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

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

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Rubi [A]  time = 0.12, antiderivative size = 297, 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)}{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)} \end {gather*}

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 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^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*}

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Mathematica [A]  time = 0.05, size = 127, normalized size = 0.43 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-\left (a^5 (2 A+3 B x)\right )-15 a^4 b x (A+2 B x)-60 a^3 A b^2 x^2+60 a^2 b^2 x^3 \log (x) (a B+A b)+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)} \end {gather*}

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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IntegrateAlgebraic [B]  time = 16.57, size = 1633, normalized size = 5.50

result too large to display

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-4*a^19*A*b^3 - 134*a^18*A*b^4*x - 6*a^19*b^3*B*x - 2152*a^17*A*b^5*x^2 - 216*
a^18*b^4*B*x^2 - 21955*a^16*A*b^6*x^3 - 3535*a^17*b^5*B*x^3 - 158798*a^15*A*b^7*x^4 - 35054*a^16*b^6*B*x^4 - 8
60431*a^14*A*b^8*x^5 - 236155*a^15*b^7*B*x^5 - 3600304*a^13*A*b^9*x^6 - 1145824*a^14*b^8*B*x^6 - 11830658*a^12
*A*b^10*x^7 - 4131458*a^13*b^9*B*x^7 - 30777328*a^11*A*b^11*x^8 - 11225068*a^12*b^10*B*x^8 - 63475808*a^10*A*b
^12*x^9 - 22938688*a^11*b^11*B*x^9 - 103276096*a^9*A*b^13*x^10 - 34421024*a^10*b^12*B*x^10 - 130921568*a^8*A*b
^14*x^11 - 35156704*a^9*b^13*B*x^11 - 126307328*a^7*A*b^15*x^12 - 17698880*a^8*b^14*B*x^12 - 88721152*a^6*A*b^
16*x^13 + 10965248*a^7*b^15*B*x^13 - 41179136*a^5*A*b^17*x^14 + 32060416*a^6*b^16*B*x^14 - 8998400*a^4*A*b^18*
x^15 + 33432064*a^5*b^17*B*x^15 + 1937408*a^3*A*b^19*x^16 + 20925440*a^4*b^18*B*x^16 + 1884160*a^2*A*b^20*x^17
+ 8249344*a^3*b^19*B*x^17 + 442368*a*A*b^21*x^18 + 1957888*a^2*b^20*B*x^18 + 24576*A*b^22*x^19 + 253952*a*b^2
1*B*x^19 + 16384*b^22*B*x^20) + Sqrt[b^2]*(4*a^20*A*b^2 + 138*a^19*A*b^3*x + 6*a^20*b^2*B*x + 2286*a^18*A*b^4*
x^2 + 222*a^19*b^3*B*x^2 + 24107*a^17*A*b^5*x^3 + 3751*a^18*b^4*B*x^3 + 180753*a^16*A*b^6*x^4 + 38589*a^17*b^5
*B*x^4 + 1019229*a^15*A*b^7*x^5 + 271209*a^16*b^6*B*x^5 + 4460735*a^14*A*b^8*x^6 + 1381979*a^15*b^7*B*x^6 + 15
430962*a^13*A*b^9*x^7 + 5277282*a^14*b^8*B*x^7 + 42607986*a^12*A*b^10*x^8 + 15356526*a^13*b^9*B*x^8 + 94253136
*a^11*A*b^11*x^9 + 34163756*a^12*b^10*B*x^9 + 166751904*a^10*A*b^12*x^10 + 57359712*a^11*b^11*B*x^10 + 2341976
64*a^9*A*b^13*x^11 + 69577728*a^10*b^12*B*x^11 + 257228896*a^8*A*b^14*x^12 + 52855584*a^9*b^13*B*x^12 + 215028
480*a^7*A*b^15*x^13 + 6733632*a^8*b^14*B*x^13 + 129900288*a^6*A*b^16*x^14 - 43025664*a^7*b^15*B*x^14 + 5017753
6*a^5*A*b^17*x^15 - 65492480*a^6*b^16*B*x^15 + 7060992*a^4*A*b^18*x^16 - 54357504*a^5*b^17*B*x^16 - 3821568*a^
3*A*b^19*x^17 - 29174784*a^4*b^18*B*x^17 - 2326528*a^2*A*b^20*x^18 - 10207232*a^3*b^19*B*x^18 - 466944*a*A*b^2
1*x^19 - 2211840*a^2*b^20*B*x^19 - 24576*A*b^22*x^20 - 270336*a*b^21*B*x^20 - 16384*b^22*B*x^21))/(3*Sqrt[b^2]
*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-4*a^14*b^2 - 104*a^13*b^3*x - 1252*a^12*b^4*x^2 - 9248*a^11*b^5*x^3 - 468
16*a^10*b^6*x^4 - 171776*a^9*b^7*x^5 - 470976*a^8*b^8*x^6 - 979968*a^7*b^9*x^7 - 1554432*a^6*b^10*x^8 - 186982
4*a^5*b^11*x^9 - 1678336*a^4*b^12*x^10 - 1089536*a^3*b^13*x^11 - 483328*a^2*b^14*x^12 - 131072*a*b^15*x^13 - 1
6384*b^16*x^14) + 3*x^3*(4*a^15*b^3 + 108*a^14*b^4*x + 1356*a^13*b^5*x^2 + 10500*a^12*b^6*x^3 + 56064*a^11*b^7
*x^4 + 218592*a^10*b^8*x^5 + 642752*a^9*b^9*x^6 + 1450944*a^8*b^10*x^7 + 2534400*a^7*b^11*x^8 + 3424256*a^6*b^
12*x^9 + 3548160*a^5*b^13*x^10 + 2767872*a^4*b^14*x^11 + 1572864*a^3*b^15*x^12 + 614400*a^2*b^16*x^13 + 147456
*a*b^17*x^14 + 16384*b^18*x^15)) + 10*a^2*A*b^3*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] + 1
0*a^3*b^2*B*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a] - 5*a^2*A*(b^2)^(3/2)*Log[-a - Sqrt[b^2
]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - 5*a^3*b*Sqrt[b^2]*B*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2
]] - 5*a^2*A*(b^2)^(3/2)*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - 5*a^3*b*Sqrt[b^2]*B*Log[a - Sq
rt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]]

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fricas [A]  time = 0.41, size = 121, normalized size = 0.41 \begin {gather*} \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 \relax (x) - 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 {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^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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giac [A]  time = 0.20, size = 190, normalized size = 0.64 \begin {gather*} \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 {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^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

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maple [A]  time = 0.06, size = 144, normalized size = 0.48 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (2 B \,b^{5} x^{6}+3 A \,b^{5} x^{5}+15 B a \,b^{4} x^{5}+60 A \,a^{2} b^{3} x^{3} \ln \relax (x )+30 A a \,b^{4} x^{4}+60 B \,a^{3} b^{2} x^{3} \ln \relax (x )+60 B \,a^{2} b^{3} x^{4}-60 A \,a^{3} b^{2} x^{2}-30 B \,a^{4} b \,x^{2}-15 A \,a^{4} b x -3 B \,a^{5} x -2 A \,a^{5}\right )}{6 \left (b x +a \right )^{5} x^{3}} \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^4,x)

[Out]

1/6*((b*x+a)^2)^(5/2)*(2*B*b^5*x^6+3*A*b^5*x^5+15*B*a*b^4*x^5+60*A*ln(x)*x^3*a^2*b^3+30*A*a*b^4*x^4+60*B*ln(x)
*x^3*a^3*b^2+60*B*a^2*b^3*x^4-60*A*a^3*b^2*x^2-30*B*a^4*b*x^2-15*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 [B]  time = 0.64, size = 557, normalized size = 1.88 \begin {gather*} 10 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B a^{3} b^{2} \log \left (2 \, b^{2} x + 2 \, a b\right ) + 10 \, \left (-1\right )^{2 \, b^{2} x + 2 \, a b} A a^{2} b^{3} \log \left (2 \, b^{2} x + 2 \, a b\right ) - 10 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B a^{3} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) - 10 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A a^{2} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + 5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a b^{3} x + 5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{4} x + 15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2} b^{2} + 15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a b^{3} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{3} x}{2 \, a} + \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{4} x}{2 \, a^{2}} + \frac {35}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{2} + \frac {35 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A b^{3}}{6 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{2}}{2 \, a^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{3}}{6 \, a^{3}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b}{2 \, a x} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{2}}{6 \, a^{2} x} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{2 \, a^{2} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b}{6 \, a^{3} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A}{3 \, a^{2} x^{3}} \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^4,x, algorithm="maxima")

[Out]

10*(-1)^(2*b^2*x + 2*a*b)*B*a^3*b^2*log(2*b^2*x + 2*a*b) + 10*(-1)^(2*b^2*x + 2*a*b)*A*a^2*b^3*log(2*b^2*x + 2
*a*b) - 10*(-1)^(2*a*b*x + 2*a^2)*B*a^3*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) - 10*(-1)^(2*a*b*x + 2*a^2)*A*a
^2*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x)) + 5*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*b^3*x + 5*sqrt(b^2*x^2 + 2*a*b
*x + a^2)*A*b^4*x + 15*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a^2*b^2 + 15*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a*b^3 + 5/
2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*b^3*x/a + 5/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^4*x/a^2 + 35/6*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*B*b^2 + 35/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*b^3/a + 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(5
/2)*B*b^2/a^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^3/a^3 - 3/2*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*b/(a*x)
- 11/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*A*b^2/(a^2*x) - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*B/(a^2*x^2) - 1/6*(
b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A*b/(a^3*x^2) - 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*A/(a^2*x^3)

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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^4} \,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^4,x)

[Out]

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

[Out]

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

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