3.7.23 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{5/2}}{x^7} \, dx\)

Optimal. Leaf size=267 \[ -\frac {A (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 a x^6}+\frac {b^5 B \log (x) \sqrt {a^2+2 a b x+b^2 x^2}}{a+b x}-\frac {5 a b^4 B \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}-\frac {5 a^2 b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {a^5 B \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a^4 b B \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {10 a^3 b^2 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {770, 78, 43} \begin {gather*} -\frac {A (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 a x^6}-\frac {a^5 B \sqrt {a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac {5 a^4 b B \sqrt {a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac {10 a^3 b^2 B \sqrt {a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}-\frac {5 a^2 b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{x^2 (a+b x)}-\frac {5 a b^4 B \sqrt {a^2+2 a b x+b^2 x^2}}{x (a+b x)}+\frac {b^5 B \log (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^7,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 127, normalized size = 0.48 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (2 a^5 (5 A+6 B x)+15 a^4 b x (4 A+5 B x)+50 a^3 b^2 x^2 (3 A+4 B x)+100 a^2 b^3 x^3 (2 A+3 B x)+150 a b^4 x^4 (A+2 B x)+60 A b^5 x^5-60 b^5 B x^6 \log (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)^(5/2))/x^7,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(60*A*b^5*x^5 + 150*a*b^4*x^4*(A + 2*B*x) + 100*a^2*b^3*x^3*(2*A + 3*B*x) + 50*a^3*b^
2*x^2*(3*A + 4*B*x) + 15*a^4*b*x*(4*A + 5*B*x) + 2*a^5*(5*A + 6*B*x) - 60*b^5*B*x^6*Log[x]))/(x^6*(a + b*x))

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 5.56, size = 2809, normalized size = 10.52 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/x^7,x]

[Out]

(8*b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-10*a^10*A*b - 110*a^9*A*b^2*x - 12*a^10*b*B*x - 550*a^8*A*b^3*x^2 - 135
*a^9*b^2*B*x^2 - 1650*a^7*A*b^4*x^3 - 695*a^8*b^3*B*x^3 - 3300*a^6*A*b^5*x^4 - 2170*a^7*b^4*B*x^4 - 4620*a^5*A
*b^6*x^5 - 4610*a^6*b^5*B*x^5 - 4610*a^4*A*b^7*x^6 - 6887*a^5*b^6*B*x^6 - 3250*a^3*A*b^8*x^7 - 7075*a^4*b^7*B*
x^7 - 1550*a^2*A*b^9*x^8 - 4700*a^3*b^8*B*x^8 - 450*a*A*b^10*x^9 - 1800*a^2*b^9*B*x^9 - 60*A*b^11*x^10 - 300*a
*b^10*B*x^10) + 8*b^5*Sqrt[b^2]*(10*a^11*A + 120*a^10*A*b*x + 12*a^11*B*x + 660*a^9*A*b^2*x^2 + 147*a^10*b*B*x
^2 + 2200*a^8*A*b^3*x^3 + 830*a^9*b^2*B*x^3 + 4950*a^7*A*b^4*x^4 + 2865*a^8*b^3*B*x^4 + 7920*a^6*A*b^5*x^5 + 6
780*a^7*b^4*B*x^5 + 9230*a^5*A*b^6*x^6 + 11497*a^6*b^5*B*x^6 + 7860*a^4*A*b^7*x^7 + 13962*a^5*b^6*B*x^7 + 4800
*a^3*A*b^8*x^8 + 11775*a^4*b^7*B*x^8 + 2000*a^2*A*b^9*x^9 + 6500*a^3*b^8*B*x^9 + 510*a*A*b^10*x^10 + 2100*a^2*
b^9*B*x^10 + 60*A*b^11*x^11 + 300*a*b^10*B*x^11))/(15*Sqrt[b^2]*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-32*a^5*b^5
- 160*a^4*b^6*x - 320*a^3*b^7*x^2 - 320*a^2*b^8*x^3 - 160*a*b^9*x^4 - 32*b^10*x^5) + 15*x^6*(32*a^6*b^6 + 192
*a^5*b^7*x + 480*a^4*b^8*x^2 + 640*a^3*b^9*x^3 + 480*a^2*b^10*x^4 + 192*a*b^11*x^5 + 32*b^12*x^6)) + (b^5*B*Lo
g[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/2 - (b^4*Sqrt[b^2]*B*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*
a*b*x + b^2*x^2]])/2 - (a^12*b^5*B*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) - (a^12*b^4*Sqrt[b^2]*
B*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^
6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) + (3*a^10*b^5*B*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x +
b^2*x^2])^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^
2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) + (3*a^10*b^4*Sqrt[b^2]*B*(-(Sqrt[b^2]*x) + Sqr
t[a^2 + 2*a*b*x + b^2*x^2])^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[
a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) - (15*a^8*b^5*B*(-(Sqrt[b^2]*
x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]
*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) - (15*a^8*b^4*Sqrt[
b^2]*B*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]]
)/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6)
+ (10*a^6*b^5*B*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x +
b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*
x^2])^6) + (10*a^6*b^4*Sqrt[b^2]*B*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*Log[a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^
2 + 2*a*b*x + b^2*x^2])^6) - (15*a^4*b^5*B*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^8*Log[a - Sqrt[b^2
]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) - (15*a^4*b^4*Sqrt[b^2]*B*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)^8*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) + (3*a^2*b^5*B*(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x +
b^2*x^2])^10*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x +
b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) + (3*a^2*b^4*Sqrt[b^2]*B*(-(Sqrt[b^2]*x) + Sq
rt[a^2 + 2*a*b*x + b^2*x^2])^10*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqr
t[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) - (b^5*B*(-(Sqrt[b^2]*x) +
Sqrt[a^2 + 2*a*b*x + b^2*x^2])^12*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6) - (b^4*Sqrt[b^2]*B*(-(S
qrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^12*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*(-a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^6)

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 121, normalized size = 0.45 \begin {gather*} \frac {60 \, B b^{5} x^{6} \log \relax (x) - 10 \, A a^{5} - 60 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{5} - 150 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{4} - 200 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{3} - 75 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{2} - 12 \, {\left (B a^{5} + 5 \, A a^{4} 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)^(5/2)/x^7,x, algorithm="fricas")

[Out]

1/60*(60*B*b^5*x^6*log(x) - 10*A*a^5 - 60*(5*B*a*b^4 + A*b^5)*x^5 - 150*(2*B*a^2*b^3 + A*a*b^4)*x^4 - 200*(B*a
^3*b^2 + A*a^2*b^3)*x^3 - 75*(B*a^4*b + 2*A*a^3*b^2)*x^2 - 12*(B*a^5 + 5*A*a^4*b)*x)/x^6

________________________________________________________________________________________

giac [A]  time = 0.17, size = 191, normalized size = 0.72 \begin {gather*} B b^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x + a\right ) - \frac {10 \, A a^{5} \mathrm {sgn}\left (b x + a\right ) + 60 \, {\left (5 \, B a b^{4} \mathrm {sgn}\left (b x + a\right ) + A b^{5} \mathrm {sgn}\left (b x + a\right )\right )} x^{5} + 150 \, {\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} + 200 \, {\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} + 75 \, {\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} + 12 \, {\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^{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)^(5/2)/x^7,x, algorithm="giac")

[Out]

B*b^5*log(abs(x))*sgn(b*x + a) - 1/60*(10*A*a^5*sgn(b*x + a) + 60*(5*B*a*b^4*sgn(b*x + a) + A*b^5*sgn(b*x + a)
)*x^5 + 150*(2*B*a^2*b^3*sgn(b*x + a) + A*a*b^4*sgn(b*x + a))*x^4 + 200*(B*a^3*b^2*sgn(b*x + a) + A*a^2*b^3*sg
n(b*x + a))*x^3 + 75*(B*a^4*b*sgn(b*x + a) + 2*A*a^3*b^2*sgn(b*x + a))*x^2 + 12*(B*a^5*sgn(b*x + a) + 5*A*a^4*
b*sgn(b*x + a))*x)/x^6

________________________________________________________________________________________

maple [A]  time = 0.06, size = 142, normalized size = 0.53 \begin {gather*} -\frac {\left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} \left (-60 B \,b^{5} x^{6} \ln \relax (x )+60 A \,b^{5} x^{5}+300 B a \,b^{4} x^{5}+150 A a \,b^{4} x^{4}+300 B \,a^{2} b^{3} x^{4}+200 A \,a^{2} b^{3} x^{3}+200 B \,a^{3} b^{2} x^{3}+150 A \,a^{3} b^{2} x^{2}+75 B \,a^{4} b \,x^{2}+60 A \,a^{4} b x +12 B \,a^{5} x +10 A \,a^{5}\right )}{60 \left (b x +a \right )^{5} x^{6}} \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^7,x)

[Out]

-1/60*((b*x+a)^2)^(5/2)*(-60*B*b^5*ln(x)*x^6+60*A*b^5*x^5+300*B*a*b^4*x^5+150*A*a*b^4*x^4+300*B*a^2*b^3*x^4+20
0*A*a^2*b^3*x^3+200*B*a^3*b^2*x^3+150*A*a^3*b^2*x^2+75*B*a^4*b*x^2+60*A*a^4*b*x+12*B*a^5*x+10*A*a^5)/(b*x+a)^5
/x^6

________________________________________________________________________________________

maxima [B]  time = 0.57, size = 554, normalized size = 2.07 \begin {gather*} \left (-1\right )^{2 \, b^{2} x + 2 \, a b} B b^{5} \log \left (2 \, b^{2} x + 2 \, a b\right ) - \left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right ) + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{6} x}{2 \, a^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{5}}{2 \, a} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{6} x}{4 \, a^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B b^{5}}{12 \, a^{3}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{5}}{15 \, a^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{6}}{6 \, a^{6}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B b^{4}}{3 \, a^{4} x} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} A b^{5}}{6 \, a^{5} x} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{3}}{15 \, a^{5} x^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{4}}{6 \, a^{6} x^{2}} - \frac {11 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b^{2}}{60 \, a^{4} x^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{3}}{6 \, a^{5} x^{3}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B b}{20 \, a^{3} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b^{2}}{6 \, a^{4} x^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} B}{5 \, a^{2} x^{5}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} A b}{6 \, a^{3} x^{5}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{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)^(5/2)/x^7,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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^7} \,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^7,x)

[Out]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/x^7, x)

________________________________________________________________________________________

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^{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)**(5/2)/x**7,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(5/2)/x**7, x)

________________________________________________________________________________________