3.49 \(\int \frac {\sqrt {c+d x} (A+B x+C x^2)}{\sqrt {e+f x}} \, dx\)

Optimal. Leaf size=246 \[ \frac {\sqrt {c+d x} \sqrt {e+f x} \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^2 f^3}-\frac {(d e-c f) \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^{5/2} f^{7/2}}-\frac {(c+d x)^{3/2} \sqrt {e+f x} (-6 B d f+7 c C f+5 C d e)}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f} \]

[Out]

-1/8*(-c*f+d*e)*(C*(c^2*f^2+2*c*d*e*f+5*d^2*e^2)+2*d*f*(4*A*d*f-B*(c*f+3*d*e)))*arctanh(f^(1/2)*(d*x+c)^(1/2)/
d^(1/2)/(f*x+e)^(1/2))/d^(5/2)/f^(7/2)-1/12*(-6*B*d*f+7*C*c*f+5*C*d*e)*(d*x+c)^(3/2)*(f*x+e)^(1/2)/d^2/f^2+1/3
*C*(d*x+c)^(5/2)*(f*x+e)^(1/2)/d^2/f+1/8*(C*(c^2*f^2+2*c*d*e*f+5*d^2*e^2)+2*d*f*(4*A*d*f-B*(c*f+3*d*e)))*(d*x+
c)^(1/2)*(f*x+e)^(1/2)/d^2/f^3

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Rubi [A]  time = 0.23, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {951, 80, 50, 63, 217, 206} \[ \frac {\sqrt {c+d x} \sqrt {e+f x} \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^2 f^3}-\frac {(d e-c f) \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right ) \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{8 d^{5/2} f^{7/2}}-\frac {(c+d x)^{3/2} \sqrt {e+f x} (-6 B d f+7 c C f+5 C d e)}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c + d*x]*(A + B*x + C*x^2))/Sqrt[e + f*x],x]

[Out]

((C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Sqrt[c + d*x]*Sqrt[e + f*x])/(8*d^2
*f^3) - ((5*C*d*e + 7*c*C*f - 6*B*d*f)*(c + d*x)^(3/2)*Sqrt[e + f*x])/(12*d^2*f^2) + (C*(c + d*x)^(5/2)*Sqrt[e
+ f*x])/(3*d^2*f) - ((d*e - c*f)*(C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Ar
cTanh[(Sqrt[f]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[e + f*x])])/(8*d^(5/2)*f^(7/2))

Rule 50

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

Rule 63

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 80

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

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 217

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

Rule 951

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

Rubi steps

\begin {align*} \int \frac {\sqrt {c+d x} \left (A+B x+C x^2\right )}{\sqrt {e+f x}} \, dx &=\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}+\frac {\int \frac {\sqrt {c+d x} \left (\frac {1}{2} \left (-5 c C d e-c^2 C f+6 A d^2 f\right )-\frac {1}{2} d (5 C d e+7 c C f-6 B d f) x\right )}{\sqrt {e+f x}} \, dx}{3 d^2 f}\\ &=-\frac {(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt {e+f x}}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}+\frac {\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \int \frac {\sqrt {c+d x}}{\sqrt {e+f x}} \, dx}{8 d^2 f^2}\\ &=\frac {\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{8 d^2 f^3}-\frac {(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt {e+f x}}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\left ((d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right )\right ) \int \frac {1}{\sqrt {c+d x} \sqrt {e+f x}} \, dx}{16 d^2 f^3}\\ &=\frac {\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{8 d^2 f^3}-\frac {(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt {e+f x}}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\left ((d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {e-\frac {c f}{d}+\frac {f x^2}{d}}} \, dx,x,\sqrt {c+d x}\right )}{8 d^3 f^3}\\ &=\frac {\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{8 d^2 f^3}-\frac {(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt {e+f x}}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {\left ((d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {f x^2}{d}} \, dx,x,\frac {\sqrt {c+d x}}{\sqrt {e+f x}}\right )}{8 d^3 f^3}\\ &=\frac {\left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \sqrt {c+d x} \sqrt {e+f x}}{8 d^2 f^3}-\frac {(5 C d e+7 c C f-6 B d f) (c+d x)^{3/2} \sqrt {e+f x}}{12 d^2 f^2}+\frac {C (c+d x)^{5/2} \sqrt {e+f x}}{3 d^2 f}-\frac {(d e-c f) \left (C \left (5 d^2 e^2+2 c d e f+c^2 f^2\right )+2 d f (4 A d f-B (3 d e+c f))\right ) \tanh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d} \sqrt {e+f x}}\right )}{8 d^{5/2} f^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 1.07, size = 225, normalized size = 0.91 \[ \frac {-d \sqrt {f} \sqrt {c+d x} (e+f x) \left (C \left (3 c^2 f^2-2 c d f (f x-2 e)+d^2 \left (-15 e^2+10 e f x-8 f^2 x^2\right )\right )-6 d f (4 A d f+B (c f-3 d e+2 d f x))\right )-3 (d e-c f)^{3/2} \sqrt {\frac {d (e+f x)}{d e-c f}} \sinh ^{-1}\left (\frac {\sqrt {f} \sqrt {c+d x}}{\sqrt {d e-c f}}\right ) \left (2 d f (4 A d f-B (c f+3 d e))+C \left (c^2 f^2+2 c d e f+5 d^2 e^2\right )\right )}{24 d^3 f^{7/2} \sqrt {e+f x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[c + d*x]*(A + B*x + C*x^2))/Sqrt[e + f*x],x]

[Out]

(-(d*Sqrt[f]*Sqrt[c + d*x]*(e + f*x)*(-6*d*f*(4*A*d*f + B*(-3*d*e + c*f + 2*d*f*x)) + C*(3*c^2*f^2 - 2*c*d*f*(
-2*e + f*x) + d^2*(-15*e^2 + 10*e*f*x - 8*f^2*x^2)))) - 3*(d*e - c*f)^(3/2)*(C*(5*d^2*e^2 + 2*c*d*e*f + c^2*f^
2) + 2*d*f*(4*A*d*f - B*(3*d*e + c*f)))*Sqrt[(d*(e + f*x))/(d*e - c*f)]*ArcSinh[(Sqrt[f]*Sqrt[c + d*x])/Sqrt[d
*e - c*f]])/(24*d^3*f^(7/2)*Sqrt[e + f*x])

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fricas [A]  time = 1.47, size = 576, normalized size = 2.34 \[ \left [-\frac {3 \, {\left (5 \, C d^{3} e^{3} - 3 \, {\left (C c d^{2} + 2 \, B d^{3}\right )} e^{2} f - {\left (C c^{2} d - 4 \, B c d^{2} - 8 \, A d^{3}\right )} e f^{2} - {\left (C c^{3} - 2 \, B c^{2} d + 8 \, A c d^{2}\right )} f^{3}\right )} \sqrt {d f} \log \left (8 \, d^{2} f^{2} x^{2} + d^{2} e^{2} + 6 \, c d e f + c^{2} f^{2} + 4 \, {\left (2 \, d f x + d e + c f\right )} \sqrt {d f} \sqrt {d x + c} \sqrt {f x + e} + 8 \, {\left (d^{2} e f + c d f^{2}\right )} x\right ) - 4 \, {\left (8 \, C d^{3} f^{3} x^{2} + 15 \, C d^{3} e^{2} f - 2 \, {\left (2 \, C c d^{2} + 9 \, B d^{3}\right )} e f^{2} - 3 \, {\left (C c^{2} d - 2 \, B c d^{2} - 8 \, A d^{3}\right )} f^{3} - 2 \, {\left (5 \, C d^{3} e f^{2} - {\left (C c d^{2} + 6 \, B d^{3}\right )} f^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{96 \, d^{3} f^{4}}, \frac {3 \, {\left (5 \, C d^{3} e^{3} - 3 \, {\left (C c d^{2} + 2 \, B d^{3}\right )} e^{2} f - {\left (C c^{2} d - 4 \, B c d^{2} - 8 \, A d^{3}\right )} e f^{2} - {\left (C c^{3} - 2 \, B c^{2} d + 8 \, A c d^{2}\right )} f^{3}\right )} \sqrt {-d f} \arctan \left (\frac {{\left (2 \, d f x + d e + c f\right )} \sqrt {-d f} \sqrt {d x + c} \sqrt {f x + e}}{2 \, {\left (d^{2} f^{2} x^{2} + c d e f + {\left (d^{2} e f + c d f^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, C d^{3} f^{3} x^{2} + 15 \, C d^{3} e^{2} f - 2 \, {\left (2 \, C c d^{2} + 9 \, B d^{3}\right )} e f^{2} - 3 \, {\left (C c^{2} d - 2 \, B c d^{2} - 8 \, A d^{3}\right )} f^{3} - 2 \, {\left (5 \, C d^{3} e f^{2} - {\left (C c d^{2} + 6 \, B d^{3}\right )} f^{3}\right )} x\right )} \sqrt {d x + c} \sqrt {f x + e}}{48 \, d^{3} f^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(5*C*d^3*e^3 - 3*(C*c*d^2 + 2*B*d^3)*e^2*f - (C*c^2*d - 4*B*c*d^2 - 8*A*d^3)*e*f^2 - (C*c^3 - 2*B*c^
2*d + 8*A*c*d^2)*f^3)*sqrt(d*f)*log(8*d^2*f^2*x^2 + d^2*e^2 + 6*c*d*e*f + c^2*f^2 + 4*(2*d*f*x + d*e + c*f)*sq
rt(d*f)*sqrt(d*x + c)*sqrt(f*x + e) + 8*(d^2*e*f + c*d*f^2)*x) - 4*(8*C*d^3*f^3*x^2 + 15*C*d^3*e^2*f - 2*(2*C*
c*d^2 + 9*B*d^3)*e*f^2 - 3*(C*c^2*d - 2*B*c*d^2 - 8*A*d^3)*f^3 - 2*(5*C*d^3*e*f^2 - (C*c*d^2 + 6*B*d^3)*f^3)*x
)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^4), 1/48*(3*(5*C*d^3*e^3 - 3*(C*c*d^2 + 2*B*d^3)*e^2*f - (C*c^2*d - 4*B*
c*d^2 - 8*A*d^3)*e*f^2 - (C*c^3 - 2*B*c^2*d + 8*A*c*d^2)*f^3)*sqrt(-d*f)*arctan(1/2*(2*d*f*x + d*e + c*f)*sqrt
(-d*f)*sqrt(d*x + c)*sqrt(f*x + e)/(d^2*f^2*x^2 + c*d*e*f + (d^2*e*f + c*d*f^2)*x)) + 2*(8*C*d^3*f^3*x^2 + 15*
C*d^3*e^2*f - 2*(2*C*c*d^2 + 9*B*d^3)*e*f^2 - 3*(C*c^2*d - 2*B*c*d^2 - 8*A*d^3)*f^3 - 2*(5*C*d^3*e*f^2 - (C*c*
d^2 + 6*B*d^3)*f^3)*x)*sqrt(d*x + c)*sqrt(f*x + e))/(d^3*f^4)]

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giac [A]  time = 1.35, size = 315, normalized size = 1.28 \[ \frac {{\left (\sqrt {{\left (d x + c\right )} d f - c d f + d^{2} e} \sqrt {d x + c} {\left (2 \, {\left (d x + c\right )} {\left (\frac {4 \, {\left (d x + c\right )} C}{d^{3} f} - \frac {7 \, C c d^{6} f^{4} - 6 \, B d^{7} f^{4} + 5 \, C d^{7} f^{3} e}{d^{9} f^{5}}\right )} + \frac {3 \, {\left (C c^{2} d^{6} f^{4} - 2 \, B c d^{7} f^{4} + 8 \, A d^{8} f^{4} + 2 \, C c d^{7} f^{3} e - 6 \, B d^{8} f^{3} e + 5 \, C d^{8} f^{2} e^{2}\right )}}{d^{9} f^{5}}\right )} - \frac {3 \, {\left (C c^{3} f^{3} - 2 \, B c^{2} d f^{3} + 8 \, A c d^{2} f^{3} + C c^{2} d f^{2} e - 4 \, B c d^{2} f^{2} e - 8 \, A d^{3} f^{2} e + 3 \, C c d^{2} f e^{2} + 6 \, B d^{3} f e^{2} - 5 \, C d^{3} e^{3}\right )} \log \left ({\left | -\sqrt {d f} \sqrt {d x + c} + \sqrt {{\left (d x + c\right )} d f - c d f + d^{2} e} \right |}\right )}{\sqrt {d f} d^{2} f^{3}}\right )} d}{24 \, {\left | d \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),x, algorithm="giac")

[Out]

1/24*(sqrt((d*x + c)*d*f - c*d*f + d^2*e)*sqrt(d*x + c)*(2*(d*x + c)*(4*(d*x + c)*C/(d^3*f) - (7*C*c*d^6*f^4 -
6*B*d^7*f^4 + 5*C*d^7*f^3*e)/(d^9*f^5)) + 3*(C*c^2*d^6*f^4 - 2*B*c*d^7*f^4 + 8*A*d^8*f^4 + 2*C*c*d^7*f^3*e -
6*B*d^8*f^3*e + 5*C*d^8*f^2*e^2)/(d^9*f^5)) - 3*(C*c^3*f^3 - 2*B*c^2*d*f^3 + 8*A*c*d^2*f^3 + C*c^2*d*f^2*e - 4
*B*c*d^2*f^2*e - 8*A*d^3*f^2*e + 3*C*c*d^2*f*e^2 + 6*B*d^3*f*e^2 - 5*C*d^3*e^3)*log(abs(-sqrt(d*f)*sqrt(d*x +
c) + sqrt((d*x + c)*d*f - c*d*f + d^2*e)))/(sqrt(d*f)*d^2*f^3))*d/abs(d)

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maple [B]  time = 0.02, size = 763, normalized size = 3.10 \[ \frac {\sqrt {d x +c}\, \sqrt {f x +e}\, \left (24 A c \,d^{2} f^{3} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )-24 A \,d^{3} e \,f^{2} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )-6 B \,c^{2} d \,f^{3} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )-12 B c \,d^{2} e \,f^{2} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+18 B \,d^{3} e^{2} f \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+3 C \,c^{3} f^{3} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+3 C \,c^{2} d e \,f^{2} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+9 C c \,d^{2} e^{2} f \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )-15 C \,d^{3} e^{3} \ln \left (\frac {2 d f x +c f +d e +2 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}}{2 \sqrt {d f}}\right )+16 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}\, C \,d^{2} f^{2} x^{2}+24 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, B \,d^{2} f^{2} x +4 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C c d \,f^{2} x -20 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C \,d^{2} e f x +48 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, A \,d^{2} f^{2}+12 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, B c d \,f^{2}-36 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, B \,d^{2} e f -6 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C \,c^{2} f^{2}-8 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C c d e f +30 \sqrt {d f}\, \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, C \,d^{2} e^{2}\right )}{48 \sqrt {\left (d x +c \right ) \left (f x +e \right )}\, \sqrt {d f}\, d^{2} f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),x)

[Out]

1/48*(d*x+c)^(1/2)*(f*x+e)^(1/2)*(16*C*x^2*d^2*f^2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2)+24*A*ln(1/2*(2*d*f*x+c*
f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*c*d^2*f^3-24*A*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f
*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*d^3*e*f^2-6*B*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(
1/2))/(d*f)^(1/2))*c^2*d*f^3-12*B*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*
c*d^2*e*f^2+18*B*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*d^3*e^2*f+24*B*(d
*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*x*d^2*f^2+3*C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))
/(d*f)^(1/2))*c^3*f^3+3*C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*c^2*d*e*
f^2+9*C*ln(1/2*(2*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*c*d^2*e^2*f-15*C*ln(1/2*(2
*d*f*x+c*f+d*e+2*((d*x+c)*(f*x+e))^(1/2)*(d*f)^(1/2))/(d*f)^(1/2))*d^3*e^3+4*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(
1/2)*x*c*d*f^2-20*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*x*d^2*e*f+48*A*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d^2
*f^2+12*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c*d*f^2-36*B*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*d^2*e*f-6*C*(d*
f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c^2*f^2-8*C*(d*f)^(1/2)*((d*x+c)*(f*x+e))^(1/2)*c*d*e*f+30*C*(d*f)^(1/2)*((d*
x+c)*(f*x+e))^(1/2)*d^2*e^2)/f^3/((d*x+c)*(f*x+e))^(1/2)/d^2/(d*f)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x^2+B*x+A)*(d*x+c)^(1/2)/(f*x+e)^(1/2),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(c*f-d*e>0)', see `assume?` for
more details)Is c*f-d*e zero or nonzero?

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mupad [B]  time = 90.55, size = 1832, normalized size = 7.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((c + d*x)^(1/2)*(A + B*x + C*x^2))/(e + f*x)^(1/2),x)

[Out]

((((c + d*x)^(1/2) - c^(1/2))*(2*A*d^2*e + 2*A*c*d*f))/(f^3*((e + f*x)^(1/2) - e^(1/2))) + ((2*A*c*f + 2*A*d*e
)*((c + d*x)^(1/2) - c^(1/2))^3)/(f^2*((e + f*x)^(1/2) - e^(1/2))^3) - (8*A*c^(1/2)*d*e^(1/2)*((c + d*x)^(1/2)
- c^(1/2))^2)/(f^2*((e + f*x)^(1/2) - e^(1/2))^2))/(((c + d*x)^(1/2) - c^(1/2))^4/((e + f*x)^(1/2) - e^(1/2))
^4 + d^2/f^2 - (2*d*((c + d*x)^(1/2) - c^(1/2))^2)/(f*((e + f*x)^(1/2) - e^(1/2))^2)) - ((((c + d*x)^(1/2) - c
^(1/2))*((C*c^3*d^3*f^3)/4 - (5*C*d^6*e^3)/4 + (C*c^2*d^4*e*f^2)/4 + (3*C*c*d^5*e^2*f)/4))/(f^9*((e + f*x)^(1/
2) - e^(1/2))) - (((c + d*x)^(1/2) - c^(1/2))^5*((33*C*d^4*e^3)/2 + (19*C*c^3*d*f^3)/2 + (275*C*c^2*d^2*e*f^2)
/2 + (313*C*c*d^3*e^2*f)/2))/(f^7*((e + f*x)^(1/2) - e^(1/2))^5) - (((c + d*x)^(1/2) - c^(1/2))^7*((19*C*c^3*f
^3)/2 + (33*C*d^3*e^3)/2 + (313*C*c*d^2*e^2*f)/2 + (275*C*c^2*d*e*f^2)/2))/(f^6*((e + f*x)^(1/2) - e^(1/2))^7)
- (((c + d*x)^(1/2) - c^(1/2))^3*((17*C*c^3*d^2*f^3)/12 - (85*C*d^5*e^3)/12 + (91*C*c^2*d^3*e*f^2)/4 + (17*C*
c*d^4*e^2*f)/4))/(f^8*((e + f*x)^(1/2) - e^(1/2))^3) + (((c + d*x)^(1/2) - c^(1/2))^11*((C*c^3*f^3)/4 - (5*C*d
^3*e^3)/4 + (3*C*c*d^2*e^2*f)/4 + (C*c^2*d*e*f^2)/4))/(d^2*f^4*((e + f*x)^(1/2) - e^(1/2))^11) - (((c + d*x)^(
1/2) - c^(1/2))^9*((17*C*c^3*f^3)/12 - (85*C*d^3*e^3)/12 + (17*C*c*d^2*e^2*f)/4 + (91*C*c^2*d*e*f^2)/4))/(d*f^
5*((e + f*x)^(1/2) - e^(1/2))^9) + (c^(1/2)*e^(1/2)*((c + d*x)^(1/2) - c^(1/2))^8*(32*C*c^2*f + 96*C*c*d*e))/(
f^4*((e + f*x)^(1/2) - e^(1/2))^8) + (c^(1/2)*e^(1/2)*(96*C*c*d^3*e + 32*C*c^2*d^2*f)*((c + d*x)^(1/2) - c^(1/
2))^4)/(f^6*((e + f*x)^(1/2) - e^(1/2))^4) + (c^(1/2)*e^(1/2)*((c + d*x)^(1/2) - c^(1/2))^6*(128*C*d^3*e^2 + 6
4*C*c^2*d*f^2 + (704*C*c*d^2*e*f)/3))/(f^6*((e + f*x)^(1/2) - e^(1/2))^6))/(((c + d*x)^(1/2) - c^(1/2))^12/((e
+ f*x)^(1/2) - e^(1/2))^12 + d^6/f^6 - (6*d*((c + d*x)^(1/2) - c^(1/2))^10)/(f*((e + f*x)^(1/2) - e^(1/2))^10
) - (6*d^5*((c + d*x)^(1/2) - c^(1/2))^2)/(f^5*((e + f*x)^(1/2) - e^(1/2))^2) + (15*d^4*((c + d*x)^(1/2) - c^(
1/2))^4)/(f^4*((e + f*x)^(1/2) - e^(1/2))^4) - (20*d^3*((c + d*x)^(1/2) - c^(1/2))^6)/(f^3*((e + f*x)^(1/2) -
e^(1/2))^6) + (15*d^2*((c + d*x)^(1/2) - c^(1/2))^8)/(f^2*((e + f*x)^(1/2) - e^(1/2))^8)) + ((((c + d*x)^(1/2)
- c^(1/2))*((B*c^2*d^2*f^2)/2 - (3*B*d^4*e^2)/2 + B*c*d^3*e*f))/(f^6*((e + f*x)^(1/2) - e^(1/2))) + (((c + d*
x)^(1/2) - c^(1/2))^3*((11*B*d^3*e^2)/2 + (7*B*c^2*d*f^2)/2 + 23*B*c*d^2*e*f))/(f^5*((e + f*x)^(1/2) - e^(1/2)
)^3) + (((c + d*x)^(1/2) - c^(1/2))^5*((7*B*c^2*f^2)/2 + (11*B*d^2*e^2)/2 + 23*B*c*d*e*f))/(f^4*((e + f*x)^(1/
2) - e^(1/2))^5) + (((c + d*x)^(1/2) - c^(1/2))^7*((B*c^2*f^2)/2 - (3*B*d^2*e^2)/2 + B*c*d*e*f))/(d*f^3*((e +
f*x)^(1/2) - e^(1/2))^7) - (c^(1/2)*e^(1/2)*((c + d*x)^(1/2) - c^(1/2))^4*(32*B*d^2*e + 16*B*c*d*f))/(f^4*((e
+ f*x)^(1/2) - e^(1/2))^4) - (8*B*c^(3/2)*e^(1/2)*((c + d*x)^(1/2) - c^(1/2))^6)/(f^2*((e + f*x)^(1/2) - e^(1/
2))^6) - (8*B*c^(3/2)*d^2*e^(1/2)*((c + d*x)^(1/2) - c^(1/2))^2)/(f^4*((e + f*x)^(1/2) - e^(1/2))^2))/(((c + d
*x)^(1/2) - c^(1/2))^8/((e + f*x)^(1/2) - e^(1/2))^8 + d^4/f^4 - (4*d*((c + d*x)^(1/2) - c^(1/2))^6)/(f*((e +
f*x)^(1/2) - e^(1/2))^6) - (4*d^3*((c + d*x)^(1/2) - c^(1/2))^2)/(f^3*((e + f*x)^(1/2) - e^(1/2))^2) + (6*d^2*
((c + d*x)^(1/2) - c^(1/2))^4)/(f^2*((e + f*x)^(1/2) - e^(1/2))^4)) + (2*A*atanh((f^(1/2)*((c + d*x)^(1/2) - c
^(1/2)))/(d^(1/2)*((e + f*x)^(1/2) - e^(1/2))))*(c*f - d*e))/(d^(1/2)*f^(3/2)) + (C*atanh((f^(1/2)*((c + d*x)^
(1/2) - c^(1/2)))/(d^(1/2)*((e + f*x)^(1/2) - e^(1/2))))*(c*f - d*e)*(c^2*f^2 + 5*d^2*e^2 + 2*c*d*e*f))/(4*d^(
5/2)*f^(7/2)) - (B*atanh((f^(1/2)*((c + d*x)^(1/2) - c^(1/2)))/(d^(1/2)*((e + f*x)^(1/2) - e^(1/2))))*(c*f - d
*e)*(c*f + 3*d*e))/(2*d^(3/2)*f^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((C*x**2+B*x+A)*(d*x+c)**(1/2)/(f*x+e)**(1/2),x)

[Out]

Timed out

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