3.7.77 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^7} \, dx\) [677]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 210, normalized size of antiderivative = 1.00, 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 = {784, 77} \begin {gather*} -\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 {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{2 x^2 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (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,x]

[Out]

-1/6*(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(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)*
Sqrt[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 77

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 784

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} \, 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*}

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Mathematica [A]
time = 0.02, size = 87, normalized size = 0.41 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (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)\right )}{60 x^6 (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,x]

[Out]

-1/60*(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)))/(x^6*(a + b*x))

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Maple [A]
time = 0.64, size = 92, normalized size = 0.44

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{2}+\left (-\frac {1}{3} A \,b^{3}-B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{4} A a \,b^{2}-\frac {3}{4} B \,a^{2} b \right ) x^{2}+\left (-\frac {3}{5} A \,a^{2} b -\frac {1}{5} B \,a^{3}\right ) x -\frac {A \,a^{3}}{6}\right )}{\left (b x +a \right ) x^{6}}\)	\(90\)
gosper	\(-\frac {\left (30 B \,b^{3} x^{4}+20 A \,b^{3} x^{3}+60 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 a^{2} b B \,x^{2}+36 A \,a^{2} b x +12 B \,a^{3} x +10 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 x^{6} \left (b x +a \right )^{3}}\)	\(92\)
default	\(-\frac {\left (30 B \,b^{3} x^{4}+20 A \,b^{3} x^{3}+60 B a \,b^{2} x^{3}+45 A a \,b^{2} x^{2}+45 a^{2} b B \,x^{2}+36 A \,a^{2} b x +12 B \,a^{3} x +10 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 x^{6} \left (b x +a \right )^{3}}\)	\(92\)

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,method=_RETURNVERBOSE)

[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 [B] Leaf count of result is larger than twice the leaf count of optimal. 375 vs. \(2 (145) = 290\).
time = 0.29, size = 375, normalized size = 1.79 \begin {gather*} -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{5}}{4 \, a^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{6}}{4 \, a^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{4}}{4 \, a^{4} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{5}}{4 \, a^{5} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{4 \, a^{5} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{4 \, a^{6} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{4 \, a^{4} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{4 \, a^{5} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{4 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{4 \, a^{4} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{5 \, a^{2} x^{5}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{30 \, a^{3} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{6 \, a^{2} x^{6}} \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,x, algorithm="maxima")

[Out]

-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^5/a^5 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^6/a^6 - 1/4*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*B*b^4/(a^4*x) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^5/(a^5*x) + 1/4*(b^2*x^2 + 2*a*
b*x + a^2)^(5/2)*B*b^3/(a^5*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^2) - 1/4*(b^2*x^2 + 2*a*b*
x + a^2)^(5/2)*B*b^2/(a^4*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^3) + 1/4*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*B*b/(a^3*x^4) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^4) - 1/5*(b^2*x^2 + 2*a*b*x + a^
2)^(5/2)*B/(a^2*x^5) + 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^5) - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2
)*A/(a^2*x^6)

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Fricas [A]
time = 2.46, size = 73, normalized size = 0.35 \begin {gather*} -\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 {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,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 = 0.00, 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^{7}}\, 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,x)

[Out]

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

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Giac [A]
time = 1.06, size = 149, normalized size = 0.71 \begin {gather*} \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 {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,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

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Mupad [B]
time = 1.18, size = 195, normalized size = 0.93 \begin {gather*} -\frac {\left (\frac {B\,a^3}{5}+\frac {3\,A\,b\,a^2}{5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^5\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{3}+B\,a\,b^2\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^3\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,x^2\,\left (a+b\,x\right )}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,x^4\,\left (a+b\,x\right )} \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,x)

[Out]

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

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