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

Optimal. Leaf size=238 \[ -\frac {2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right )}{4 c^{7/2}}+\frac {e \sqrt {b x+c x^2} \left (2 c e x \left (-4 b c (A e+B d)+8 A c^2 d+5 b^2 B e\right )+12 b^2 c e (A e+3 B d)-8 b c^2 d (3 A e+2 B d)+32 A c^3 d^2-15 b^3 B e^2\right )}{4 b^2 c^3} \]

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Rubi [A]  time = 0.22, antiderivative size = 238, 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, 779, 620, 206} \begin {gather*} \frac {e \sqrt {b x+c x^2} \left (2 c e x \left (-4 b c (A e+B d)+8 A c^2 d+5 b^2 B e\right )+12 b^2 c e (A e+3 B d)-8 b c^2 d (3 A e+2 B d)+32 A c^3 d^2-15 b^3 B e^2\right )}{4 b^2 c^3}+\frac {3 e \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right )}{4 c^{7/2}}-\frac {2 (d+e x)^2 \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)^2*(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*(32*A*c^
3*d^2 - 15*b^3*B*e^2 + 12*b^2*c*e*(3*B*d + A*e) - 8*b*c^2*d*(2*B*d + 3*A*e) + 2*c*e*(8*A*c^2*d + 5*b^2*B*e - 4
*b*c*(B*d + A*e))*x)*Sqrt[b*x + c*x^2])/(4*b^2*c^3) + (3*e*(4*A*c*e*(2*c*d - b*e) + B*(8*c^2*d^2 - 12*b*c*d*e
+ 5*b^2*e^2))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(4*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 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)^3}{\left (b x+c x^2\right )^{3/2}} \, dx &=-\frac {2 (d+e x)^2 \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 {(d+e x) \left (\frac {1}{2} b (b B+4 A c) d e+\frac {1}{2} e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right )}{\sqrt {b x+c x^2}} \, dx}{b^2 c}\\ &=-\frac {2 (d+e x)^2 \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 (32 A c^3 d^2-15 b^3 B e^2+12 b^2 c e (3 B d+A e)-8 b c^2 d (2 B d+3 A e)+2 c e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right ) \sqrt {b x+c x^2}}{4 b^2 c^3}+\frac {\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2-12 b c d e+5 b^2 e^2\right )\right )\right ) \int \frac {1}{\sqrt {b x+c x^2}} \, dx}{8 c^3}\\ &=-\frac {2 (d+e x)^2 \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 (32 A c^3 d^2-15 b^3 B e^2+12 b^2 c e (3 B d+A e)-8 b c^2 d (2 B d+3 A e)+2 c e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right ) \sqrt {b x+c x^2}}{4 b^2 c^3}+\frac {\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2-12 b c d e+5 b^2 e^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {b x+c x^2}}\right )}{4 c^3}\\ &=-\frac {2 (d+e x)^2 \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 (32 A c^3 d^2-15 b^3 B e^2+12 b^2 c e (3 B d+A e)-8 b c^2 d (2 B d+3 A e)+2 c e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right ) \sqrt {b x+c x^2}}{4 b^2 c^3}+\frac {3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2-12 b c d e+5 b^2 e^2\right )\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 229, normalized size = 0.96 \begin {gather*} \frac {3 b^{5/2} e \sqrt {x} \sqrt {\frac {c x}{b}+1} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right ) \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right )+\sqrt {c} \left (4 A c \left (3 b^3 e^3 x+b^2 c e^2 x (e x-6 d)-2 b c^2 d^2 (d-3 e x)-4 c^3 d^3 x\right )+b B x \left (-15 b^3 e^3+b^2 c e^2 (36 d-5 e x)+2 b c^2 e \left (-12 d^2+6 d e x+e^2 x^2\right )+8 c^3 d^3\right )\right )}{4 b^2 c^{7/2} \sqrt {x (b+c x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*(4*A*c*(-4*c^3*d^3*x + 3*b^3*e^3*x - 2*b*c^2*d^2*(d - 3*e*x) + b^2*c*e^2*x*(-6*d + e*x)) + b*B*x*(8*c
^3*d^3 - 15*b^3*e^3 + b^2*c*e^2*(36*d - 5*e*x) + 2*b*c^2*e*(-12*d^2 + 6*d*e*x + e^2*x^2))) + 3*b^(5/2)*e*(4*A*
c*e*(2*c*d - b*e) + B*(8*c^2*d^2 - 12*b*c*d*e + 5*b^2*e^2))*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x]
)/Sqrt[b]])/(4*b^2*c^(7/2)*Sqrt[x*(b + c*x)])

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IntegrateAlgebraic [A]  time = 0.79, size = 287, normalized size = 1.21 \begin {gather*} \frac {\sqrt {b x+c x^2} \left (12 A b^3 c e^3 x-24 A b^2 c^2 d e^2 x+4 A b^2 c^2 e^3 x^2-8 A b c^3 d^3+24 A b c^3 d^2 e x-16 A c^4 d^3 x-15 b^4 B e^3 x+36 b^3 B c d e^2 x-5 b^3 B c e^3 x^2-24 b^2 B c^2 d^2 e x+12 b^2 B c^2 d e^2 x^2+2 b^2 B c^2 e^3 x^3+8 b B c^3 d^3 x\right )}{4 b^2 c^3 x (b+c x)}-\frac {3 \log \left (-2 c^{7/2} \sqrt {b x+c x^2}+b c^3+2 c^4 x\right ) \left (-4 A b c e^3+8 A c^2 d e^2+5 b^2 B e^3-12 b B c d e^2+8 B c^2 d^2 e\right )}{8 c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x + c*x^2]*(-8*A*b*c^3*d^3 + 8*b*B*c^3*d^3*x - 16*A*c^4*d^3*x - 24*b^2*B*c^2*d^2*e*x + 24*A*b*c^3*d^2*
e*x + 36*b^3*B*c*d*e^2*x - 24*A*b^2*c^2*d*e^2*x - 15*b^4*B*e^3*x + 12*A*b^3*c*e^3*x + 12*b^2*B*c^2*d*e^2*x^2 -
5*b^3*B*c*e^3*x^2 + 4*A*b^2*c^2*e^3*x^2 + 2*b^2*B*c^2*e^3*x^3))/(4*b^2*c^3*x*(b + c*x)) - (3*(8*B*c^2*d^2*e -
12*b*B*c*d*e^2 + 8*A*c^2*d*e^2 + 5*b^2*B*e^3 - 4*A*b*c*e^3)*Log[b*c^3 + 2*c^4*x - 2*c^(7/2)*Sqrt[b*x + c*x^2]
])/(8*c^(7/2))

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fricas [A]  time = 0.44, size = 694, normalized size = 2.92 \begin {gather*} \left [\frac {3 \, {\left ({\left (8 \, B b^{2} c^{3} d^{2} e - 4 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} + {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x^{2} + {\left (8 \, B b^{3} c^{2} d^{2} e - 4 \, {\left (3 \, B b^{4} c - 2 \, A b^{3} c^{2}\right )} d e^{2} + {\left (5 \, B b^{5} - 4 \, A b^{4} c\right )} e^{3}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, B b^{2} c^{3} e^{3} x^{3} - 8 \, A b c^{4} d^{3} + {\left (12 \, B b^{2} c^{3} d e^{2} - {\left (5 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} e^{3}\right )} x^{2} + {\left (8 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 24 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d^{2} e + 12 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} - 3 \, {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (b^{2} c^{5} x^{2} + b^{3} c^{4} x\right )}}, -\frac {3 \, {\left ({\left (8 \, B b^{2} c^{3} d^{2} e - 4 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} + {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x^{2} + {\left (8 \, B b^{3} c^{2} d^{2} e - 4 \, {\left (3 \, B b^{4} c - 2 \, A b^{3} c^{2}\right )} d e^{2} + {\left (5 \, B b^{5} - 4 \, A b^{4} c\right )} e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, B b^{2} c^{3} e^{3} x^{3} - 8 \, A b c^{4} d^{3} + {\left (12 \, B b^{2} c^{3} d e^{2} - {\left (5 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} e^{3}\right )} x^{2} + {\left (8 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 24 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d^{2} e + 12 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} - 3 \, {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (b^{2} c^{5} x^{2} + b^{3} c^{4} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.31, size = 251, normalized size = 1.05 \begin {gather*} -\frac {\frac {8 \, A d^{3}}{b} - {\left ({\left (\frac {2 \, B x e^{3}}{c} + \frac {12 \, B b^{2} c^{2} d e^{2} - 5 \, B b^{3} c e^{3} + 4 \, A b^{2} c^{2} e^{3}}{b^{2} c^{3}}\right )} x + \frac {8 \, B b c^{3} d^{3} - 16 \, A c^{4} d^{3} - 24 \, B b^{2} c^{2} d^{2} e + 24 \, A b c^{3} d^{2} e + 36 \, B b^{3} c d e^{2} - 24 \, A b^{2} c^{2} d e^{2} - 15 \, B b^{4} e^{3} + 12 \, A b^{3} c e^{3}}{b^{2} c^{3}}\right )} x}{4 \, \sqrt {c x^{2} + b x}} - \frac {3 \, {\left (8 \, B c^{2} d^{2} e - 12 \, B b c d e^{2} + 8 \, A c^{2} d e^{2} + 5 \, B b^{2} e^{3} - 4 \, A b c e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*A*d^3/b - ((2*B*x*e^3/c + (12*B*b^2*c^2*d*e^2 - 5*B*b^3*c*e^3 + 4*A*b^2*c^2*e^3)/(b^2*c^3))*x + (8*B*b
*c^3*d^3 - 16*A*c^4*d^3 - 24*B*b^2*c^2*d^2*e + 24*A*b*c^3*d^2*e + 36*B*b^3*c*d*e^2 - 24*A*b^2*c^2*d*e^2 - 15*B
*b^4*e^3 + 12*A*b^3*c*e^3)/(b^2*c^3))*x)/sqrt(c*x^2 + b*x) - 3/8*(8*B*c^2*d^2*e - 12*B*b*c*d*e^2 + 8*A*c^2*d*e
^2 + 5*B*b^2*e^3 - 4*A*b*c*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.06, size = 450, normalized size = 1.89 \begin {gather*} \frac {B \,e^{3} x^{3}}{2 \sqrt {c \,x^{2}+b x}\, c}+\frac {A \,e^{3} x^{2}}{\sqrt {c \,x^{2}+b x}\, c}-\frac {5 B b \,e^{3} x^{2}}{4 \sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {3 B d \,e^{2} x^{2}}{\sqrt {c \,x^{2}+b x}\, c}+\frac {3 A b \,e^{3} x}{\sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {6 A \,d^{2} e x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {6 A d \,e^{2} x}{\sqrt {c \,x^{2}+b x}\, c}-\frac {15 B \,b^{2} e^{3} x}{4 \sqrt {c \,x^{2}+b x}\, c^{3}}+\frac {9 B b d \,e^{2} x}{\sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {2 B \,d^{3} x}{\sqrt {c \,x^{2}+b x}\, b}-\frac {6 B \,d^{2} e x}{\sqrt {c \,x^{2}+b x}\, c}-\frac {3 A b \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {5}{2}}}+\frac {3 A d \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}+\frac {15 B \,b^{2} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {7}{2}}}-\frac {9 B b d \,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 {3 B \,d^{2} 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^{3}}{\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)^3/(c*x^2+b*x)^(3/2),x)

[Out]

1/2*B*e^3*x^3/c/(c*x^2+b*x)^(1/2)-5/4*B*e^3*b/c^2*x^2/(c*x^2+b*x)^(1/2)-15/4*B*e^3*b^2/c^3/(c*x^2+b*x)^(1/2)*x
+15/8*B*e^3*b^2/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))+x^2/c/(c*x^2+b*x)^(1/2)*A*e^3+3*x^2/c/(c*x^2
+b*x)^(1/2)*B*d*e^2+3*b/c^2/(c*x^2+b*x)^(1/2)*x*A*e^3+9*b/c^2/(c*x^2+b*x)^(1/2)*x*B*d*e^2-3/2*b/c^(5/2)*ln((c*
x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))*A*e^3-9/2*b/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d*e^2-6/c/
(c*x^2+b*x)^(1/2)*x*A*d*e^2-6/c/(c*x^2+b*x)^(1/2)*x*B*d^2*e+3/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2)
)*A*d*e^2+3/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x)^(1/2))*B*d^2*e+6/b/(c*x^2+b*x)^(1/2)*x*A*d^2*e+2/b/(c*x
^2+b*x)^(1/2)*x*B*d^3-2*A*d^3*(2*c*x+b)/b^2/(c*x^2+b*x)^(1/2)

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maxima [A]  time = 0.72, size = 367, normalized size = 1.54 \begin {gather*} \frac {B e^{3} x^{3}}{2 \, \sqrt {c x^{2} + b x} c} - \frac {5 \, B b e^{3} x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {2 \, B d^{3} x}{\sqrt {c x^{2} + b x} b} - \frac {4 \, A c d^{3} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {6 \, A d^{2} e x}{\sqrt {c x^{2} + b x} b} - \frac {15 \, B b^{2} e^{3} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {15 \, B b^{2} e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} - \frac {2 \, A d^{3}}{\sqrt {c x^{2} + b x} b} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} x^{2}}{\sqrt {c x^{2} + b x} c} + \frac {3 \, {\left (3 \, B d e^{2} + A e^{3}\right )} b x}{\sqrt {c x^{2} + b x} c^{2}} - \frac {6 \, {\left (B d^{2} e + A d e^{2}\right )} x}{\sqrt {c x^{2} + b x} c} - \frac {3 \, {\left (3 \, B d e^{2} + A e^{3}\right )} b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} + \frac {3 \, {\left (B d^{2} e + A d 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)^3/(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

1/2*B*e^3*x^3/(sqrt(c*x^2 + b*x)*c) - 5/4*B*b*e^3*x^2/(sqrt(c*x^2 + b*x)*c^2) + 2*B*d^3*x/(sqrt(c*x^2 + b*x)*b
) - 4*A*c*d^3*x/(sqrt(c*x^2 + b*x)*b^2) + 6*A*d^2*e*x/(sqrt(c*x^2 + b*x)*b) - 15/4*B*b^2*e^3*x/(sqrt(c*x^2 + b
*x)*c^3) + 15/8*B*b^2*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/2) - 2*A*d^3/(sqrt(c*x^2 + b*x)*b)
+ (3*B*d*e^2 + A*e^3)*x^2/(sqrt(c*x^2 + b*x)*c) + 3*(3*B*d*e^2 + A*e^3)*b*x/(sqrt(c*x^2 + b*x)*c^2) - 6*(B*d^
2*e + A*d*e^2)*x/(sqrt(c*x^2 + b*x)*c) - 3/2*(3*B*d*e^2 + A*e^3)*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)
)/c^(5/2) + 3*(B*d^2*e + A*d*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 = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^3}{{\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)^3)/(b*x + c*x^2)^(3/2),x)

[Out]

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

[Out]

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

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