### 3.1210 $$\int \frac{(a+b x+c x^2)^{3/2}}{(b d+2 c d x)^4} \, dx$$

Optimal. Leaf size=107 $-\frac{\sqrt{a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{5/2} d^4}-\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}$

[Out]

-Sqrt[a + b*x + c*x^2]/(8*c^2*d^4*(b + 2*c*x)) - (a + b*x + c*x^2)^(3/2)/(6*c*d^4*(b + 2*c*x)^3) + ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]/(16*c^(5/2)*d^4)

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Rubi [A]  time = 0.0501314, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.115, Rules used = {684, 621, 206} $-\frac{\sqrt{a+b x+c x^2}}{8 c^2 d^4 (b+2 c x)}+\frac{\tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{5/2} d^4}-\frac{\left (a+b x+c x^2\right )^{3/2}}{6 c d^4 (b+2 c x)^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4,x]

[Out]

-Sqrt[a + b*x + c*x^2]/(8*c^2*d^4*(b + 2*c*x)) - (a + b*x + c*x^2)^(3/2)/(6*c*d^4*(b + 2*c*x)^3) + ArcTanh[(b
+ 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]/(16*c^(5/2)*d^4)

Rule 684

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

Rule 621

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

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])

Rubi steps

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

Mathematica [C]  time = 0.0467636, size = 95, normalized size = 0.89 $\frac{\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{48 c^2 d^4 (b+2 c x)^3 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)^(3/2)/(b*d + 2*c*d*x)^4,x]

[Out]

((b^2 - 4*a*c)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-3/2, -3/2, -1/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(48*c^2
*d^4*(b + 2*c*x)^3*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)])

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Maple [B]  time = 0.192, size = 629, normalized size = 5.9 \begin{align*} -{\frac{1}{12\,{c}^{3}{d}^{4} \left ( 4\,ac-{b}^{2} \right ) } \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{5}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-3}}-{\frac{2}{3\,c{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{5}{2}}} \left ( x+{\frac{b}{2\,c}} \right ) ^{-1}}+{\frac{2\,x}{3\,{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}}}+{\frac{b}{3\,c{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}} \right ) ^{{\frac{3}{2}}}}+{\frac{ax}{{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}}}-{\frac{{b}^{2}x}{4\,c{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}}}+{\frac{ab}{2\,c{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}}}-{\frac{{b}^{3}}{8\,{c}^{2}{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}}}+{\frac{{a}^{2}}{{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\ln \left ( \left ( x+{\frac{b}{2\,c}} \right ) \sqrt{c}+\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{b}^{2}a}{2\,{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\ln \left ( \left ( x+{\frac{b}{2\,c}} \right ) \sqrt{c}+\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}} \right ){c}^{-{\frac{3}{2}}}}+{\frac{{b}^{4}}{16\,{d}^{4} \left ( 4\,ac-{b}^{2} \right ) ^{2}}\ln \left ( \left ( x+{\frac{b}{2\,c}} \right ) \sqrt{c}+\sqrt{ \left ( x+{\frac{b}{2\,c}} \right ) ^{2}c+{\frac{4\,ac-{b}^{2}}{4\,c}}} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x)

[Out]

-1/12/d^4/c^3/(4*a*c-b^2)/(x+1/2*b/c)^3*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)-2/3/d^4/c/(4*a*c-b^2)^2/(x+1
/2*b/c)*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(5/2)+2/3/d^4/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3
/2)*x+1/3/d^4/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(3/2)*b+1/d^4/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c
+1/4*(4*a*c-b^2)/c)^(1/2)*x*a-1/4/d^4/c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*x*b^2+1/2/d^4/
c/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2)*b*a-1/8/d^4/c^2/(4*a*c-b^2)^2*((x+1/2*b/c)^2*c+1/4*(
4*a*c-b^2)/c)^(1/2)*b^3+1/d^4/c^(1/2)/(4*a*c-b^2)^2*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)
^(1/2))*a^2-1/2/d^4/c^(3/2)/(4*a*c-b^2)^2*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*b^
2*a+1/16/d^4/c^(5/2)/(4*a*c-b^2)^2*ln((x+1/2*b/c)*c^(1/2)+((x+1/2*b/c)^2*c+1/4*(4*a*c-b^2)/c)^(1/2))*b^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.30189, size = 776, normalized size = 7.25 \begin{align*} \left [\frac{3 \,{\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\right )} \sqrt{c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) - 4 \,{\left (16 \, c^{3} x^{2} + 16 \, b c^{2} x + 3 \, b^{2} c + 4 \, a c^{2}\right )} \sqrt{c x^{2} + b x + a}}{96 \,{\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}}, -\frac{3 \,{\left (8 \, c^{3} x^{3} + 12 \, b c^{2} x^{2} + 6 \, b^{2} c x + b^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \,{\left (16 \, c^{3} x^{2} + 16 \, b c^{2} x + 3 \, b^{2} c + 4 \, a c^{2}\right )} \sqrt{c x^{2} + b x + a}}{48 \,{\left (8 \, c^{6} d^{4} x^{3} + 12 \, b c^{5} d^{4} x^{2} + 6 \, b^{2} c^{4} d^{4} x + b^{3} c^{3} d^{4}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x, algorithm="fricas")

[Out]

[1/96*(3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^2*c*x + b^3)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 +
b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(16*c^3*x^2 + 16*b*c^2*x + 3*b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))
/(8*c^6*d^4*x^3 + 12*b*c^5*d^4*x^2 + 6*b^2*c^4*d^4*x + b^3*c^3*d^4), -1/48*(3*(8*c^3*x^3 + 12*b*c^2*x^2 + 6*b^
2*c*x + b^3)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(16*c
^3*x^2 + 16*b*c^2*x + 3*b^2*c + 4*a*c^2)*sqrt(c*x^2 + b*x + a))/(8*c^6*d^4*x^3 + 12*b*c^5*d^4*x^2 + 6*b^2*c^4*
d^4*x + b^3*c^3*d^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac{b x \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx + \int \frac{c x^{2} \sqrt{a + b x + c x^{2}}}{b^{4} + 8 b^{3} c x + 24 b^{2} c^{2} x^{2} + 32 b c^{3} x^{3} + 16 c^{4} x^{4}}\, dx}{d^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(3/2)/(2*c*d*x+b*d)**4,x)

[Out]

(Integral(a*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4), x)
+ Integral(b*x*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**4*x**4)
, x) + Integral(c*x**2*sqrt(a + b*x + c*x**2)/(b**4 + 8*b**3*c*x + 24*b**2*c**2*x**2 + 32*b*c**3*x**3 + 16*c**
4*x**4), x))/d**4

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(3/2)/(2*c*d*x+b*d)^4,x, algorithm="giac")

[Out]

Exception raised: TypeError