3}{b}^{2}-30\,B{x}^{2}{a}^{4}b-15\,A{a}^{4}bx-3\,B{a}^{5}x-2\,A{a}^{5}}{6\, \left ( bx+a \right ) ^{5}{x}^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\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^4,x)

[Out]

1/6*((b*x+a)^2)^(5/2)*(2*B*b^5*x^6+3*A*x^5*b^5+15*B*x^5*a*b^4+60*A*ln(x)*x^3*a^2*b^3+30*A*x^4*a*b^4+60*B*ln(x)
*x^3*a^3*b^2+60*B*x^4*a^2*b^3-60*A*x^2*a^3*b^2-30*B*x^2*a^4*b-15*A*a^4*b*x-3*B*a^5*x-2*A*a^5)/(b*x+a)^5/x^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.30604, size = 261, normalized size = 0.88 \begin{align*} \frac{2 \, B b^{5} x^{6} - 2 \, A a^{5} + 3 \,{\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} + 30 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} + 60 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} \log \left (x\right ) - 30 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 3 \,{\left (B a^{5} + 5 \, A a^{4} b\right )} x}{6 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

1/6*(2*B*b^5*x^6 - 2*A*a^5 + 3*(5*B*a*b^4 + A*b^5)*x^5 + 30*(2*B*a^2*b^3 + A*a*b^4)*x^4 + 60*(B*a^3*b^2 + A*a^
2*b^3)*x^3*log(x) - 30*(B*a^4*b + 2*A*a^3*b^2)*x^2 - 3*(B*a^5 + 5*A*a^4*b)*x)/x^3

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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^{4}}\, 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**4,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**4, x)

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Giac [A]  time = 1.18592, size = 257, normalized size = 0.87 \begin{align*} \frac{1}{3} \, B b^{5} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{5}{2} \, B a b^{4} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, A b^{5} x^{2} \mathrm{sgn}\left (b x + a\right ) + 10 \, B a^{2} b^{3} x \mathrm{sgn}\left (b x + a\right ) + 5 \, A a b^{4} x \mathrm{sgn}\left (b x + a\right ) + 10 \,{\left (B a^{3} b^{2} \mathrm{sgn}\left (b x + a\right ) + A a^{2} b^{3} \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac{2 \, A a^{5} \mathrm{sgn}\left (b x + a\right ) + 30 \,{\left (B a^{4} b \mathrm{sgn}\left (b x + a\right ) + 2 \, A a^{3} b^{2} \mathrm{sgn}\left (b x + a\right )\right )} x^{2} + 3 \,{\left (B a^{5} \mathrm{sgn}\left (b x + a\right ) + 5 \, A a^{4} b \mathrm{sgn}\left (b x + a\right )\right )} x}{6 \, x^{3}} \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^4,x, algorithm="giac")

[Out]

1/3*B*b^5*x^3*sgn(b*x + a) + 5/2*B*a*b^4*x^2*sgn(b*x + a) + 1/2*A*b^5*x^2*sgn(b*x + a) + 10*B*a^2*b^3*x*sgn(b*
x + a) + 5*A*a*b^4*x*sgn(b*x + a) + 10*(B*a^3*b^2*sgn(b*x + a) + A*a^2*b^3*sgn(b*x + a))*log(abs(x)) - 1/6*(2*
A*a^5*sgn(b*x + a) + 30*(B*a^4*b*sgn(b*x + a) + 2*A*a^3*b^2*sgn(b*x + a))*x^2 + 3*(B*a^5*sgn(b*x + a) + 5*A*a^
4*b*sgn(b*x + a))*x)/x^3