3.286 \(\int (d+e x)^2 \sqrt {b x+c x^2} \, dx\)

Optimal. Leaf size=162 \[ -\frac {b^2 \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}+\frac {(b+2 c x) \sqrt {b x+c x^2} \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{64 c^3}+\frac {5 e \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{24 c^2}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)}{4 c} \]

[Out]

5/24*e*(-b*e+2*c*d)*(c*x^2+b*x)^(3/2)/c^2+1/4*e*(e*x+d)*(c*x^2+b*x)^(3/2)/c-1/64*b^2*(5*b^2*e^2-16*b*c*d*e+16*
c^2*d^2)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+1/64*(5*b^2*e^2-16*b*c*d*e+16*c^2*d^2)*(2*c*x+b)*(c*x^2+
b*x)^(1/2)/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {742, 640, 612, 620, 206} \[ \frac {(b+2 c x) \sqrt {b x+c x^2} \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{64 c^3}-\frac {b^2 \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}+\frac {5 e \left (b x+c x^2\right )^{3/2} (2 c d-b e)}{24 c^2}+\frac {e \left (b x+c x^2\right )^{3/2} (d+e x)}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*Sqrt[b*x + c*x^2],x]

[Out]

((16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) + (5*e*(2*c*d - b*e)*(b*x + c*x
^2)^(3/2))/(24*c^2) + (e*(d + e*x)*(b*x + c*x^2)^(3/2))/(4*c) - (b^2*(16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*Arc
Tanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*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 612

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

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 640

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

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rubi steps

\begin {align*} \int (d+e x)^2 \sqrt {b x+c x^2} \, dx &=\frac {e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}+\frac {\int \left (\frac {1}{2} d (8 c d-3 b e)+\frac {5}{2} e (2 c d-b e) x\right ) \sqrt {b x+c x^2} \, dx}{4 c}\\ &=\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}+\frac {\left (c d (8 c d-3 b e)-\frac {5}{2} b e (2 c d-b e)\right ) \int \sqrt {b x+c x^2} \, dx}{8 c^2}\\ &=\frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}+\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{128 c^3}\\ &=\frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}+\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{64 c^3}\\ &=\frac {\left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) (b+2 c x) \sqrt {b x+c x^2}}{64 c^3}+\frac {5 e (2 c d-b e) \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac {e (d+e x) \left (b x+c x^2\right )^{3/2}}{4 c}-\frac {b^2 \left (16 c^2 d^2-16 b c d e+5 b^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{64 c^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.28, size = 164, normalized size = 1.01 \[ \frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (15 b^3 e^2-2 b^2 c e (24 d+5 e x)+8 b c^2 \left (6 d^2+4 d e x+e^2 x^2\right )+16 c^3 x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )-\frac {3 b^{3/2} \left (5 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {x} \sqrt {\frac {c x}{b}+1}}\right )}{192 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3*e^2 - 2*b^2*c*e*(24*d + 5*e*x) + 8*b*c^2*(6*d^2 + 4*d*e*x + e^2*x^2) + 16*
c^3*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2)) - (3*b^(3/2)*(16*c^2*d^2 - 16*b*c*d*e + 5*b^2*e^2)*ArcSinh[(Sqrt[c]*Sqrt[
x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/2))

________________________________________________________________________________________

fricas [A]  time = 0.98, size = 337, normalized size = 2.08 \[ \left [\frac {3 \, {\left (16 \, b^{2} c^{2} d^{2} - 16 \, b^{3} c d e + 5 \, b^{4} e^{2}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (48 \, c^{4} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 48 \, b^{2} c^{2} d e + 15 \, b^{3} c e^{2} + 8 \, {\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - 5 \, b^{2} c^{2} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{384 \, c^{4}}, \frac {3 \, {\left (16 \, b^{2} c^{2} d^{2} - 16 \, b^{3} c d e + 5 \, b^{4} e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (48 \, c^{4} e^{2} x^{3} + 48 \, b c^{3} d^{2} - 48 \, b^{2} c^{2} d e + 15 \, b^{3} c e^{2} + 8 \, {\left (16 \, c^{4} d e + b c^{3} e^{2}\right )} x^{2} + 2 \, {\left (48 \, c^{4} d^{2} + 16 \, b c^{3} d e - 5 \, b^{2} c^{2} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{192 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(3*(16*b^2*c^2*d^2 - 16*b^3*c*d*e + 5*b^4*e^2)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2
*(48*c^4*e^2*x^3 + 48*b*c^3*d^2 - 48*b^2*c^2*d*e + 15*b^3*c*e^2 + 8*(16*c^4*d*e + b*c^3*e^2)*x^2 + 2*(48*c^4*d
^2 + 16*b*c^3*d*e - 5*b^2*c^2*e^2)*x)*sqrt(c*x^2 + b*x))/c^4, 1/192*(3*(16*b^2*c^2*d^2 - 16*b^3*c*d*e + 5*b^4*
e^2)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*c^4*e^2*x^3 + 48*b*c^3*d^2 - 48*b^2*c^2*d*e + 15*
b^3*c*e^2 + 8*(16*c^4*d*e + b*c^3*e^2)*x^2 + 2*(48*c^4*d^2 + 16*b*c^3*d*e - 5*b^2*c^2*e^2)*x)*sqrt(c*x^2 + b*x
))/c^4]

________________________________________________________________________________________

giac [A]  time = 0.24, size = 172, normalized size = 1.06 \[ \frac {1}{192} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, {\left (6 \, x e^{2} + \frac {16 \, c^{3} d e + b c^{2} e^{2}}{c^{3}}\right )} x + \frac {48 \, c^{3} d^{2} + 16 \, b c^{2} d e - 5 \, b^{2} c e^{2}}{c^{3}}\right )} x + \frac {3 \, {\left (16 \, b c^{2} d^{2} - 16 \, b^{2} c d e + 5 \, b^{3} e^{2}\right )}}{c^{3}}\right )} + \frac {{\left (16 \, b^{2} c^{2} d^{2} - 16 \, b^{3} c d e + 5 \, b^{4} e^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*x*e^2 + (16*c^3*d*e + b*c^2*e^2)/c^3)*x + (48*c^3*d^2 + 16*b*c^2*d*e - 5*b^2*
c*e^2)/c^3)*x + 3*(16*b*c^2*d^2 - 16*b^2*c*d*e + 5*b^3*e^2)/c^3) + 1/128*(16*b^2*c^2*d^2 - 16*b^3*c*d*e + 5*b^
4*e^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 287, normalized size = 1.77 \[ -\frac {5 b^{4} e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{128 c^{\frac {7}{2}}}+\frac {b^{3} d e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}-\frac {b^{2} d^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{2} e^{2} x}{32 c^{2}}-\frac {\sqrt {c \,x^{2}+b x}\, b d e x}{2 c}+\frac {\sqrt {c \,x^{2}+b x}\, d^{2} x}{2}+\frac {5 \sqrt {c \,x^{2}+b x}\, b^{3} e^{2}}{64 c^{3}}-\frac {\sqrt {c \,x^{2}+b x}\, b^{2} d e}{4 c^{2}}+\frac {\sqrt {c \,x^{2}+b x}\, b \,d^{2}}{4 c}+\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}} e^{2} x}{4 c}-\frac {5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b \,e^{2}}{24 c^{2}}+\frac {2 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} d e}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^2*x*(c*x^2+b*x)^(3/2)/c-5/24*e^2*b/c^2*(c*x^2+b*x)^(3/2)+5/32*e^2*b^2/c^2*x*(c*x^2+b*x)^(1/2)+5/64*e^2*b
^3/c^3*(c*x^2+b*x)^(1/2)-5/128*e^2*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+2/3*d*e*(c*x^2+b*x)^(
3/2)/c-1/2*d*e*b/c*x*(c*x^2+b*x)^(1/2)-1/4*d*e*b^2/c^2*(c*x^2+b*x)^(1/2)+1/8*d*e*b^3/c^(5/2)*ln((c*x+1/2*b)/c^
(1/2)+(c*x^2+b*x)^(1/2))+1/2*d^2*x*(c*x^2+b*x)^(1/2)+1/4*d^2/c*(c*x^2+b*x)^(1/2)*b-1/8*d^2*b^2/c^(3/2)*ln((c*x
+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.40, size = 283, normalized size = 1.75 \[ \frac {1}{2} \, \sqrt {c x^{2} + b x} d^{2} x - \frac {\sqrt {c x^{2} + b x} b d e x}{2 \, c} + \frac {5 \, \sqrt {c x^{2} + b x} b^{2} e^{2} x}{32 \, c^{2}} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} e^{2} x}{4 \, c} - \frac {b^{2} d^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {3}{2}}} + \frac {b^{3} d e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} - \frac {5 \, b^{4} e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{128 \, c^{\frac {7}{2}}} + \frac {\sqrt {c x^{2} + b x} b d^{2}}{4 \, c} - \frac {\sqrt {c x^{2} + b x} b^{2} d e}{4 \, c^{2}} + \frac {2 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} d e}{3 \, c} + \frac {5 \, \sqrt {c x^{2} + b x} b^{3} e^{2}}{64 \, c^{3}} - \frac {5 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b e^{2}}{24 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(c*x^2 + b*x)*d^2*x - 1/2*sqrt(c*x^2 + b*x)*b*d*e*x/c + 5/32*sqrt(c*x^2 + b*x)*b^2*e^2*x/c^2 + 1/4*(c*
x^2 + b*x)^(3/2)*e^2*x/c - 1/8*b^2*d^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + 1/8*b^3*d*e*log(
2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 5/128*b^4*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/
c^(7/2) + 1/4*sqrt(c*x^2 + b*x)*b*d^2/c - 1/4*sqrt(c*x^2 + b*x)*b^2*d*e/c^2 + 2/3*(c*x^2 + b*x)^(3/2)*d*e/c +
5/64*sqrt(c*x^2 + b*x)*b^3*e^2/c^3 - 5/24*(c*x^2 + b*x)^(3/2)*b*e^2/c^2

________________________________________________________________________________________

mupad [B]  time = 0.69, size = 231, normalized size = 1.43 \[ d^2\,\sqrt {c\,x^2+b\,x}\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )-\frac {5\,b\,e^2\,\left (\frac {b^3\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{16\,c^{5/2}}+\frac {\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{24\,c^2}\right )}{8\,c}-\frac {b^2\,d^2\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{8\,c^{3/2}}+\frac {e^2\,x\,{\left (c\,x^2+b\,x\right )}^{3/2}}{4\,c}+\frac {b^3\,d\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x}\right )}{8\,c^{5/2}}+\frac {d\,e\,\sqrt {c\,x^2+b\,x}\,\left (-3\,b^2+2\,b\,c\,x+8\,c^2\,x^2\right )}{12\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*(b*x + c*x^2)^(1/2)*(x/2 + b/(4*c)) - (5*b*e^2*((b^3*log((b + 2*c*x)/c^(1/2) + 2*(b*x + c*x^2)^(1/2)))/(16
*c^(5/2)) + ((b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(24*c^2)))/(8*c) - (b^2*d^2*log((b/2 + c*x)/c^
(1/2) + (b*x + c*x^2)^(1/2)))/(8*c^(3/2)) + (e^2*x*(b*x + c*x^2)^(3/2))/(4*c) + (b^3*d*e*log((b + 2*c*x)/c^(1/
2) + 2*(b*x + c*x^2)^(1/2)))/(8*c^(5/2)) + (d*e*(b*x + c*x^2)^(1/2)*(8*c^2*x^2 - 3*b^2 + 2*b*c*x))/(12*c^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________