3.7.22 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=293 \[ -\frac {5 a^2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{x^2 (a+b x)}+\frac {b^4 \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (5 a B+A b)}{a+b x}-\frac {5 a b^3 \sqrt {a^2+2 a b x+b^2 x^2} (2 a B+A b)}{x (a+b x)}+\frac {b^5 B x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {a^4 \sqrt {a^2+2 a b x+b^2 x^2} (a B+5 A b)}{4 x^4 (a+b x)}-\frac {5 a^3 b \sqrt {a^2+2 a b x+b^2 x^2} (a B+2 A b)}{3 x^3 (a+b x)} \]

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Rubi [A]  time = 0.12, antiderivative size = 293, 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 = {770, 76} \begin {gather*} -\frac {a^4 \sqrt {a^2+2 a b x+b^2 x^2} (a B+5 A b)}{4 x^4 (a+b x)}-\frac {5 a^3 b \sqrt {a^2+2 a b x+b^2 x^2} (a B+2 A b)}{3 x^3 (a+b x)}-\frac {5 a^2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{x^2 (a+b x)}-\frac {5 a b^3 \sqrt {a^2+2 a b x+b^2 x^2} (2 a B+A b)}{x (a+b x)}+\frac {b^4 \log (x) \sqrt {a^2+2 a b x+b^2 x^2} (5 a B+A b)}{a+b x}-\frac {a^5 A \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}+\frac {b^5 B x \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^6,x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 127, normalized size = 0.43 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (3 a^5 (4 A+5 B x)+25 a^4 b x (3 A+4 B x)+100 a^3 b^2 x^2 (2 A+3 B x)+300 a^2 b^3 x^3 (A+2 B x)-60 b^4 x^5 \log (x) (5 a B+A b)+300 a A b^4 x^4-60 b^5 B x^6\right )}{60 x^5 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^6,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(300*a*A*b^4*x^4 - 60*b^5*B*x^6 + 300*a^2*b^3*x^3*(A + 2*B*x) + 100*a^3*b^2*x^2*(2*A
+ 3*B*x) + 25*a^4*b*x*(3*A + 4*B*x) + 3*a^5*(4*A + 5*B*x) - 60*b^4*(A*b + 5*a*B)*x^5*Log[x]))/(x^5*(a + b*x))

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IntegrateAlgebraic [B]  time = 3.71, size = 911, normalized size = 3.11 \begin {gather*} A \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 b x a+b^2 x^2}}{a}\right ) b^5+5 a B \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 b x a+b^2 x^2}}{a}\right ) b^4-\frac {1}{2} A \sqrt {b^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b^4-\frac {1}{2} A \sqrt {b^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b^4-\frac {5}{2} a \sqrt {b^2} B \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b^3-\frac {5}{2} a \sqrt {b^2} B \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) b^3+\frac {-4 \sqrt {a^2+2 b x a+b^2 x^2} \left (-60 b^{10} B x^{10}-270 a b^9 B x^9+300 a A b^9 x^8+120 a^2 b^8 B x^8+1500 a^2 A b^8 x^7+2280 a^3 b^7 B x^7+3200 a^3 A b^7 x^6+4720 a^4 b^6 B x^6+3875 a^4 A b^6 x^5+4585 a^5 b^5 B x^5+3012 a^5 A b^5 x^4+2460 a^6 b^4 B x^4+1598 a^6 A b^4 x^3+790 a^7 b^3 B x^3+572 a^7 A b^3 x^2+160 a^8 b^2 B x^2+123 a^8 A b^2 x+15 a^9 b B x+12 a^9 A b\right ) b^4-4 \sqrt {b^2} \left (60 b^{10} B x^{11}+330 a b^9 B x^{10}-300 a A b^9 x^9+150 a^2 b^8 B x^9-1800 a^2 A b^8 x^8-2400 a^3 b^7 B x^8-4700 a^3 A b^7 x^7-7000 a^4 b^6 B x^7-7075 a^4 A b^6 x^6-9305 a^5 b^5 B x^6-6887 a^5 A b^5 x^5-7045 a^6 b^4 B x^5-4610 a^6 A b^4 x^4-3250 a^7 b^3 B x^4-2170 a^7 A b^3 x^3-950 a^8 b^2 B x^3-695 a^8 A b^2 x^2-175 a^9 b B x^2-135 a^9 A b x-15 a^{10} B x-12 a^{10} A\right ) b^4}{15 \sqrt {b^2} \sqrt {a^2+2 b x a+b^2 x^2} \left (-16 x^4 b^8-64 a x^3 b^7-96 a^2 x^2 b^6-64 a^3 x b^5-16 a^4 b^4\right ) x^5+15 \left (16 x^5 b^{10}+80 a x^4 b^9+160 a^2 x^3 b^8+160 a^3 x^2 b^7+80 a^4 x b^6+16 a^5 b^5\right ) x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^6,x]

[Out]

(-4*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(12*a^9*A*b + 123*a^8*A*b^2*x + 15*a^9*b*B*x + 572*a^7*A*b^3*x^2 + 160*a
^8*b^2*B*x^2 + 1598*a^6*A*b^4*x^3 + 790*a^7*b^3*B*x^3 + 3012*a^5*A*b^5*x^4 + 2460*a^6*b^4*B*x^4 + 3875*a^4*A*b
^6*x^5 + 4585*a^5*b^5*B*x^5 + 3200*a^3*A*b^7*x^6 + 4720*a^4*b^6*B*x^6 + 1500*a^2*A*b^8*x^7 + 2280*a^3*b^7*B*x^
7 + 300*a*A*b^9*x^8 + 120*a^2*b^8*B*x^8 - 270*a*b^9*B*x^9 - 60*b^10*B*x^10) - 4*b^4*Sqrt[b^2]*(-12*a^10*A - 13
5*a^9*A*b*x - 15*a^10*B*x - 695*a^8*A*b^2*x^2 - 175*a^9*b*B*x^2 - 2170*a^7*A*b^3*x^3 - 950*a^8*b^2*B*x^3 - 461
0*a^6*A*b^4*x^4 - 3250*a^7*b^3*B*x^4 - 6887*a^5*A*b^5*x^5 - 7045*a^6*b^4*B*x^5 - 7075*a^4*A*b^6*x^6 - 9305*a^5
*b^5*B*x^6 - 4700*a^3*A*b^7*x^7 - 7000*a^4*b^6*B*x^7 - 1800*a^2*A*b^8*x^8 - 2400*a^3*b^7*B*x^8 - 300*a*A*b^9*x
^9 + 150*a^2*b^8*B*x^9 + 330*a*b^9*B*x^10 + 60*b^10*B*x^11))/(15*Sqrt[b^2]*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(
-16*a^4*b^4 - 64*a^3*b^5*x - 96*a^2*b^6*x^2 - 64*a*b^7*x^3 - 16*b^8*x^4) + 15*x^5*(16*a^5*b^5 + 80*a^4*b^6*x +
160*a^3*b^7*x^2 + 160*a^2*b^8*x^3 + 80*a*b^9*x^4 + 16*b^10*x^5)) + A*b^5*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 +
2*a*b*x + b^2*x^2]/a] + 5*a*b^4*B*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] - (A*b^4*Sqrt[b^
2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (5*a*b^3*Sqrt[b^2]*B*Log[-a - Sqrt[b^2]*x + Sqrt
[a^2 + 2*a*b*x + b^2*x^2]])/2 - (A*b^4*Sqrt[b^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (5*
a*b^3*Sqrt[b^2]*B*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2

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fricas [A]  time = 0.41, size = 121, normalized size = 0.41 \begin {gather*} \frac {60 \, B b^{5} x^{6} + 60 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} \log \relax (x) - 12 \, A a^{5} - 300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} - 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x}{60 \, x^{5}} \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)^(5/2)/x^6,x, algorithm="fricas")

[Out]

1/60*(60*B*b^5*x^6 + 60*(5*B*a*b^4 + A*b^5)*x^5*log(x) - 12*A*a^5 - 300*(2*B*a^2*b^3 + A*a*b^4)*x^4 - 300*(B*a
^3*b^2 + A*a^2*b^3)*x^3 - 100*(B*a^4*b + 2*A*a^3*b^2)*x^2 - 15*(B*a^5 + 5*A*a^4*b)*x)/x^5

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giac [A]  time = 0.19, size = 188, normalized size = 0.64 \begin {gather*} B b^{5} x \mathrm {sgn}\left (b x + a\right ) + {\left (5 \, B a b^{4} \mathrm {sgn}\left (b x + a\right ) + A b^{5} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right ) - \frac {12 \, A a^{5} \mathrm {sgn}\left (b x + a\right ) + 300 \, {\left (2 \, B a^{2} b^{3} \mathrm {sgn}\left (b x + a\right ) + A a b^{4} \mathrm {sgn}\left (b x + a\right )\right )} x^{4} + 300 \, {\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 )} x^{3} + 100 \, {\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} + 15 \, {\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}{60 \, x^{5}} \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)^(5/2)/x^6,x, algorithm="giac")

[Out]

B*b^5*x*sgn(b*x + a) + (5*B*a*b^4*sgn(b*x + a) + A*b^5*sgn(b*x + a))*log(abs(x)) - 1/60*(12*A*a^5*sgn(b*x + a)
+ 300*(2*B*a^2*b^3*sgn(b*x + a) + A*a*b^4*sgn(b*x + a))*x^4 + 300*(B*a^3*b^2*sgn(b*x + a) + A*a^2*b^3*sgn(b*x
+ a))*x^3 + 100*(B*a^4*b*sgn(b*x + a) + 2*A*a^3*b^2*sgn(b*x + a))*x^2 + 15*(B*a^5*sgn(b*x + a) + 5*A*a^4*b*sg
n(b*x + a))*x)/x^5

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maple [A]  time = 0.07, size = 144, normalized size = 0.49 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (60 A \,b^{5} x^{5} \ln \relax (x )+300 B a \,b^{4} x^{5} \ln \relax (x )+60 B \,b^{5} x^{6}-300 A a \,b^{4} x^{4}-600 B \,a^{2} b^{3} x^{4}-300 A \,a^{2} b^{3} x^{3}-300 B \,a^{3} b^{2} x^{3}-200 A \,a^{3} b^{2} x^{2}-100 B \,a^{4} b \,x^{2}-75 A \,a^{4} b x -15 B \,a^{5} x -12 A \,a^{5}\right )}{60 \left (b x +a \right )^{5} x^{5}} \end {gather*}

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^6,x)

[Out]

1/60*((b*x+a)^2)^(5/2)*(60*A*ln(x)*x^5*b^5+300*B*ln(x)*x^5*a*b^4+60*B*b^5*x^6-300*A*a*b^4*x^4-600*B*a^2*b^3*x^
4-300*A*a^2*b^3*x^3-300*B*a^3*b^2*x^3-200*A*a^3*b^2*x^2-100*B*a^4*b*x^2-75*A*a^4*b*x-15*B*a^5*x-12*A*a^5)/(b*x
+a)^5/x^5

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maxima [B]  time = 0.64, size = 673, normalized size = 2.30 \begin {gather*} 5 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a b^{4} \log \left (2 \, b^{2} x + 2 \, a b\right ) + \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A b^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a b^{4} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{5} x}{2 \, a} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{6} x}{2 \, a^{2}} + \frac {15}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{4} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{5}}{2 \, a} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{5} x}{4 \, a^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{6} x}{4 \, a^{4}} + \frac {35 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{4}}{12 \, a^{2}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{5}}{12 \, a^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{3 \, a^{4}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{15 \, a^{5}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{3}}{3 \, a^{3} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{4}}{3 \, a^{4} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{2}}{3 \, a^{4} x^{2}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{3}}{15 \, a^{5} x^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b}{12 \, a^{3} x^{3}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{2}}{60 \, a^{4} x^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{4 \, a^{2} x^{4}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b}{20 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A}{5 \, a^{2} x^{5}} \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)^(5/2)/x^6,x, algorithm="maxima")

[Out]

5*(-1)^(2*b^2*x + 2*a*b)*B*a*b^4*log(2*b^2*x + 2*a*b) + (-1)^(2*b^2*x + 2*a*b)*A*b^5*log(2*b^2*x + 2*a*b) - 5*
(-1)^(2*a*b*x + 2*a^2)*B*a*b^4*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) - (-1)^(2*a*b*x + 2*a^2)*A*b^5*log(2*a*b*x/a
bs(x) + 2*a^2/abs(x)) + 5/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^5*x/a + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^6*
x/a^2 + 15/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^4 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b^5/a + 5/4*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*B*b^5*x/a^3 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^6*x/a^4 + 35/12*(b^2*x^2 + 2*a*b*x
+ a^2)^(3/2)*B*b^4/a^2 + 7/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^5/a^3 + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*
B*b^4/a^4 - 2/15*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^5/a^5 - 2/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b^3/(a^3*x)
- 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^4/(a^4*x) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b^2/(a^4*x^2) + 2
/15*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^3/(a^5*x^2) + 1/12*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B*b/(a^3*x^3) - 11/
60*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b^2/(a^4*x^3) - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B/(a^2*x^4) + 3/20*(b
^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b/(a^3*x^4) - 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A/(a^2*x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{x^6} \,d x \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)^(5/2))/x^6,x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x^6, x)

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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 {5}{2}}}{x^{6}}\, 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)**(5/2)/x**6,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**6, x)

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