3.4.15 \(\int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx\) [315]

Optimal. Leaf size=149 \[ \frac {e (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {e \left (64 c^2 d^2-54 b c d e+15 b^2 e^2+10 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{24 c^3}+\frac {(2 c d-b e) \left (8 c^2 d^2-8 b c d e+5 b^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}} \]

[Out]

1/8*(-b*e+2*c*d)*(5*b^2*e^2-8*b*c*d*e+8*c^2*d^2)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)+1/3*e*(e*x+d)^2*
(c*x^2+b*x)^(1/2)/c+1/24*e*(64*c^2*d^2-54*b*c*d*e+15*b^2*e^2+10*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x)^(1/2)/c^3

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Rubi [A]
time = 0.10, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {756, 793, 634, 212} \begin {gather*} \frac {(2 c d-b e) \left (5 b^2 e^2-8 b c d e+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}+\frac {e \sqrt {b x+c x^2} \left (15 b^2 e^2+10 c e x (2 c d-b e)-54 b c d e+64 c^2 d^2\right )}{24 c^3}+\frac {e \sqrt {b x+c x^2} (d+e x)^2}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/Sqrt[b*x + c*x^2],x]

[Out]

(e*(d + e*x)^2*Sqrt[b*x + c*x^2])/(3*c) + (e*(64*c^2*d^2 - 54*b*c*d*e + 15*b^2*e^2 + 10*c*e*(2*c*d - b*e)*x)*S
qrt[b*x + c*x^2])/(24*c^3) + ((2*c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e + 5*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x +
c*x^2]])/(8*c^(7/2))

Rule 212

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

Rule 634

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 756

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]

Rule 793

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]

Rubi steps

\begin {align*} \int \frac {(d+e x)^3}{\sqrt {b x+c x^2}} \, dx &=\frac {e (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {\int \frac {(d+e x) \left (\frac {1}{2} d (6 c d-b e)+\frac {5}{2} e (2 c d-b e) x\right )}{\sqrt {b x+c x^2}} \, dx}{3 c}\\ &=\frac {e (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {e \left (64 c^2 d^2-54 b c d e+15 b^2 e^2+10 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{24 c^3}+\frac {\left ((2 c d-b e) \left (8 c^2 d^2-8 b c d e+5 b^2 e^2\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{16 c^3}\\ &=\frac {e (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {e \left (64 c^2 d^2-54 b c d e+15 b^2 e^2+10 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{24 c^3}+\frac {\left ((2 c d-b e) \left (8 c^2 d^2-8 b c d e+5 b^2 e^2\right )\right ) \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{8 c^3}\\ &=\frac {e (d+e x)^2 \sqrt {b x+c x^2}}{3 c}+\frac {e \left (64 c^2 d^2-54 b c d e+15 b^2 e^2+10 c e (2 c d-b e) x\right ) \sqrt {b x+c x^2}}{24 c^3}+\frac {(2 c d-b e) \left (8 c^2 d^2-8 b c d e+5 b^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 156, normalized size = 1.05 \begin {gather*} \frac {\sqrt {c} e x (b+c x) \left (15 b^2 e^2-2 b c e (27 d+5 e x)+4 c^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+3 \left (-16 c^3 d^3+24 b c^2 d^2 e-18 b^2 c d e^2+5 b^3 e^3\right ) \sqrt {x} \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{24 c^{7/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[c]*e*x*(b + c*x)*(15*b^2*e^2 - 2*b*c*e*(27*d + 5*e*x) + 4*c^2*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 3*(-16*c
^3*d^3 + 24*b*c^2*d^2*e - 18*b^2*c*d*e^2 + 5*b^3*e^3)*Sqrt[x]*Sqrt[b + c*x]*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b +
c*x]])/(24*c^(7/2)*Sqrt[x*(b + c*x)])

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Maple [A]
time = 0.44, size = 261, normalized size = 1.75

method	result	size
risch	\(\frac {\left (8 c^{2} e^{2} x^{2}-10 b c \,e^{2} x +36 c^{2} d x e +15 b^{2} e^{2}-54 b c d e +72 d^{2} c^{2}\right ) e x \left (c x +b \right )}{24 c^{3} \sqrt {x \left (c x +b \right )}}-\frac {5 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) b^{3} e^{3}}{16 c^{\frac {7}{2}}}+\frac {9 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) b^{2} d \,e^{2}}{8 c^{\frac {5}{2}}}-\frac {3 \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right ) b \,d^{2} e}{2 c^{\frac {3}{2}}}+\frac {d^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}\)	\(209\)
default	\(e^{3} \left (\frac {x^{2} \sqrt {c \,x^{2}+b x}}{3 c}-\frac {5 b \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )}{6 c}\right )+3 d \,e^{2} \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )+3 d^{2} e \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {d^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}\)	\(261\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+b*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

e^3*(1/3*x^2*(c*x^2+b*x)^(1/2)/c-5/6*b/c*(1/2*x*(c*x^2+b*x)^(1/2)/c-3/4*b/c*((c*x^2+b*x)^(1/2)/c-1/2*b/c^(3/2)
*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2)))))+3*d*e^2*(1/2*x*(c*x^2+b*x)^(1/2)/c-3/4*b/c*((c*x^2+b*x)^(1/2)/c-
1/2*b/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))))+3*d^2*e*((c*x^2+b*x)^(1/2)/c-1/2*b/c^(3/2)*ln((1/2*b
+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2)))+d^3*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))/c^(1/2)

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Maxima [A]
time = 0.31, size = 255, normalized size = 1.71 \begin {gather*} \frac {d^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{\sqrt {c}} - \frac {3 \, b d^{2} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x} x^{2} e^{3}}{3 \, c} + \frac {3 \, \sqrt {c x^{2} + b x} d x e^{2}}{2 \, c} + \frac {3 \, \sqrt {c x^{2} + b x} d^{2} e}{c} + \frac {9 \, b^{2} d e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} - \frac {5 \, \sqrt {c x^{2} + b x} b x e^{3}}{12 \, c^{2}} - \frac {9 \, \sqrt {c x^{2} + b x} b d e^{2}}{4 \, c^{2}} - \frac {5 \, b^{3} e^{3} \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^{2} e^{3}}{8 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

d^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) - 3/2*b*d^2*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqr
t(c))/c^(3/2) + 1/3*sqrt(c*x^2 + b*x)*x^2*e^3/c + 3/2*sqrt(c*x^2 + b*x)*d*x*e^2/c + 3*sqrt(c*x^2 + b*x)*d^2*e/
c + 9/8*b^2*d*e^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) - 5/12*sqrt(c*x^2 + b*x)*b*x*e^3/c^2 -
9/4*sqrt(c*x^2 + b*x)*b*d*e^2/c^2 - 5/16*b^3*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) + 5/8*sq
rt(c*x^2 + b*x)*b^2*e^3/c^3

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Fricas [A]
time = 1.47, size = 280, normalized size = 1.88 \begin {gather*} \left [-\frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (72 \, c^{3} d^{2} e + {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c\right )} e^{3} + 18 \, {\left (2 \, c^{3} d x - 3 \, b c^{2} d\right )} e^{2}\right )} \sqrt {c x^{2} + b x}}{48 \, c^{4}}, -\frac {3 \, {\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (72 \, c^{3} d^{2} e + {\left (8 \, c^{3} x^{2} - 10 \, b c^{2} x + 15 \, b^{2} c\right )} e^{3} + 18 \, {\left (2 \, c^{3} d x - 3 \, b c^{2} d\right )} e^{2}\right )} \sqrt {c x^{2} + b x}}{24 \, c^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 18*b^2*c*d*e^2 - 5*b^3*e^3)*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*
x)*sqrt(c)) - 2*(72*c^3*d^2*e + (8*c^3*x^2 - 10*b*c^2*x + 15*b^2*c)*e^3 + 18*(2*c^3*d*x - 3*b*c^2*d)*e^2)*sqrt
(c*x^2 + b*x))/c^4, -1/24*(3*(16*c^3*d^3 - 24*b*c^2*d^2*e + 18*b^2*c*d*e^2 - 5*b^3*e^3)*sqrt(-c)*arctan(sqrt(c
*x^2 + b*x)*sqrt(-c)/(c*x)) - (72*c^3*d^2*e + (8*c^3*x^2 - 10*b*c^2*x + 15*b^2*c)*e^3 + 18*(2*c^3*d*x - 3*b*c^
2*d)*e^2)*sqrt(c*x^2 + b*x))/c^4]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{3}}{\sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+b*x)**(1/2),x)

[Out]

Integral((d + e*x)**3/sqrt(x*(b + c*x)), x)

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Giac [A]
time = 1.56, size = 147, normalized size = 0.99 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, x {\left (\frac {4 \, x e^{3}}{c} + \frac {18 \, c^{2} d e^{2} - 5 \, b c e^{3}}{c^{3}}\right )} + \frac {3 \, {\left (24 \, c^{2} d^{2} e - 18 \, b c d e^{2} + 5 \, b^{2} e^{3}\right )}}{c^{3}}\right )} - \frac {{\left (16 \, c^{3} d^{3} - 24 \, b c^{2} d^{2} e + 18 \, b^{2} c d e^{2} - 5 \, b^{3} e^{3}\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*x*(4*x*e^3/c + (18*c^2*d*e^2 - 5*b*c*e^3)/c^3) + 3*(24*c^2*d^2*e - 18*b*c*d*e^2 + 5*
b^2*e^3)/c^3) - 1/16*(16*c^3*d^3 - 24*b*c^2*d^2*e + 18*b^2*c*d*e^2 - 5*b^3*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c
*x^2 + b*x))*sqrt(c) - b))/c^(7/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{\sqrt {c\,x^2+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(b*x + c*x^2)^(1/2),x)

[Out]

int((d + e*x)^3/(b*x + c*x^2)^(1/2), x)

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