3.9.32 \(\int \frac {A+B x}{\sqrt {x} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [832]

Optimal. Leaf size=258 \[ \frac {5 (7 A b+a B) \sqrt {x}}{64 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) \sqrt {x}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(7 A b+a B) \sqrt {x}}{24 a^2 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (7 A b+a B) \sqrt {x}}{96 a^3 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (7 A b+a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{9/2} b^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

5/64*(7*A*b+B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/a^(9/2)/b^(3/2)/((b*x+a)^2)^(1/2)+5/64*(7*A*b+B*a)*x^
(1/2)/a^4/b/((b*x+a)^2)^(1/2)+1/4*(A*b-B*a)*x^(1/2)/a/b/(b*x+a)^3/((b*x+a)^2)^(1/2)+1/24*(7*A*b+B*a)*x^(1/2)/a
^2/b/(b*x+a)^2/((b*x+a)^2)^(1/2)+5/96*(7*A*b+B*a)*x^(1/2)/a^3/b/(b*x+a)/((b*x+a)^2)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {784, 79, 44, 65, 211} \begin {gather*} \frac {\sqrt {x} (A b-a B)}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\sqrt {x} (a B+7 A b)}{24 a^2 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (a+b x) (a B+7 A b) \text {ArcTan}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{9/2} b^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 \sqrt {x} (a B+7 A b)}{64 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 \sqrt {x} (a B+7 A b)}{96 a^3 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(7*A*b + a*B)*Sqrt[x])/(64*a^4*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*Sqrt[x])/(4*a*b*(a + b*x)^3*
Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((7*A*b + a*B)*Sqrt[x])/(24*a^2*b*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
+ (5*(7*A*b + a*B)*Sqrt[x])/(96*a^3*b*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (5*(7*A*b + a*B)*(a + b*x)*Ar
cTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(64*a^(9/2)*b^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

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] || L
tQ[p, n]))))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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}{\sqrt {x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {A+B x}{\sqrt {x} \left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) \sqrt {x}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 (7 A b+a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )^4} \, dx}{8 a \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) \sqrt {x}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(7 A b+a B) \sqrt {x}}{24 a^2 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 b (7 A b+a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )^3} \, dx}{48 a^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(A b-a B) \sqrt {x}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(7 A b+a B) \sqrt {x}}{24 a^2 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (7 A b+a B) \sqrt {x}}{96 a^3 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 (7 A b+a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )^2} \, dx}{64 a^3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 (7 A b+a B) \sqrt {x}}{64 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) \sqrt {x}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(7 A b+a B) \sqrt {x}}{24 a^2 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (7 A b+a B) \sqrt {x}}{96 a^3 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 (7 A b+a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\sqrt {x} \left (a b+b^2 x\right )} \, dx}{128 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 (7 A b+a B) \sqrt {x}}{64 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) \sqrt {x}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(7 A b+a B) \sqrt {x}}{24 a^2 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (7 A b+a B) \sqrt {x}}{96 a^3 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (5 (7 A b+a B) \left (a b+b^2 x\right )\right ) \text {Subst}\left (\int \frac {1}{a b+b^2 x^2} \, dx,x,\sqrt {x}\right )}{64 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {5 (7 A b+a B) \sqrt {x}}{64 a^4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) \sqrt {x}}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(7 A b+a B) \sqrt {x}}{24 a^2 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (7 A b+a B) \sqrt {x}}{96 a^3 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 (7 A b+a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{64 a^{9/2} b^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 145, normalized size = 0.56 \begin {gather*} \frac {\sqrt {a} \sqrt {b} \sqrt {x} \left (-15 a^4 B+105 A b^4 x^3+5 a b^3 x^2 (77 A+3 B x)+a^2 b^2 x (511 A+55 B x)+a^3 b (279 A+73 B x)\right )+15 (7 A b+a B) (a+b x)^4 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{192 a^{9/2} b^{3/2} (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*Sqrt[b]*Sqrt[x]*(-15*a^4*B + 105*A*b^4*x^3 + 5*a*b^3*x^2*(77*A + 3*B*x) + a^2*b^2*x*(511*A + 55*B*x)
+ a^3*b*(279*A + 73*B*x)) + 15*(7*A*b + a*B)*(a + b*x)^4*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(192*a^(9/2)*b^(3/
2)*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(356\) vs. \(2(175)=350\).
time = 0.76, size = 357, normalized size = 1.38

method	result	size
default	\(\frac {\left (105 A \sqrt {a b}\, x^{\frac {7}{2}} b^{4}+15 B \sqrt {a b}\, x^{\frac {7}{2}} a \,b^{3}+385 A \sqrt {a b}\, x^{\frac {5}{2}} a \,b^{3}+105 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) b^{5} x^{4}+55 B \sqrt {a b}\, x^{\frac {5}{2}} a^{2} b^{2}+15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{4}+420 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a \,b^{4} x^{3}+60 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{3}+511 A \sqrt {a b}\, x^{\frac {3}{2}} a^{2} b^{2}+630 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{2} b^{3} x^{2}+73 B \sqrt {a b}\, x^{\frac {3}{2}} a^{3} b +90 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x^{2}+420 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{3} b^{2} x +60 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b x +279 A \sqrt {a b}\, \sqrt {x}\, a^{3} b +105 A \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{4} b -15 B \sqrt {a b}\, \sqrt {x}\, a^{4}+15 B \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) a^{5}\right ) \left (b x +a \right )}{192 \sqrt {a b}\, b \,a^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(357\)

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^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/192*(105*A*(a*b)^(1/2)*x^(7/2)*b^4+15*B*(a*b)^(1/2)*x^(7/2)*a*b^3+385*A*(a*b)^(1/2)*x^(5/2)*a*b^3+105*A*arct
an(b*x^(1/2)/(a*b)^(1/2))*b^5*x^4+55*B*(a*b)^(1/2)*x^(5/2)*a^2*b^2+15*B*arctan(b*x^(1/2)/(a*b)^(1/2))*a*b^4*x^
4+420*A*arctan(b*x^(1/2)/(a*b)^(1/2))*a*b^4*x^3+60*B*arctan(b*x^(1/2)/(a*b)^(1/2))*a^2*b^3*x^3+511*A*(a*b)^(1/
2)*x^(3/2)*a^2*b^2+630*A*arctan(b*x^(1/2)/(a*b)^(1/2))*a^2*b^3*x^2+73*B*(a*b)^(1/2)*x^(3/2)*a^3*b+90*B*arctan(
b*x^(1/2)/(a*b)^(1/2))*a^3*b^2*x^2+420*A*arctan(b*x^(1/2)/(a*b)^(1/2))*a^3*b^2*x+60*B*arctan(b*x^(1/2)/(a*b)^(
1/2))*a^4*b*x+279*A*(a*b)^(1/2)*x^(1/2)*a^3*b+105*A*arctan(b*x^(1/2)/(a*b)^(1/2))*a^4*b-15*B*(a*b)^(1/2)*x^(1/
2)*a^4+15*B*arctan(b*x^(1/2)/(a*b)^(1/2))*a^5)*(b*x+a)/(a*b)^(1/2)/b/a^4/((b*x+a)^2)^(5/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (175) = 350\).
time = 0.58, size = 390, normalized size = 1.51 \begin {gather*} -\frac {5 \, {\left ({\left (3 \, B a b^{5} + 7 \, A b^{6}\right )} x^{2} - 21 \, {\left (B a^{2} b^{4} + 9 \, A a b^{5}\right )} x\right )} x^{\frac {9}{2}} + 10 \, {\left ({\left (3 \, B a^{2} b^{4} + 7 \, A a b^{5}\right )} x^{2} - 63 \, {\left (B a^{3} b^{3} + 9 \, A a^{2} b^{4}\right )} x\right )} x^{\frac {7}{2}} - 20 \, {\left (2 \, {\left (3 \, B a^{3} b^{3} + 7 \, A a^{2} b^{4}\right )} x^{2} + 77 \, {\left (B a^{4} b^{2} + 9 \, A a^{3} b^{3}\right )} x\right )} x^{\frac {5}{2}} - 2 \, {\left (85 \, {\left (3 \, B a^{4} b^{2} + 7 \, A a^{3} b^{3}\right )} x^{2} + 973 \, {\left (B a^{5} b + 9 \, A a^{4} b^{2}\right )} x\right )} x^{\frac {3}{2}} - {\left (253 \, {\left (3 \, B a^{5} b + 7 \, A a^{4} b^{2}\right )} x^{2} + 1315 \, {\left (B a^{6} + 9 \, A a^{5} b\right )} x\right )} \sqrt {x} - \frac {1280 \, {\left (A a^{5} b x^{2} + 3 \, A a^{6} x\right )}}{\sqrt {x}}}{1920 \, {\left (a^{6} b^{5} x^{5} + 5 \, a^{7} b^{4} x^{4} + 10 \, a^{8} b^{3} x^{3} + 10 \, a^{9} b^{2} x^{2} + 5 \, a^{10} b x + a^{11}\right )}} + \frac {5 \, {\left (B a + 7 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{4} b} + \frac {{\left (3 \, B a b + 7 \, A b^{2}\right )} x^{\frac {3}{2}} - 30 \, {\left (B a^{2} + 7 \, A a b\right )} \sqrt {x}}{384 \, a^{6} b} \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^(1/2),x, algorithm="maxima")

[Out]

-1/1920*(5*((3*B*a*b^5 + 7*A*b^6)*x^2 - 21*(B*a^2*b^4 + 9*A*a*b^5)*x)*x^(9/2) + 10*((3*B*a^2*b^4 + 7*A*a*b^5)*
x^2 - 63*(B*a^3*b^3 + 9*A*a^2*b^4)*x)*x^(7/2) - 20*(2*(3*B*a^3*b^3 + 7*A*a^2*b^4)*x^2 + 77*(B*a^4*b^2 + 9*A*a^
3*b^3)*x)*x^(5/2) - 2*(85*(3*B*a^4*b^2 + 7*A*a^3*b^3)*x^2 + 973*(B*a^5*b + 9*A*a^4*b^2)*x)*x^(3/2) - (253*(3*B
*a^5*b + 7*A*a^4*b^2)*x^2 + 1315*(B*a^6 + 9*A*a^5*b)*x)*sqrt(x) - 1280*(A*a^5*b*x^2 + 3*A*a^6*x)/sqrt(x))/(a^6
*b^5*x^5 + 5*a^7*b^4*x^4 + 10*a^8*b^3*x^3 + 10*a^9*b^2*x^2 + 5*a^10*b*x + a^11) + 5/64*(B*a + 7*A*b)*arctan(b*
sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4*b) + 1/384*((3*B*a*b + 7*A*b^2)*x^(3/2) - 30*(B*a^2 + 7*A*a*b)*sqrt(x))/(a^6
*b)

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Fricas [A]
time = 3.90, size = 523, normalized size = 2.03 \begin {gather*} \left [-\frac {15 \, {\left (B a^{5} + 7 \, A a^{4} b + {\left (B a b^{4} + 7 \, A b^{5}\right )} x^{4} + 4 \, {\left (B a^{2} b^{3} + 7 \, A a b^{4}\right )} x^{3} + 6 \, {\left (B a^{3} b^{2} + 7 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (B a^{4} b + 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (15 \, B a^{5} b - 279 \, A a^{4} b^{2} - 15 \, {\left (B a^{2} b^{4} + 7 \, A a b^{5}\right )} x^{3} - 55 \, {\left (B a^{3} b^{3} + 7 \, A a^{2} b^{4}\right )} x^{2} - 73 \, {\left (B a^{4} b^{2} + 7 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{384 \, {\left (a^{5} b^{6} x^{4} + 4 \, a^{6} b^{5} x^{3} + 6 \, a^{7} b^{4} x^{2} + 4 \, a^{8} b^{3} x + a^{9} b^{2}\right )}}, -\frac {15 \, {\left (B a^{5} + 7 \, A a^{4} b + {\left (B a b^{4} + 7 \, A b^{5}\right )} x^{4} + 4 \, {\left (B a^{2} b^{3} + 7 \, A a b^{4}\right )} x^{3} + 6 \, {\left (B a^{3} b^{2} + 7 \, A a^{2} b^{3}\right )} x^{2} + 4 \, {\left (B a^{4} b + 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (15 \, B a^{5} b - 279 \, A a^{4} b^{2} - 15 \, {\left (B a^{2} b^{4} + 7 \, A a b^{5}\right )} x^{3} - 55 \, {\left (B a^{3} b^{3} + 7 \, A a^{2} b^{4}\right )} x^{2} - 73 \, {\left (B a^{4} b^{2} + 7 \, A a^{3} b^{3}\right )} x\right )} \sqrt {x}}{192 \, {\left (a^{5} b^{6} x^{4} + 4 \, a^{6} b^{5} x^{3} + 6 \, a^{7} b^{4} x^{2} + 4 \, a^{8} b^{3} x + a^{9} b^{2}\right )}}\right ] \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^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(B*a^5 + 7*A*a^4*b + (B*a*b^4 + 7*A*b^5)*x^4 + 4*(B*a^2*b^3 + 7*A*a*b^4)*x^3 + 6*(B*a^3*b^2 + 7*A*
a^2*b^3)*x^2 + 4*(B*a^4*b + 7*A*a^3*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(15
*B*a^5*b - 279*A*a^4*b^2 - 15*(B*a^2*b^4 + 7*A*a*b^5)*x^3 - 55*(B*a^3*b^3 + 7*A*a^2*b^4)*x^2 - 73*(B*a^4*b^2 +
7*A*a^3*b^3)*x)*sqrt(x))/(a^5*b^6*x^4 + 4*a^6*b^5*x^3 + 6*a^7*b^4*x^2 + 4*a^8*b^3*x + a^9*b^2), -1/192*(15*(B
*a^5 + 7*A*a^4*b + (B*a*b^4 + 7*A*b^5)*x^4 + 4*(B*a^2*b^3 + 7*A*a*b^4)*x^3 + 6*(B*a^3*b^2 + 7*A*a^2*b^3)*x^2 +
4*(B*a^4*b + 7*A*a^3*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (15*B*a^5*b - 279*A*a^4*b^2 - 15*(B*a^
2*b^4 + 7*A*a*b^5)*x^3 - 55*(B*a^3*b^3 + 7*A*a^2*b^4)*x^2 - 73*(B*a^4*b^2 + 7*A*a^3*b^3)*x)*sqrt(x))/(a^5*b^6*
x^4 + 4*a^6*b^5*x^3 + 6*a^7*b^4*x^2 + 4*a^8*b^3*x + a^9*b^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{\sqrt {x} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, 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**(1/2),x)

[Out]

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

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Giac [A]
time = 0.98, size = 147, normalized size = 0.57 \begin {gather*} \frac {5 \, {\left (B a + 7 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{64 \, \sqrt {a b} a^{4} b \mathrm {sgn}\left (b x + a\right )} + \frac {15 \, B a b^{3} x^{\frac {7}{2}} + 105 \, A b^{4} x^{\frac {7}{2}} + 55 \, B a^{2} b^{2} x^{\frac {5}{2}} + 385 \, A a b^{3} x^{\frac {5}{2}} + 73 \, B a^{3} b x^{\frac {3}{2}} + 511 \, A a^{2} b^{2} x^{\frac {3}{2}} - 15 \, B a^{4} \sqrt {x} + 279 \, A a^{3} b \sqrt {x}}{192 \, {\left (b x + a\right )}^{4} a^{4} b \mathrm {sgn}\left (b x + a\right )} \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^(1/2),x, algorithm="giac")

[Out]

5/64*(B*a + 7*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4*b*sgn(b*x + a)) + 1/192*(15*B*a*b^3*x^(7/2) + 10
5*A*b^4*x^(7/2) + 55*B*a^2*b^2*x^(5/2) + 385*A*a*b^3*x^(5/2) + 73*B*a^3*b*x^(3/2) + 511*A*a^2*b^2*x^(3/2) - 15
*B*a^4*sqrt(x) + 279*A*a^3*b*sqrt(x))/((b*x + a)^4*a^4*b*sgn(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)

[Out]

int((A + B*x)/(x^(1/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)

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