3.11.55 \(\int \frac {(A+B x) (d+e x)^2}{(b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=151 \[ \frac {e \sqrt {b x+c x^2} \left (-2 b c (A e+B d)+4 A c^2 d+3 b^2 B e\right )}{b^2 c^2}-\frac {2 (d+e x) \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) (2 A c e-3 b B e+4 B c d)}{c^{5/2}} \]

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Rubi [A]  time = 0.15, antiderivative size = 151, 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 = {818, 640, 620, 206} \begin {gather*} \frac {e \sqrt {b x+c x^2} \left (-2 b c (A e+B d)+4 A c^2 d+3 b^2 B e\right )}{b^2 c^2}-\frac {2 (d+e x) \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) (2 A c e-3 b B e+4 B c d)}{c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(b^2*c*Sqrt[b*x + c*x^2]) + (e*(4*A*c^2*d
+ 3*b^2*B*e - 2*b*c*(B*d + A*e))*Sqrt[b*x + c*x^2])/(b^2*c^2) + (e*(4*B*c*d - 3*b*B*e + 2*A*c*e)*ArcTanh[(Sqr
t[c]*x)/Sqrt[b*x + c*x^2]])/c^(5/2)

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 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 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 150, normalized size = 0.99 \begin {gather*} \frac {\sqrt {c} \left (b B x \left (3 b^2 e^2+b c e (e x-4 d)+2 c^2 d^2\right )-2 A c \left (b^2 e^2 x+b c d (d-2 e x)+2 c^2 d^2 x\right )\right )-b^{5/2} e \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right ) (-2 A c e+3 b B e-4 B c d)}{b^2 c^{5/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(-2*A*c*(2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x)) + b*B*x*(2*c^2*d^2 + 3*b^2*e^2 + b*c*e*(-4*d +
e*x))) - b^(5/2)*e*(-4*B*c*d + 3*b*B*e - 2*A*c*e)*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]]
)/(b^2*c^(5/2)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.55, size = 179, normalized size = 1.19 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (-2 A b^2 c e^2 x-2 A b c^2 d^2+4 A b c^2 d e x-4 A c^3 d^2 x+3 b^3 B e^2 x-4 b^2 B c d e x+b^2 B c e^2 x^2+2 b B c^2 d^2 x\right )}{b^2 c^2 x (b+c x)}+\frac {\log \left (-2 c^{5/2} \sqrt {b x+c x^2}+b c^2+2 c^3 x\right ) \left (-2 A c e^2+3 b B e^2-4 B c d e\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[b*x + c*x^2]*(-2*A*b*c^2*d^2 + 2*b*B*c^2*d^2*x - 4*A*c^3*d^2*x - 4*b^2*B*c*d*e*x + 4*A*b*c^2*d*e*x + 3*b
^3*B*e^2*x - 2*A*b^2*c*e^2*x + b^2*B*c*e^2*x^2))/(b^2*c^2*x*(b + c*x)) + ((-4*B*c*d*e + 3*b*B*e^2 - 2*A*c*e^2)
*Log[b*c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[b*x + c*x^2]])/(2*c^(5/2))

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fricas [A]  time = 0.43, size = 446, normalized size = 2.95 \begin {gather*} \left [\frac {{\left ({\left (4 \, B b^{2} c^{2} d e - {\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x^{2} + {\left (4 \, B b^{3} c d e - {\left (3 \, B b^{4} - 2 \, A b^{3} c\right )} e^{2}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (B b^{2} c^{2} e^{2} x^{2} - 2 \, A b c^{3} d^{2} + {\left (2 \, {\left (B b c^{3} - 2 \, A c^{4}\right )} d^{2} - 4 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} d e + {\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (b^{2} c^{4} x^{2} + b^{3} c^{3} x\right )}}, -\frac {{\left ({\left (4 \, B b^{2} c^{2} d e - {\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x^{2} + {\left (4 \, B b^{3} c d e - {\left (3 \, B b^{4} - 2 \, A b^{3} c\right )} e^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (B b^{2} c^{2} e^{2} x^{2} - 2 \, A b c^{3} d^{2} + {\left (2 \, {\left (B b c^{3} - 2 \, A c^{4}\right )} d^{2} - 4 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} d e + {\left (3 \, B b^{3} c - 2 \, A b^{2} c^{2}\right )} e^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{b^{2} c^{4} x^{2} + b^{3} c^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 155, normalized size = 1.03 \begin {gather*} -\frac {\frac {2 \, A d^{2}}{b} - {\left (\frac {B x e^{2}}{c} + \frac {2 \, B b c^{2} d^{2} - 4 \, A c^{3} d^{2} - 4 \, B b^{2} c d e + 4 \, A b c^{2} d e + 3 \, B b^{3} e^{2} - 2 \, A b^{2} c e^{2}}{b^{2} c^{2}}\right )} x}{\sqrt {c x^{2} + b x}} - \frac {{\left (4 \, B c d e - 3 \, B b e^{2} + 2 \, A c e^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*A*d^2/b - (B*x*e^2/c + (2*B*b*c^2*d^2 - 4*A*c^3*d^2 - 4*B*b^2*c*d*e + 4*A*b*c^2*d*e + 3*B*b^3*e^2 - 2*A*b^
2*c*e^2)/(b^2*c^2))*x)/sqrt(c*x^2 + b*x) - 1/2*(4*B*c*d*e - 3*B*b*e^2 + 2*A*c*e^2)*log(abs(-2*(sqrt(c)*x - sqr
t(c*x^2 + b*x))*sqrt(c) - b))/c^(5/2)

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maple [A]  time = 0.06, size = 252, normalized size = 1.67 \begin {gather*} \frac {B \,e^{2} x^{2}}{\sqrt {c \,x^{2}+b x}\, c}+\frac {4 A d e x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {2 A \,e^{2} x}{\sqrt {c \,x^{2}+b x}\, c}+\frac {3 B b \,e^{2} x}{\sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {2 B \,d^{2} x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {4 B d e x}{\sqrt {c \,x^{2}+b x}\, c}+\frac {A \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}-\frac {3 B b \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {5}{2}}}+\frac {2 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 {2 \left (2 c x +b \right ) A \,d^{2}}{\sqrt {c \,x^{2}+b x}\, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 227, normalized size = 1.50 \begin {gather*} \frac {B e^{2} x^{2}}{\sqrt {c x^{2} + b x} c} + \frac {2 \, B d^{2} x}{\sqrt {c x^{2} + b x} b} - \frac {4 \, A c d^{2} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {4 \, A d e x}{\sqrt {c x^{2} + b x} b} + \frac {3 \, B b e^{2} x}{\sqrt {c x^{2} + b x} c^{2}} - \frac {3 \, B b e^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} - \frac {2 \, A d^{2}}{\sqrt {c x^{2} + b x} b} - \frac {2 \, {\left (2 \, B d e + A e^{2}\right )} x}{\sqrt {c x^{2} + b x} c} + \frac {{\left (2 \, B d e + A e^{2}\right )} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((A + B*x)*(d + e*x)^2)/(b*x + c*x^2)^(3/2), 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 (d + e x\right )^{2}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*(d + e*x)**2/(x*(b + c*x))**(3/2), x)

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