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

Optimal. Leaf size=210 \[ -\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 x^5 (a+b x)}-\frac{3 a b \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{4 x^4 (a+b x)}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{3 x^3 (a+b x)}-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac{b^3 B \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0789072, antiderivative size = 210, 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} \[ -\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{5 x^5 (a+b x)}-\frac{3 a b \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{4 x^4 (a+b x)}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{3 x^3 (a+b x)}-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac{b^3 B \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^7,x]

[Out]

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

Mathematica [A]  time = 0.0279849, size = 87, normalized size = 0.41 \[ -\frac{\sqrt{(a+b x)^2} \left (9 a^2 b x (4 A+5 B x)+2 a^3 (5 A+6 B x)+15 a b^2 x^2 (3 A+4 B x)+10 b^3 x^3 (2 A+3 B x)\right )}{60 x^6 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^7,x]

[Out]

-(Sqrt[(a + b*x)^2]*(10*b^3*x^3*(2*A + 3*B*x) + 15*a*b^2*x^2*(3*A + 4*B*x) + 9*a^2*b*x*(4*A + 5*B*x) + 2*a^3*(
5*A + 6*B*x)))/(60*x^6*(a + b*x))

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Maple [A]  time = 0.005, size = 92, normalized size = 0.4 \begin{align*} -{\frac{30\,B{x}^{4}{b}^{3}+20\,A{b}^{3}{x}^{3}+60\,B{x}^{3}a{b}^{2}+45\,A{x}^{2}a{b}^{2}+45\,B{x}^{2}{a}^{2}b+36\,A{a}^{2}bx+12\,{a}^{3}Bx+10\,A{a}^{3}}{60\,{x}^{6} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{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)^(3/2)/x^7,x)

[Out]

-1/60*(30*B*b^3*x^4+20*A*b^3*x^3+60*B*a*b^2*x^3+45*A*a*b^2*x^2+45*B*a^2*b*x^2+36*A*a^2*b*x+12*B*a^3*x+10*A*a^3
)*((b*x+a)^2)^(3/2)/x^6/(b*x+a)^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)^(3/2)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32112, size = 165, normalized size = 0.79 \begin{align*} -\frac{30 \, B b^{3} x^{4} + 10 \, A a^{3} + 20 \,{\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} + 12 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x}{60 \, x^{6}} \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)^(3/2)/x^7,x, algorithm="fricas")

[Out]

-1/60*(30*B*b^3*x^4 + 10*A*a^3 + 20*(3*B*a*b^2 + A*b^3)*x^3 + 45*(B*a^2*b + A*a*b^2)*x^2 + 12*(B*a^3 + 3*A*a^2
*b)*x)/x^6

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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{3}{2}}}{x^{7}}\, 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)**(3/2)/x**7,x)

[Out]

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

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Giac [A]  time = 1.32415, size = 201, normalized size = 0.96 \begin{align*} \frac{{\left (3 \, B a b^{5} - A b^{6}\right )} \mathrm{sgn}\left (b x + a\right )}{60 \, a^{3}} - \frac{30 \, B b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + 60 \, B a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 20 \, A b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 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 ) + 12 \, B a^{3} x \mathrm{sgn}\left (b x + a\right ) + 36 \, A a^{2} b x \mathrm{sgn}\left (b x + a\right ) + 10 \, A a^{3} \mathrm{sgn}\left (b x + a\right )}{60 \, x^{6}} \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)^(3/2)/x^7,x, algorithm="giac")

[Out]

1/60*(3*B*a*b^5 - A*b^6)*sgn(b*x + a)/a^3 - 1/60*(30*B*b^3*x^4*sgn(b*x + a) + 60*B*a*b^2*x^3*sgn(b*x + a) + 20
*A*b^3*x^3*sgn(b*x + a) + 45*B*a^2*b*x^2*sgn(b*x + a) + 45*A*a*b^2*x^2*sgn(b*x + a) + 12*B*a^3*x*sgn(b*x + a)
+ 36*A*a^2*b*x*sgn(b*x + a) + 10*A*a^3*sgn(b*x + a))/x^6