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

Optimal. Leaf size=255 \[ -\frac {5 \sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{8 x}+\frac {5 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {c}}-\frac {5 \left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {a}}-\frac {5 \left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{12 x^2}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3} \]

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Rubi [A]  time = 0.32, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {812, 843, 621, 206, 724} \begin {gather*} \frac {5 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {c}}-\frac {5 \sqrt {a+b x+c x^2} \left (-x \left (4 a B c+4 A b c+b^2 B\right )+A \left (4 a c+b^2\right )+4 a b B\right )}{8 x}-\frac {5 \left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {a}}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac {5 \left (a+b x+c x^2\right )^{3/2} (2 a B-x (2 A c+b B)+A b)}{12 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(4*a*b*B + A*(b^2 + 4*a*c) - (b^2*B + 4*A*b*c + 4*a*B*c)*x)*Sqrt[a + b*x + c*x^2])/(8*x) - (5*(A*b + 2*a*B
- (b*B + 2*A*c)*x)*(a + b*x + c*x^2)^(3/2))/(12*x^2) - ((A - B*x)*(a + b*x + c*x^2)^(5/2))/(3*x^3) - (5*(2*a*
B*(3*b^2 + 4*a*c) + A*(b^3 + 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*Sqrt[a]) +
(5*(b^3*B + 6*A*b^2*c + 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(16*S
qrt[c])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^{5/2}}{x^4} \, dx &=-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac {5}{18} \int \frac {(-3 (A b+2 a B)-3 (b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{x^3} \, dx\\ &=-\frac {5 (A b+2 a B-(b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{12 x^2}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}+\frac {5}{48} \int \frac {\left (6 \left (4 a b B+A \left (b^2+4 a c\right )\right )+6 \left (b^2 B+4 A b c+4 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{x^2} \, dx\\ &=-\frac {5 \left (4 a b B+A \left (b^2+4 a c\right )-\left (b^2 B+4 A b c+4 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{8 x}-\frac {5 (A b+2 a B-(b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{12 x^2}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac {5}{96} \int \frac {-6 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right )-6 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right ) x}{x \sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {5 \left (4 a b B+A \left (b^2+4 a c\right )-\left (b^2 B+4 A b c+4 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{8 x}-\frac {5 (A b+2 a B-(b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{12 x^2}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}+\frac {1}{16} \left (5 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx+\frac {1}{16} \left (5 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right )\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {5 \left (4 a b B+A \left (b^2+4 a c\right )-\left (b^2 B+4 A b c+4 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{8 x}-\frac {5 (A b+2 a B-(b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{12 x^2}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}+\frac {1}{8} \left (5 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )-\frac {1}{8} \left (5 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x}{\sqrt {a+b x+c x^2}}\right )\\ &=-\frac {5 \left (4 a b B+A \left (b^2+4 a c\right )-\left (b^2 B+4 A b c+4 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{8 x}-\frac {5 (A b+2 a B-(b B+2 A c) x) \left (a+b x+c x^2\right )^{3/2}}{12 x^2}-\frac {(A-B x) \left (a+b x+c x^2\right )^{5/2}}{3 x^3}-\frac {5 \left (2 a B \left (3 b^2+4 a c\right )+A \left (b^3+12 a b c\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {a}}+\frac {5 \left (b^3 B+6 A b^2 c+12 a b B c+8 a A c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 236, normalized size = 0.93 \begin {gather*} \frac {1}{48} \left (\frac {2 \sqrt {a+x (b+c x)} \left (-4 a^2 (2 A+3 B x)-2 a x (A (13 b+28 c x)+B x (27 b-28 c x))+x^2 \left (A \left (-33 b^2+54 b c x+12 c^2 x^2\right )+B x \left (33 b^2+26 b c x+8 c^2 x^2\right )\right )\right )}{x^3}+\frac {15 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}-\frac {15 \left (A \left (12 a b c+b^3\right )+2 a B \left (4 a c+3 b^2\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a + x*(b + c*x)]*(-4*a^2*(2*A + 3*B*x) - 2*a*x*(B*x*(27*b - 28*c*x) + A*(13*b + 28*c*x)) + x^2*(B*x*(
33*b^2 + 26*b*c*x + 8*c^2*x^2) + A*(-33*b^2 + 54*b*c*x + 12*c^2*x^2))))/x^3 - (15*(2*a*B*(3*b^2 + 4*a*c) + A*(
b^3 + 12*a*b*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/Sqrt[a] + (15*(b^3*B + 6*A*b^2*c + 12
*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/Sqrt[c])/48

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IntegrateAlgebraic [A]  time = 2.53, size = 246, normalized size = 0.96 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-8 a^2 A-12 a^2 B x-26 a A b x-56 a A c x^2-54 a b B x^2+56 a B c x^3-33 A b^2 x^2+54 A b c x^3+12 A c^2 x^4+33 b^2 B x^3+26 b B c x^4+8 B c^2 x^5\right )}{24 x^3}-\frac {5 \left (8 a^2 B c+12 a A b c+6 a b^2 B+A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{8 \sqrt {a}}-\frac {5 \left (8 a A c^2+12 a b B c+6 A b^2 c+b^3 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{16 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-8*a^2*A - 26*a*A*b*x - 12*a^2*B*x - 33*A*b^2*x^2 - 54*a*b*B*x^2 - 56*a*A*c*x^2 + 33*b
^2*B*x^3 + 54*A*b*c*x^3 + 56*a*B*c*x^3 + 26*b*B*c*x^4 + 12*A*c^2*x^4 + 8*B*c^2*x^5))/(24*x^3) - (5*(A*b^3 + 6*
a*b^2*B + 12*a*A*b*c + 8*a^2*B*c)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(8*Sqrt[a]) - (5*(b
^3*B + 6*A*b^2*c + 12*a*b*B*c + 8*a*A*c^2)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(16*Sqrt[c])

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fricas [A]  time = 3.27, size = 1293, normalized size = 5.07

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="fricas")

[Out]

[1/96*(15*(B*a*b^3 + 8*A*a^2*c^2 + 6*(2*B*a^2*b + A*a*b^2)*c)*sqrt(c)*x^3*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*s
qrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 15*(4*(2*B*a^2 + 3*A*a*b)*c^2 + (6*B*a*b^2 + A*b^3)*c)*sqr
t(a)*x^3*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) + 4*(8*
B*a*c^3*x^5 - 8*A*a^3*c + 2*(13*B*a*b*c^2 + 6*A*a*c^3)*x^4 + (33*B*a*b^2*c + 2*(28*B*a^2 + 27*A*a*b)*c^2)*x^3
- 2*(6*B*a^3 + 13*A*a^2*b)*c*x - (56*A*a^2*c^2 + 3*(18*B*a^2*b + 11*A*a*b^2)*c)*x^2)*sqrt(c*x^2 + b*x + a))/(a
*c*x^3), -1/96*(30*(B*a*b^3 + 8*A*a^2*c^2 + 6*(2*B*a^2*b + A*a*b^2)*c)*sqrt(-c)*x^3*arctan(1/2*sqrt(c*x^2 + b*
x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 15*(4*(2*B*a^2 + 3*A*a*b)*c^2 + (6*B*a*b^2 + A*b^3)*c)*
sqrt(a)*x^3*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 4*
(8*B*a*c^3*x^5 - 8*A*a^3*c + 2*(13*B*a*b*c^2 + 6*A*a*c^3)*x^4 + (33*B*a*b^2*c + 2*(28*B*a^2 + 27*A*a*b)*c^2)*x
^3 - 2*(6*B*a^3 + 13*A*a^2*b)*c*x - (56*A*a^2*c^2 + 3*(18*B*a^2*b + 11*A*a*b^2)*c)*x^2)*sqrt(c*x^2 + b*x + a))
/(a*c*x^3), 1/96*(30*(4*(2*B*a^2 + 3*A*a*b)*c^2 + (6*B*a*b^2 + A*b^3)*c)*sqrt(-a)*x^3*arctan(1/2*sqrt(c*x^2 +
b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) + 15*(B*a*b^3 + 8*A*a^2*c^2 + 6*(2*B*a^2*b + A*a*b^2)*c
)*sqrt(c)*x^3*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(8*B*a
*c^3*x^5 - 8*A*a^3*c + 2*(13*B*a*b*c^2 + 6*A*a*c^3)*x^4 + (33*B*a*b^2*c + 2*(28*B*a^2 + 27*A*a*b)*c^2)*x^3 - 2
*(6*B*a^3 + 13*A*a^2*b)*c*x - (56*A*a^2*c^2 + 3*(18*B*a^2*b + 11*A*a*b^2)*c)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*
x^3), 1/48*(15*(4*(2*B*a^2 + 3*A*a*b)*c^2 + (6*B*a*b^2 + A*b^3)*c)*sqrt(-a)*x^3*arctan(1/2*sqrt(c*x^2 + b*x +
a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 15*(B*a*b^3 + 8*A*a^2*c^2 + 6*(2*B*a^2*b + A*a*b^2)*c)*sqrt
(-c)*x^3*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(8*B*a*c^3*x^5 - 8
*A*a^3*c + 2*(13*B*a*b*c^2 + 6*A*a*c^3)*x^4 + (33*B*a*b^2*c + 2*(28*B*a^2 + 27*A*a*b)*c^2)*x^3 - 2*(6*B*a^3 +
13*A*a^2*b)*c*x - (56*A*a^2*c^2 + 3*(18*B*a^2*b + 11*A*a*b^2)*c)*x^2)*sqrt(c*x^2 + b*x + a))/(a*c*x^3)]

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giac [B]  time = 0.62, size = 784, normalized size = 3.07 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, B c^{2} x + \frac {13 \, B b c^{3} + 6 \, A c^{4}}{c^{2}}\right )} x + \frac {33 \, B b^{2} c^{2} + 56 \, B a c^{3} + 54 \, A b c^{3}}{c^{2}}\right )} + \frac {5 \, {\left (6 \, B a b^{2} + A b^{3} + 8 \, B a^{2} c + 12 \, A a b c\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a}} - \frac {5 \, {\left (B b^{3} + 12 \, B a b c + 6 \, A b^{2} c + 8 \, A a c^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, \sqrt {c}} + \frac {54 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} B a b^{2} \sqrt {c} + 33 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} A b^{3} \sqrt {c} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} B a^{2} c^{\frac {3}{2}} + 108 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} A a b c^{\frac {3}{2}} + 144 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} B a^{2} b c + 144 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} A a b^{2} c + 144 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} A a^{2} c^{2} - 96 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} B a^{2} b^{2} \sqrt {c} - 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a b^{3} \sqrt {c} - 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a^{2} b c^{\frac {3}{2}} - 240 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} B a^{3} b c - 144 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a^{2} b^{2} c - 192 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a^{3} c^{2} + 42 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{3} b^{2} \sqrt {c} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{2} b^{3} \sqrt {c} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{4} c^{\frac {3}{2}} + 36 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{3} b c^{\frac {3}{2}} + 96 \, B a^{4} b c + 48 \, A a^{3} b^{2} c + 112 \, A a^{4} c^{2}}{24 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )}^{3} \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*B*c^2*x + (13*B*b*c^3 + 6*A*c^4)/c^2)*x + (33*B*b^2*c^2 + 56*B*a*c^3 + 54*A*b
*c^3)/c^2) + 5/8*(6*B*a*b^2 + A*b^3 + 8*B*a^2*c + 12*A*a*b*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt
(-a))/sqrt(-a) - 5/16*(B*b^3 + 12*B*a*b*c + 6*A*b^2*c + 8*A*a*c^2)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a
))*sqrt(c) + b))/sqrt(c) + 1/24*(54*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a*b^2*sqrt(c) + 33*(sqrt(c)*x - sq
rt(c*x^2 + b*x + a))^5*A*b^3*sqrt(c) + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*c^(3/2) + 108*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^5*A*a*b*c^(3/2) + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^2*b*c + 144*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^4*A*a*b^2*c + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^2*c^2 - 96*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^3*B*a^2*b^2*sqrt(c) - 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*b^3*sqrt(c) - 48*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^2*b*c^(3/2) - 240*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^3*b*c - 1
44*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^2*b^2*c - 192*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^3*c^2 + 4
2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^3*b^2*sqrt(c) + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^3*sqr
t(c) - 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^4*c^(3/2) + 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*b*c
^(3/2) + 96*B*a^4*b*c + 48*A*a^3*b^2*c + 112*A*a^4*c^2)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^3*sqrt(c)
)

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maple [B]  time = 0.07, size = 840, normalized size = 3.29 \begin {gather*} \frac {5 A a \,c^{\frac {3}{2}} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}-\frac {15 A \sqrt {a}\, b c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{4}-\frac {5 A \,b^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 \sqrt {a}}+\frac {15 A \,b^{2} \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8}-\frac {5 B \,a^{\frac {3}{2}} c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2}+\frac {15 B a b \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4}-\frac {15 B \sqrt {a}\, b^{2} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{8}+\frac {5 B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 \sqrt {c}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} c x}{8 a}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,c^{2} x}{2}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, B b c x}{2}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3}}{8 a}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,c^{2} x}{3 a}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2} c x}{24 a^{2}}+5 \sqrt {c \,x^{2}+b x +a}\, A b c +\frac {5 \sqrt {c \,x^{2}+b x +a}\, B a c}{2}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b c x}{4 a}+\frac {25 \sqrt {c \,x^{2}+b x +a}\, B \,b^{2}}{8}+\frac {25 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b c}{12 a}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{3}}{24 a^{2}}+\frac {4 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A \,c^{2} x}{3 a^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A \,b^{2} c x}{8 a^{3}}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2}}{4 a}+\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B b c x}{4 a^{2}}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B c}{6}+\frac {17 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A b c}{12 a^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} A \,b^{3}}{8 a^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B c}{2 a}+\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} B \,b^{2}}{4 a^{2}}-\frac {4 \left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} A c}{3 a^{2} x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} A \,b^{2}}{8 a^{3} x}-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} B b}{4 a^{2} x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} A b}{12 a^{2} x^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} B}{2 a \,x^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {7}{2}} A}{3 a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*B/a^2*b*c*(c*x^2+b*x+a)^(5/2)*x+5/24*A/a^2*b^2*c*(c*x^2+b*x+a)^(3/2)*x+1/8*A/a^3*b^2*c*(c*x^2+b*x+a)^(5/2)
*x+5/8*A/a*b^2*(c*x^2+b*x+a)^(1/2)*x*c+5/4*B/a*b*c*(c*x^2+b*x+a)^(3/2)*x+5/6*B*c*(c*x^2+b*x+a)^(3/2)+25/8*B*b^
2*(c*x^2+b*x+a)^(1/2)-3/4*B/a^2*b/x*(c*x^2+b*x+a)^(7/2)+5/2*B*b*(c*x^2+b*x+a)^(1/2)*x*c+5*A*b*c*(c*x^2+b*x+a)^
(1/2)-5/16*A/a^(1/2)*b^3*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/3*A/a/x^3*(c*x^2+b*x+a)^(7/2)+5/2*A*c
^(3/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+5/2*A*c^2*(c*x^2+b*x+a)^(1/2)*x+5/8*A/a*b^3*(c*x^2+b*x+a)
^(1/2)+15/8*A*b^2*c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/8*A/a^3*b^3*(c*x^2+b*x+a)^(5/2)+5/24*A
/a^2*b^3*(c*x^2+b*x+a)^(3/2)+5/2*B*c*a*(c*x^2+b*x+a)^(1/2)+1/2*B*c/a*(c*x^2+b*x+a)^(5/2)-5/2*B*c*a^(3/2)*ln((b
*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/2*B/a/x^2*(c*x^2+b*x+a)^(7/2)-15/8*B*a^(1/2)*b^2*ln((b*x+2*a+2*(c*x
^2+b*x+a)^(1/2)*a^(1/2))/x)+3/4*B/a^2*b^2*(c*x^2+b*x+a)^(5/2)+5/4*B/a*b^2*(c*x^2+b*x+a)^(3/2)+5/16*B*b^3/c^(1/
2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+15/4*B*a*b*c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-
15/4*A*a^(1/2)*b*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+5/3*A*c^2/a*(c*x^2+b*x+a)^(3/2)*x-4/3*A*c/a^2
/x*(c*x^2+b*x+a)^(7/2)+25/12*A/a*b*c*(c*x^2+b*x+a)^(3/2)-1/12*A/a^2*b/x^2*(c*x^2+b*x+a)^(7/2)-1/8*A/a^3*b^2/x*
(c*x^2+b*x+a)^(7/2)+4/3*A*c^2/a^2*(c*x^2+b*x+a)^(5/2)*x+17/12*A/a^2*b*c*(c*x^2+b*x+a)^(5/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(5/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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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 (c\,x^2+b\,x+a\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 + b*x + c*x^2)^(5/2))/x^4,x)

[Out]

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

[Out]

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

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