3.7.81 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^{11}} \, dx\) [681]

Optimal. Leaf size=210 \[ -\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {3 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)} \]

[Out]

-1/10*a^3*A*((b*x+a)^2)^(1/2)/x^10/(b*x+a)-1/9*a^2*(3*A*b+B*a)*((b*x+a)^2)^(1/2)/x^9/(b*x+a)-3/8*a*b*(A*b+B*a)
*((b*x+a)^2)^(1/2)/x^8/(b*x+a)-1/7*b^2*(A*b+3*B*a)*((b*x+a)^2)^(1/2)/x^7/(b*x+a)-1/6*b^3*B*((b*x+a)^2)^(1/2)/x
^6/(b*x+a)

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Rubi [A]
time = 0.06, 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)}{9 x^9 (a+b x)}-\frac {3 a b \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{8 x^8 (a+b x)}-\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{7 x^7 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (a+b x)}-\frac {a^3 A \sqrt {a^2+2 a b x+b^2 x^2}}{10 x^{10} (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^11,x]

[Out]

-1/10*(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(x^10*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/(9*x^9*(a + b*x)) - (3*a*b*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*x^8*(a + b*x)) - (b^2*(A*b + 3*a*B
)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*x^7*(a + b*x)) - (b^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*x^6*(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^{11}} \, 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^{11}} \, 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^{11}}+\frac {a^2 b^3 (3 A b+a B)}{x^{10}}+\frac {3 a b^4 (A b+a B)}{x^9}+\frac {b^5 (A b+3 a B)}{x^8}+\frac {b^6 B}{x^7}\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}}{10 x^{10} (a+b x)}-\frac {a^2 (3 A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{9 x^9 (a+b x)}-\frac {3 a b (A b+a B) \sqrt {a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac {b^2 (A b+3 a B) \sqrt {a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{6 x^6 (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 (60 b^3 x^3 (6 A+7 B x)+135 a b^2 x^2 (7 A+8 B x)+105 a^2 b x (8 A+9 B x)+28 a^3 (9 A+10 B x)\right )}{2520 x^{10} (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^11,x]

[Out]

-1/2520*(Sqrt[(a + b*x)^2]*(60*b^3*x^3*(6*A + 7*B*x) + 135*a*b^2*x^2*(7*A + 8*B*x) + 105*a^2*b*x*(8*A + 9*B*x)
+ 28*a^3*(9*A + 10*B*x)))/(x^10*(a + b*x))

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Maple [A]
time = 0.69, size = 92, normalized size = 0.44

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {B \,b^{3} x^{4}}{6}+\left (-\frac {1}{7} A \,b^{3}-\frac {3}{7} B a \,b^{2}\right ) x^{3}+\left (-\frac {3}{8} A a \,b^{2}-\frac {3}{8} B \,a^{2} b \right ) x^{2}+\left (-\frac {1}{3} A \,a^{2} b -\frac {1}{9} B \,a^{3}\right ) x -\frac {A \,a^{3}}{10}\right )}{\left (b x +a \right ) x^{10}}\)	\(90\)
gosper	\(-\frac {\left (420 B \,b^{3} x^{4}+360 A \,b^{3} x^{3}+1080 B a \,b^{2} x^{3}+945 A a \,b^{2} x^{2}+945 a^{2} b B \,x^{2}+840 A \,a^{2} b x +280 B \,a^{3} x +252 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 x^{10} \left (b x +a \right )^{3}}\)	\(92\)
default	\(-\frac {\left (420 B \,b^{3} x^{4}+360 A \,b^{3} x^{3}+1080 B a \,b^{2} x^{3}+945 A a \,b^{2} x^{2}+945 a^{2} b B \,x^{2}+840 A \,a^{2} b x +280 B \,a^{3} x +252 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{2520 x^{10} \left (b x +a \right )^{3}}\)	\(92\)

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^11,x,method=_RETURNVERBOSE)

[Out]

-1/2520*(420*B*b^3*x^4+360*A*b^3*x^3+1080*B*a*b^2*x^3+945*A*a*b^2*x^2+945*B*a^2*b*x^2+840*A*a^2*b*x+280*B*a^3*
x+252*A*a^3)*((b*x+a)^2)^(3/2)/x^10/(b*x+a)^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 615 vs. \(2 (145) = 290\).
time = 0.29, size = 615, normalized size = 2.93 \begin {gather*} -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{9}}{4 \, a^{9}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{10}}{4 \, a^{10}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{8}}{4 \, a^{8} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{9}}{4 \, a^{9} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{7}}{4 \, a^{9} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{8}}{4 \, a^{10} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{6}}{4 \, a^{8} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{7}}{4 \, a^{9} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{4 \, a^{7} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{4 \, a^{8} x^{4}} - \frac {125 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{504 \, a^{6} x^{5}} + \frac {209 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{840 \, a^{7} x^{5}} + \frac {121 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{504 \, a^{5} x^{6}} - \frac {41 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{168 \, a^{6} x^{6}} - \frac {37 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{168 \, a^{4} x^{7}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{56 \, a^{5} x^{7}} + \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{72 \, a^{3} x^{8}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{24 \, a^{4} x^{8}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B}{9 \, a^{2} x^{9}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b}{6 \, a^{3} x^{9}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A}{10 \, a^{2} x^{10}} \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^11,x, algorithm="maxima")

[Out]

-1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^9/a^9 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^10/a^10 - 1/4*(b^2*x^
2 + 2*a*b*x + a^2)^(3/2)*B*b^8/(a^8*x) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^9/(a^9*x) + 1/4*(b^2*x^2 + 2*
a*b*x + a^2)^(5/2)*B*b^7/(a^9*x^2) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^8/(a^10*x^2) - 1/4*(b^2*x^2 + 2*a
*b*x + a^2)^(5/2)*B*b^6/(a^8*x^3) + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^7/(a^9*x^3) + 1/4*(b^2*x^2 + 2*a*b
*x + a^2)^(5/2)*B*b^5/(a^7*x^4) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^6/(a^8*x^4) - 125/504*(b^2*x^2 + 2*a
*b*x + a^2)^(5/2)*B*b^4/(a^6*x^5) + 209/840*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/(a^7*x^5) + 121/504*(b^2*x^2
+ 2*a*b*x + a^2)^(5/2)*B*b^3/(a^5*x^6) - 41/168*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^6*x^6) - 37/168*(b^2
*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^2/(a^4*x^7) + 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/(a^5*x^7) + 13/72*(b
^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a^3*x^8) - 5/24*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^4*x^8) - 1/9*(b^2*
x^2 + 2*a*b*x + a^2)^(5/2)*B/(a^2*x^9) + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b/(a^3*x^9) - 1/10*(b^2*x^2 + 2
*a*b*x + a^2)^(5/2)*A/(a^2*x^10)

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Fricas [A]
time = 2.54, size = 73, normalized size = 0.35 \begin {gather*} -\frac {420 \, B b^{3} x^{4} + 252 \, A a^{3} + 360 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 945 \, {\left (B a^{2} b + A a b^{2}\right )} x^{2} + 280 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x}{2520 \, x^{10}} \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^11,x, algorithm="fricas")

[Out]

-1/2520*(420*B*b^3*x^4 + 252*A*a^3 + 360*(3*B*a*b^2 + A*b^3)*x^3 + 945*(B*a^2*b + A*a*b^2)*x^2 + 280*(B*a^3 +
3*A*a^2*b)*x)/x^10

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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^{11}}\, 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**11,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**11, x)

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Giac [A]
time = 1.76, size = 149, normalized size = 0.71 \begin {gather*} \frac {{\left (5 \, B a b^{9} - 3 \, A b^{10}\right )} \mathrm {sgn}\left (b x + a\right )}{2520 \, a^{7}} - \frac {420 \, B b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 1080 \, B a b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 360 \, A b^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + 945 \, B a^{2} b x^{2} \mathrm {sgn}\left (b x + a\right ) + 945 \, A a b^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + 280 \, B a^{3} x \mathrm {sgn}\left (b x + a\right ) + 840 \, A a^{2} b x \mathrm {sgn}\left (b x + a\right ) + 252 \, A a^{3} \mathrm {sgn}\left (b x + a\right )}{2520 \, x^{10}} \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^11,x, algorithm="giac")

[Out]

1/2520*(5*B*a*b^9 - 3*A*b^10)*sgn(b*x + a)/a^7 - 1/2520*(420*B*b^3*x^4*sgn(b*x + a) + 1080*B*a*b^2*x^3*sgn(b*x
+ a) + 360*A*b^3*x^3*sgn(b*x + a) + 945*B*a^2*b*x^2*sgn(b*x + a) + 945*A*a*b^2*x^2*sgn(b*x + a) + 280*B*a^3*x
*sgn(b*x + a) + 840*A*a^2*b*x*sgn(b*x + a) + 252*A*a^3*sgn(b*x + a))/x^10

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Mupad [B]
time = 1.18, size = 196, normalized size = 0.93 \begin {gather*} -\frac {\left (\frac {B\,a^3}{9}+\frac {A\,b\,a^2}{3}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^9\,\left (a+b\,x\right )}-\frac {\left (\frac {A\,b^3}{7}+\frac {3\,B\,a\,b^2}{7}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{x^7\,\left (a+b\,x\right )}-\frac {A\,a^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{10\,x^{10}\,\left (a+b\,x\right )}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,x^6\,\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}}{8\,x^8\,\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^11,x)

[Out]

- (((B*a^3)/9 + (A*a^2*b)/3)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^9*(a + b*x)) - (((A*b^3)/7 + (3*B*a*b^2)/7)*(
a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(x^7*(a + b*x)) - (A*a^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(10*x^10*(a + b*x))
- (B*b^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*x^6*(a + b*x)) - (3*a*b*(A*b + B*a)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/
2))/(8*x^8*(a + b*x))

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