3.1252 \(\int \frac{(b d+2 c d x)^4}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=96 \[ 16 c^{3/2} d^4 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )-\frac{8 c d^4 (b+2 c x)}{\sqrt{a+b x+c x^2}}-\frac{2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0499499, antiderivative size = 96, 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 = {686, 621, 206} \[ 16 c^{3/2} d^4 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )-\frac{8 c d^4 (b+2 c x)}{\sqrt{a+b x+c x^2}}-\frac{2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(5/2),x]

[Out]

(-2*d^4*(b + 2*c*x)^3)/(3*(a + b*x + c*x^2)^(3/2)) - (8*c*d^4*(b + 2*c*x))/Sqrt[a + b*x + c*x^2] + 16*c^(3/2)*
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 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 )^{5/2}} \, dx &=-\frac{2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}}+\left (4 c d^2\right ) \int \frac{(b d+2 c d x)^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx\\ &=-\frac{2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{8 c d^4 (b+2 c x)}{\sqrt{a+b x+c x^2}}+\left (16 c^2 d^4\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx\\ &=-\frac{2 d^4 (b+2 c x)^3}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{8 c d^4 (b+2 c x)}{\sqrt{a+b x+c x^2}}+\left (32 c^2 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}{3 \left (a+b x+c x^2\right )^{3/2}}-\frac{8 c d^4 (b+2 c x)}{\sqrt{a+b x+c x^2}}+16 c^{3/2} 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 = 0.348217, size = 142, normalized size = 1.48 \[ d^4 \left (\frac{16 c^{3/2} \sqrt{a+x (b+c x)} \sinh ^{-1}\left (\frac{b+2 c x}{\sqrt{c} \sqrt{4 a-\frac{b^2}{c}}}\right )}{\sqrt{4 a-\frac{b^2}{c}} \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}}}-\frac{2 (b+2 c x) \left (4 c \left (3 a+4 c x^2\right )+b^2+16 b c x\right )}{3 (a+x (b+c x))^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^4/(a + b*x + c*x^2)^(5/2),x]

[Out]

d^4*((-2*(b + 2*c*x)*(b^2 + 16*b*c*x + 4*c*(3*a + 4*c*x^2)))/(3*(a + x*(b + c*x))^(3/2)) + (16*c^(3/2)*Sqrt[a
+ x*(b + c*x)]*ArcSinh[(b + 2*c*x)/(Sqrt[4*a - b^2/c]*Sqrt[c])])/(Sqrt[4*a - b^2/c]*Sqrt[(c*(a + x*(b + c*x)))
/(-b^2 + 4*a*c)]))

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Maple [B]  time = 0.053, size = 531, normalized size = 5.5 \begin{align*} -8\,{\frac{{c}^{2}{d}^{4}{b}^{2}ax}{ \left ( 4\,ac-{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{3/2}}}-64\,{\frac{{c}^{3}{d}^{4}{b}^{2}ax}{ \left ( 4\,ac-{b}^{2} \right ) ^{2}\sqrt{c{x}^{2}+bx+a}}}-24\,{\frac{{d}^{4}b{c}^{2}{x}^{2}}{ \left ( c{x}^{2}+bx+a \right ) ^{3/2}}}-18\,{\frac{{d}^{4}{b}^{2}cx}{ \left ( c{x}^{2}+bx+a \right ) ^{3/2}}}+8\,{\frac{{b}^{5}{d}^{4}c}{ \left ( 4\,ac-{b}^{2} \right ) ^{2}\sqrt{c{x}^{2}+bx+a}}}-16\,{\frac{c{d}^{4}ba}{ \left ( c{x}^{2}+bx+a \right ) ^{3/2}}}+8\,{\frac{c{d}^{4}{b}^{3}}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}+{\frac{{d}^{4}{b}^{3}}{3} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}+2\,{\frac{c{d}^{4}{b}^{4}x}{ \left ( 4\,ac-{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{3/2}}}+16\,{\frac{{c}^{2}{d}^{4}{b}^{4}x}{ \left ( 4\,ac-{b}^{2} \right ) ^{2}\sqrt{c{x}^{2}+bx+a}}}-4\,{\frac{c{d}^{4}{b}^{3}a}{ \left ( 4\,ac-{b}^{2} \right ) \left ( c{x}^{2}+bx+a \right ) ^{3/2}}}-32\,{\frac{{c}^{2}{d}^{4}{b}^{3}a}{ \left ( 4\,ac-{b}^{2} \right ) ^{2}\sqrt{c{x}^{2}+bx+a}}}+16\,{\frac{{c}^{2}{d}^{4}{b}^{2}x}{ \left ( 4\,ac-{b}^{2} \right ) \sqrt{c{x}^{2}+bx+a}}}-{\frac{16\,{d}^{4}{c}^{3}{x}^{3}}{3} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}+{\frac{{b}^{5}{d}^{4}}{4\,ac-{b}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}}-16\,{\frac{{d}^{4}{c}^{2}x}{\sqrt{c{x}^{2}+bx+a}}}+8\,{\frac{c{d}^{4}b}{\sqrt{c{x}^{2}+bx+a}}}+16\,{d}^{4}{c}^{3/2}\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)^(5/2),x)

[Out]

-8*d^4*c^2*b^2*a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x-64*d^4*c^3*b^2*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-24*d^4
*c^2*b*x^2/(c*x^2+b*x+a)^(3/2)-18*d^4*c*b^2*x/(c*x^2+b*x+a)^(3/2)+8*d^4*c*b^5/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2
)-16*d^4*c*b*a/(c*x^2+b*x+a)^(3/2)+8*d^4*c*b^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+1/3*d^4*b^3/(c*x^2+b*x+a)^(3/2)
+2*d^4*c*b^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)*x+16*d^4*c^2*b^4/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)*x-4*d^4*c*b^3*
a/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-32*d^4*c^2*b^3*a/(4*a*c-b^2)^2/(c*x^2+b*x+a)^(1/2)+16*d^4*c^2*b^2/(4*a*c-b^2
)/(c*x^2+b*x+a)^(1/2)*x-16/3*d^4*c^3*x^3/(c*x^2+b*x+a)^(3/2)+d^4*b^5/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)-16*d^4*c^
2*x/(c*x^2+b*x+a)^(1/2)+8*d^4*c*b/(c*x^2+b*x+a)^(1/2)+16*d^4*c^(3/2)*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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 8.80002, size = 979, normalized size = 10.2 \begin{align*} \left [\frac{2 \,{\left (12 \,{\left (c^{3} d^{4} x^{4} + 2 \, b c^{2} d^{4} x^{3} + 2 \, a b c d^{4} x + a^{2} c d^{4} +{\left (b^{2} c + 2 \, a c^{2}\right )} d^{4} x^{2}\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 ) -{\left (32 \, c^{3} d^{4} x^{3} + 48 \, b c^{2} d^{4} x^{2} + 6 \,{\left (3 \, b^{2} c + 4 \, a c^{2}\right )} d^{4} x +{\left (b^{3} + 12 \, a b c\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}\right )}}{3 \,{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}}, -\frac{2 \,{\left (24 \,{\left (c^{3} d^{4} x^{4} + 2 \, b c^{2} d^{4} x^{3} + 2 \, a b c d^{4} x + a^{2} c d^{4} +{\left (b^{2} c + 2 \, a c^{2}\right )} d^{4} x^{2}\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 (32 \, c^{3} d^{4} x^{3} + 48 \, b c^{2} d^{4} x^{2} + 6 \,{\left (3 \, b^{2} c + 4 \, a c^{2}\right )} d^{4} x +{\left (b^{3} + 12 \, a b c\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a}\right )}}{3 \,{\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x +{\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}}\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)^(5/2),x, algorithm="fricas")

[Out]

[2/3*(12*(c^3*d^4*x^4 + 2*b*c^2*d^4*x^3 + 2*a*b*c*d^4*x + a^2*c*d^4 + (b^2*c + 2*a*c^2)*d^4*x^2)*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) - (32*c^3*d^4*x^3 + 48*b*c^2*
d^4*x^2 + 6*(3*b^2*c + 4*a*c^2)*d^4*x + (b^3 + 12*a*b*c)*d^4)*sqrt(c*x^2 + b*x + a))/(c^2*x^4 + 2*b*c*x^3 + 2*
a*b*x + (b^2 + 2*a*c)*x^2 + a^2), -2/3*(24*(c^3*d^4*x^4 + 2*b*c^2*d^4*x^3 + 2*a*b*c*d^4*x + a^2*c*d^4 + (b^2*c
+ 2*a*c^2)*d^4*x^2)*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)) +
(32*c^3*d^4*x^3 + 48*b*c^2*d^4*x^2 + 6*(3*b^2*c + 4*a*c^2)*d^4*x + (b^3 + 12*a*b*c)*d^4)*sqrt(c*x^2 + b*x + a
))/(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.17013, size = 458, normalized size = 4.77 \begin{align*} -\frac{8 \, d^{4} \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 (8 \,{\left (\frac{2 \,{\left (b^{4} c^{3} d^{4} - 8 \, a b^{2} c^{4} d^{4} + 16 \, a^{2} c^{5} d^{4}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (b^{5} c^{2} d^{4} - 8 \, a b^{3} c^{3} d^{4} + 16 \, a^{2} b c^{4} d^{4}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (3 \, b^{6} c d^{4} - 20 \, a b^{4} c^{2} d^{4} + 16 \, a^{2} b^{2} c^{3} d^{4} + 64 \, a^{3} c^{4} d^{4}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{b^{7} d^{4} + 4 \, a b^{5} c d^{4} - 80 \, a^{2} b^{3} c^{2} d^{4} + 192 \, a^{3} b c^{3} d^{4}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}}{3 \,{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}} \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)^(5/2),x, algorithm="giac")

[Out]

-8*d^4*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/sqrt(c) - 1/3*(2*(8*(2*(b^4*c^3*d^4 - 8*a*
b^2*c^4*d^4 + 16*a^2*c^5*d^4)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(b^5*c^2*d^4 - 8*a*b^3*c^3*d^4 + 16*a
^2*b*c^4*d^4)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 3*(3*b^6*c*d^4 - 20*a*b^4*c^2*d^4 + 16*a^2*b^2*c^3*d^4
+ 64*a^3*c^4*d^4)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + (b^7*d^4 + 4*a*b^5*c*d^4 - 80*a^2*b^3*c^2*d^4 + 1
92*a^3*b*c^3*d^4)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2)