3.1192 \(\int \frac {(A+B x) (d+e x)^2}{\sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=189 \[ \frac {\sqrt {b x+c x^2} \left (2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)+B \left (15 b^2 e^2-36 b c d e+16 c^2 d^2\right )\right )}{24 c^3}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (6 b^2 c e (A e+2 B d)-8 b c^2 d (2 A e+B d)+16 A c^3 d^2-5 b^3 B e^2\right )}{8 c^{7/2}}+\frac {B \sqrt {b x+c x^2} (d+e x)^2}{3 c} \]

[Out]

1/8*(16*A*c^3*d^2-5*b^3*B*e^2+6*b^2*c*e*(A*e+2*B*d)-8*b*c^2*d*(2*A*e+B*d))*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2)
)/c^(7/2)+1/3*B*(e*x+d)^2*(c*x^2+b*x)^(1/2)/c+1/24*(6*A*c*e*(-3*b*e+8*c*d)+B*(15*b^2*e^2-36*b*c*d*e+16*c^2*d^2
)+2*c*e*(6*A*c*e-5*B*b*e+4*B*c*d)*x)*(c*x^2+b*x)^(1/2)/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {832, 779, 620, 206} \[ \frac {\sqrt {b x+c x^2} \left (2 c e x (6 A c e-5 b B e+4 B c d)+6 A c e (8 c d-3 b e)+B \left (15 b^2 e^2-36 b c d e+16 c^2 d^2\right )\right )}{24 c^3}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (6 b^2 c e (A e+2 B d)-8 b c^2 d (2 A e+B d)+16 A c^3 d^2-5 b^3 B e^2\right )}{8 c^{7/2}}+\frac {B \sqrt {b x+c x^2} (d+e x)^2}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(d + e*x)^2*Sqrt[b*x + c*x^2])/(3*c) + ((6*A*c*e*(8*c*d - 3*b*e) + B*(16*c^2*d^2 - 36*b*c*d*e + 15*b^2*e^2)
+ 2*c*e*(4*B*c*d - 5*b*B*e + 6*A*c*e)*x)*Sqrt[b*x + c*x^2])/(24*c^3) + ((16*A*c^3*d^2 - 5*b^3*B*e^2 + 6*b^2*c
*e*(2*B*d + A*e) - 8*b*c^2*d*(B*d + 2*A*e))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(8*c^(7/2))

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 620

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

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 832

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

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^2}{\sqrt {b x+c x^2}} \, dx &=\frac {B (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {\int \frac {(d+e x) \left (-\frac {1}{2} (b B-6 A c) d+\frac {1}{2} (4 B c d-5 b B e+6 A c e) x\right )}{\sqrt {b x+c x^2}} \, dx}{3 c}\\ &=\frac {B (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {\left (6 A c e (8 c d-3 b e)+B \left (16 c^2 d^2-36 b c d e+15 b^2 e^2\right )+2 c e (4 B c d-5 b B e+6 A c e) x\right ) \sqrt {b x+c x^2}}{24 c^3}+\frac {\left (16 A c^3 d^2-5 b^3 B e^2+6 b^2 c e (2 B d+A e)-8 b c^2 d (B d+2 A e)\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{16 c^3}\\ &=\frac {B (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {\left (6 A c e (8 c d-3 b e)+B \left (16 c^2 d^2-36 b c d e+15 b^2 e^2\right )+2 c e (4 B c d-5 b B e+6 A c e) x\right ) \sqrt {b x+c x^2}}{24 c^3}+\frac {\left (16 A c^3 d^2-5 b^3 B e^2+6 b^2 c e (2 B d+A e)-8 b c^2 d (B d+2 A e)\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{8 c^3}\\ &=\frac {B (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {\left (6 A c e (8 c d-3 b e)+B \left (16 c^2 d^2-36 b c d e+15 b^2 e^2\right )+2 c e (4 B c d-5 b B e+6 A c e) x\right ) \sqrt {b x+c x^2}}{24 c^3}+\frac {\left (16 A c^3 d^2-5 b^3 B e^2+6 b^2 c e (2 B d+A e)-8 b c^2 d (B d+2 A e)\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 190, normalized size = 1.01 \[ \frac {\sqrt {c} x (b+c x) \left (6 A c e (-3 b e+8 c d+2 c e x)+B \left (15 b^2 e^2-2 b c e (18 d+5 e x)+8 c^2 \left (3 d^2+3 d e x+e^2 x^2\right )\right )\right )-3 \sqrt {b} \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right ) \left (-6 b^2 c e (A e+2 B d)+8 b c^2 d (2 A e+B d)-16 A c^3 d^2+5 b^3 B e^2\right )}{24 c^{7/2} \sqrt {x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(b + c*x)*(6*A*c*e*(8*c*d - 3*b*e + 2*c*e*x) + B*(15*b^2*e^2 - 2*b*c*e*(18*d + 5*e*x) + 8*c^2*(3*d^
2 + 3*d*e*x + e^2*x^2))) - 3*Sqrt[b]*(-16*A*c^3*d^2 + 5*b^3*B*e^2 - 6*b^2*c*e*(2*B*d + A*e) + 8*b*c^2*d*(B*d +
2*A*e))*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(24*c^(7/2)*Sqrt[x*(b + c*x)])

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 391, normalized size = 2.07 \[ \left [-\frac {3 \, {\left (8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \, {\left (3 \, B b^{2} c - 4 \, A b c^{2}\right )} d e + {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} e^{2} x^{2} + 24 \, B c^{3} d^{2} - 12 \, {\left (3 \, B b c^{2} - 4 \, A c^{3}\right )} d e + 3 \, {\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} e^{2} + 2 \, {\left (12 \, B c^{3} d e - {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, \frac {3 \, {\left (8 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \, {\left (3 \, B b^{2} c - 4 \, A b c^{2}\right )} d e + {\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, B c^{3} e^{2} x^{2} + 24 \, B c^{3} d^{2} - 12 \, {\left (3 \, B b c^{2} - 4 \, A c^{3}\right )} d e + 3 \, {\left (5 \, B b^{2} c - 6 \, A b c^{2}\right )} e^{2} + 2 \, {\left (12 \, B c^{3} d e - {\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(3*(8*(B*b*c^2 - 2*A*c^3)*d^2 - 4*(3*B*b^2*c - 4*A*b*c^2)*d*e + (5*B*b^3 - 6*A*b^2*c)*e^2)*sqrt(c)*log(
2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^3*e^2*x^2 + 24*B*c^3*d^2 - 12*(3*B*b*c^2 - 4*A*c^3)*d*e +
3*(5*B*b^2*c - 6*A*b*c^2)*e^2 + 2*(12*B*c^3*d*e - (5*B*b*c^2 - 6*A*c^3)*e^2)*x)*sqrt(c*x^2 + b*x))/c^4, 1/24*(
3*(8*(B*b*c^2 - 2*A*c^3)*d^2 - 4*(3*B*b^2*c - 4*A*b*c^2)*d*e + (5*B*b^3 - 6*A*b^2*c)*e^2)*sqrt(-c)*arctan(sqrt
(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^3*e^2*x^2 + 24*B*c^3*d^2 - 12*(3*B*b*c^2 - 4*A*c^3)*d*e + 3*(5*B*b^2*c
- 6*A*b*c^2)*e^2 + 2*(12*B*c^3*d*e - (5*B*b*c^2 - 6*A*c^3)*e^2)*x)*sqrt(c*x^2 + b*x))/c^4]

________________________________________________________________________________________

giac [A]  time = 0.28, size = 196, normalized size = 1.04 \[ \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (\frac {4 \, B x e^{2}}{c} + \frac {12 \, B c^{2} d e - 5 \, B b c e^{2} + 6 \, A c^{2} e^{2}}{c^{3}}\right )} x + \frac {3 \, {\left (8 \, B c^{2} d^{2} - 12 \, B b c d e + 16 \, A c^{2} d e + 5 \, B b^{2} e^{2} - 6 \, A b c e^{2}\right )}}{c^{3}}\right )} + \frac {{\left (8 \, B b c^{2} d^{2} - 16 \, A c^{3} d^{2} - 12 \, B b^{2} c d e + 16 \, A b c^{2} d e + 5 \, B b^{3} e^{2} - 6 \, A b^{2} c e^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*(4*B*x*e^2/c + (12*B*c^2*d*e - 5*B*b*c*e^2 + 6*A*c^2*e^2)/c^3)*x + 3*(8*B*c^2*d^2 -
12*B*b*c*d*e + 16*A*c^2*d*e + 5*B*b^2*e^2 - 6*A*b*c*e^2)/c^3) + 1/16*(8*B*b*c^2*d^2 - 16*A*c^3*d^2 - 12*B*b^2*
c*d*e + 16*A*b*c^2*d*e + 5*B*b^3*e^2 - 6*A*b^2*c*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))
/c^(7/2)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 395, normalized size = 2.09 \[ \frac {\sqrt {c \,x^{2}+b x}\, B \,e^{2} x^{2}}{3 c}+\frac {3 A \,b^{2} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}-\frac {A b d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}+\frac {A \,d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}-\frac {5 B \,b^{3} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {7}{2}}}+\frac {3 B \,b^{2} d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{4 c^{\frac {5}{2}}}-\frac {B b \,d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+b x}\, A \,e^{2} x}{2 c}-\frac {5 \sqrt {c \,x^{2}+b x}\, B b \,e^{2} x}{12 c^{2}}+\frac {\sqrt {c \,x^{2}+b x}\, B d e x}{c}-\frac {3 \sqrt {c \,x^{2}+b x}\, A b \,e^{2}}{4 c^{2}}+\frac {2 \sqrt {c \,x^{2}+b x}\, A d e}{c}+\frac {5 \sqrt {c \,x^{2}+b x}\, B \,b^{2} e^{2}}{8 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x}\, B b d e}{2 c^{2}}+\frac {\sqrt {c \,x^{2}+b x}\, B \,d^{2}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.68, size = 299, normalized size = 1.58 \[ \frac {\sqrt {c x^{2} + b x} B e^{2} x^{2}}{3 \, c} - \frac {5 \, \sqrt {c x^{2} + b x} B b e^{2} x}{12 \, c^{2}} + \frac {A d^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{\sqrt {c}} - \frac {5 \, B b^{3} e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {7}{2}}} + \frac {5 \, \sqrt {c x^{2} + b x} B b^{2} e^{2}}{8 \, c^{3}} + \frac {{\left (2 \, B d e + A e^{2}\right )} \sqrt {c x^{2} + b x} x}{2 \, c} + \frac {3 \, {\left (2 \, B d e + A e^{2}\right )} b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} - \frac {{\left (B d^{2} + 2 \, A d e\right )} b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} - \frac {3 \, {\left (2 \, B d e + A e^{2}\right )} \sqrt {c x^{2} + b x} b}{4 \, c^{2}} + \frac {{\left (B d^{2} + 2 \, A d e\right )} \sqrt {c x^{2} + b x}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(c*x^2 + b*x)*B*e^2*x^2/c - 5/12*sqrt(c*x^2 + b*x)*B*b*e^2*x/c^2 + A*d^2*log(2*c*x + b + 2*sqrt(c*x^2
+ b*x)*sqrt(c))/sqrt(c) - 5/16*B*b^3*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 5/8*sqrt(c*x^2
+ b*x)*B*b^2*e^2/c^3 + 1/2*(2*B*d*e + A*e^2)*sqrt(c*x^2 + b*x)*x/c + 3/8*(2*B*d*e + A*e^2)*b^2*log(2*c*x + b
+ 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 1/2*(B*d^2 + 2*A*d*e)*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/
c^(3/2) - 3/4*(2*B*d*e + A*e^2)*sqrt(c*x^2 + b*x)*b/c^2 + (B*d^2 + 2*A*d*e)*sqrt(c*x^2 + b*x)/c

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^2}{\sqrt {c\,x^2+b\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \left (d + e x\right )^{2}}{\sqrt {x \left (b + c x\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(d + e*x)**2/sqrt(x*(b + c*x)), x)

________________________________________________________________________________________