### 3.1242 $$\int \frac{(b d+2 c d x)^4}{(a+b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=102 $6 \sqrt{c} d^4 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )+12 c d^4 (b+2 c x) \sqrt{a+b x+c x^2}-\frac{2 d^4 (b+2 c x)^3}{\sqrt{a+b x+c x^2}}$

[Out]

(-2*d^4*(b + 2*c*x)^3)/Sqrt[a + b*x + c*x^2] + 12*c*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2] + 6*Sqrt[c]*(b^2 - 4
*a*c)*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]

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Rubi [A]  time = 0.0533681, antiderivative size = 102, normalized size of antiderivative = 1., 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 = {686, 692, 621, 206} $6 \sqrt{c} d^4 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )+12 c d^4 (b+2 c x) \sqrt{a+b x+c x^2}-\frac{2 d^4 (b+2 c x)^3}{\sqrt{a+b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*d^4*(b + 2*c*x)^3)/Sqrt[a + b*x + c*x^2] + 12*c*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2] + 6*Sqrt[c]*(b^2 - 4
*a*c)*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]

Rule 686

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*(d + e*x)^(m - 1)*
(a + b*x + c*x^2)^(p + 1))/(b*(p + 1)), x] - Dist[(d*e*(m - 1))/(b*(p + 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] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 692

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

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{(b d+2 c d x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 d^4 (b+2 c x)^3}{\sqrt{a+b x+c x^2}}+\left (12 c d^2\right ) \int \frac{(b d+2 c d x)^2}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d^4 (b+2 c x)^3}{\sqrt{a+b x+c x^2}}+12 c d^4 (b+2 c x) \sqrt{a+b x+c x^2}+\left (6 c \left (b^2-4 a c\right ) d^4\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d^4 (b+2 c x)^3}{\sqrt{a+b x+c x^2}}+12 c d^4 (b+2 c x) \sqrt{a+b x+c x^2}+\left (12 c \left (b^2-4 a c\right ) d^4\right ) \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 )\\ &=-\frac{2 d^4 (b+2 c x)^3}{\sqrt{a+b x+c x^2}}+12 c d^4 (b+2 c x) \sqrt{a+b x+c x^2}+6 \sqrt{c} \left (b^2-4 a c\right ) d^4 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.410818, size = 139, normalized size = 1.36 $d^4 \left (-\frac{6 c^{3/2} (a+x (b+c x))^{3/2} \sinh ^{-1}\left (\frac{b+2 c x}{\sqrt{c} \sqrt{4 a-\frac{b^2}{c}}}\right )}{\sqrt{4 a-\frac{b^2}{c}} \left (\frac{c (a+x (b+c x))}{4 a c-b^2}\right )^{3/2}}-\frac{2 (b+2 c x) \left (-2 c \left (3 a+c x^2\right )+b^2-2 b c x\right )}{\sqrt{a+x (b+c x)}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

d^4*((-2*(b + 2*c*x)*(b^2 - 2*b*c*x - 2*c*(3*a + c*x^2)))/Sqrt[a + x*(b + c*x)] - (6*c^(3/2)*(a + x*(b + c*x))
^(3/2)*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])])/(Sqrt[4*a - b^2/c]*((c*(a + x*(b + c*x)))/(-b^2 + 4*a
*c))^(3/2)))

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Maple [B]  time = 0.052, size = 340, normalized size = 3.3 \begin{align*} -6\,{\frac{c{d}^{4}{b}^{4}x}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+12\,{\frac{c{d}^{4}{b}^{3}a}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+24\,{\frac{{c}^{2}{d}^{4}{b}^{2}ax}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+8\,{\frac{{d}^{4}{c}^{3}{x}^{3}}{\sqrt{c{x}^{2}+bx+a}}}-3\,{\frac{{d}^{4}{b}^{5}}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-5\,{\frac{{d}^{4}{b}^{3}}{\sqrt{c{x}^{2}+bx+a}}}+12\,{\frac{{d}^{4}b{c}^{2}{x}^{2}}{\sqrt{c{x}^{2}+bx+a}}}-6\,{\frac{{d}^{4}{b}^{2}cx}{\sqrt{c{x}^{2}+bx+a}}}+6\,{d}^{4}\sqrt{c}{b}^{2}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) +12\,{\frac{c{d}^{4}ba}{\sqrt{c{x}^{2}+bx+a}}}+24\,{\frac{{d}^{4}a{c}^{2}x}{\sqrt{c{x}^{2}+bx+a}}}-24\,{d}^{4}{c}^{3/2}a\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*d^4*c*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+12*d^4*c*b^3*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+24*d^4*c^2*b^2*a
/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+8*d^4*c^3*x^3/(c*x^2+b*x+a)^(1/2)-3*d^4*b^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)
-5*d^4*b^3/(c*x^2+b*x+a)^(1/2)+12*d^4*c^2*b*x^2/(c*x^2+b*x+a)^(1/2)-6*d^4*c*b^2*x/(c*x^2+b*x+a)^(1/2)+6*d^4*c^
(1/2)*b^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+12*d^4*c*b*a/(c*x^2+b*x+a)^(1/2)+24*d^4*c^2*a*x/(c*x^2+b
*x+a)^(1/2)-24*d^4*c^(3/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.59705, size = 788, normalized size = 7.73 \begin{align*} \left [-\frac{3 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} x^{2} +{\left (b^{3} - 4 \, a b c\right )} d^{4} x +{\left (a b^{2} - 4 \, a^{2} c\right )} d^{4}\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 ) - 2 \,{\left (4 \, c^{3} d^{4} x^{3} + 6 \, b c^{2} d^{4} x^{2} + 12 \, a c^{2} d^{4} x -{\left (b^{3} - 6 \, a b c\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}}{c x^{2} + b x + a}, -\frac{2 \,{\left (3 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} x^{2} +{\left (b^{3} - 4 \, a b c\right )} d^{4} x +{\left (a b^{2} - 4 \, a^{2} c\right )} d^{4}\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 ) -{\left (4 \, c^{3} d^{4} x^{3} + 6 \, b c^{2} d^{4} x^{2} + 12 \, a c^{2} d^{4} x -{\left (b^{3} - 6 \, a b c\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}\right )}}{c x^{2} + b x + a}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(3*((b^2*c - 4*a*c^2)*d^4*x^2 + (b^3 - 4*a*b*c)*d^4*x + (a*b^2 - 4*a^2*c)*d^4)*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) - 2*(4*c^3*d^4*x^3 + 6*b*c^2*d^4*x^2 + 12*a*c
^2*d^4*x - (b^3 - 6*a*b*c)*d^4)*sqrt(c*x^2 + b*x + a))/(c*x^2 + b*x + a), -2*(3*((b^2*c - 4*a*c^2)*d^4*x^2 + (
b^3 - 4*a*b*c)*d^4*x + (a*b^2 - 4*a^2*c)*d^4)*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)) - (4*c^3*d^4*x^3 + 6*b*c^2*d^4*x^2 + 12*a*c^2*d^4*x - (b^3 - 6*a*b*c)*d^4)*sqrt(c*x^2
+ b*x + a))/(c*x^2 + b*x + a)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.21195, size = 336, normalized size = 3.29 \begin{align*} -\frac{6 \,{\left (b^{2} c d^{4} - 4 \, a c^{2} d^{4}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{\sqrt{c}} + \frac{2 \,{\left (2 \,{\left ({\left (\frac{2 \,{\left (b^{2} c^{5} d^{4} - 4 \, a c^{6} d^{4}\right )} x}{b^{2} c^{2} - 4 \, a c^{3}} + \frac{3 \,{\left (b^{3} c^{4} d^{4} - 4 \, a b c^{5} d^{4}\right )}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x + \frac{6 \,{\left (a b^{2} c^{4} d^{4} - 4 \, a^{2} c^{5} d^{4}\right )}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x - \frac{b^{5} c^{2} d^{4} - 10 \, a b^{3} c^{3} d^{4} + 24 \, a^{2} b c^{4} d^{4}}{b^{2} c^{2} - 4 \, a c^{3}}\right )}}{\sqrt{c x^{2} + b x + a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*(b^2*c*d^4 - 4*a*c^2*d^4)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/sqrt(c) + 2*(2*((2*(
b^2*c^5*d^4 - 4*a*c^6*d^4)*x/(b^2*c^2 - 4*a*c^3) + 3*(b^3*c^4*d^4 - 4*a*b*c^5*d^4)/(b^2*c^2 - 4*a*c^3))*x + 6*
(a*b^2*c^4*d^4 - 4*a^2*c^5*d^4)/(b^2*c^2 - 4*a*c^3))*x - (b^5*c^2*d^4 - 10*a*b^3*c^3*d^4 + 24*a^2*b*c^4*d^4)/(
b^2*c^2 - 4*a*c^3))/sqrt(c*x^2 + b*x + a)