3.1300 \(\int \frac{(b d+2 c d x)^{5/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=131 \[ \frac{6 c d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{6 c d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{d (b d+2 c d x)^{3/2}}{a+b x+c x^2} \]

[Out]

-((d*(b*d + 2*c*d*x)^(3/2))/(a + b*x + c*x^2)) + (6*c*d^(5/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*
Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (6*c*d^(5/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2
- 4*a*c)^(1/4)

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Rubi [A]  time = 0.0988734, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {686, 694, 329, 298, 203, 206} \[ \frac{6 c d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{6 c d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{d (b d+2 c d x)^{3/2}}{a+b x+c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(b*d + 2*c*d*x)^(3/2))/(a + b*x + c*x^2)) + (6*c*d^(5/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*
Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (6*c*d^(5/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2
- 4*a*c)^(1/4)

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 694

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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)^{5/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\left (3 c d^2\right ) \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=-\frac{d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\frac{1}{2} (3 d) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}} \, dx,x,b d+2 c d x\right )\\ &=-\frac{d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+(3 d) \operatorname{Subst}\left (\int \frac{x^2}{a-\frac{b^2}{4 c}+\frac{x^4}{4 c d^2}} \, dx,x,\sqrt{d (b+2 c x)}\right )\\ &=-\frac{d (b d+2 c d x)^{3/2}}{a+b x+c x^2}-\left (6 c d^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d-x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )+\left (6 c d^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d+x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )\\ &=-\frac{d (b d+2 c d x)^{3/2}}{a+b x+c x^2}+\frac{6 c d^{5/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac{6 c d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\sqrt [4]{b^2-4 a c}}\\ \end{align*}

Mathematica [C]  time = 0.0850247, size = 83, normalized size = 0.63 \[ -\frac{4 d (d (b+2 c x))^{3/2} \left (4 c (a+x (b+c x)) \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )-4 a c+b^2\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d*(d*(b + 2*c*x))^(3/2)*(b^2 - 4*a*c + 4*c*(a + x*(b + c*x))*Hypergeometric2F1[3/4, 2, 7/4, (b + 2*c*x)^2/
(b^2 - 4*a*c)]))/((b^2 - 4*a*c)*(a + x*(b + c*x)))

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Maple [B]  time = 0.222, size = 327, normalized size = 2.5 \begin{align*} -4\,{\frac{c{d}^{3} \left ( 2\,cdx+bd \right ) ^{3/2}}{4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2}}}+{\frac{3\,c{d}^{3}\sqrt{2}}{2}\ln \left ({ \left ( 2\,cdx+bd-\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) \left ( 2\,cdx+bd+\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}\sqrt{2\,cdx+bd}\sqrt{2}+\sqrt{4\,ac{d}^{2}-{b}^{2}{d}^{2}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+3\,{\frac{c{d}^{3}\sqrt{2}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) }-3\,{\frac{c{d}^{3}\sqrt{2}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}\arctan \left ( -{\frac{\sqrt{2}\sqrt{2\,cdx+bd}}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}+1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x)

[Out]

-4*c*d^3*(2*c*d*x+b*d)^(3/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)+3/2*c*d^3*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)
*ln((2*c*d*x+b*d-(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2))/(2*c*d*x+b*d
+(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)*2^(1/2)+(4*a*c*d^2-b^2*d^2)^(1/2)))+3*c*d^3*2^(1/2)/(4*a*c*d^2-
b^2*d^2)^(1/4)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-3*c*d^3*2^(1/2)/(4*a*c*d^2-b^2*
d^2)^(1/4)*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)

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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)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.97312, size = 805, normalized size = 6.15 \begin{align*} -\frac{12 \, \left (\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \arctan \left (-\frac{\left (\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}} \sqrt{2 \, c d x + b d} c^{3} d^{7} - \sqrt{2 \, c^{7} d^{15} x + b c^{6} d^{15} + \sqrt{\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}}{\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{10}} \left (\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}}{c^{4} d^{10}}\right ) + 3 \, \left (\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt{2 \, c d x + b d} c^{3} d^{7} + 27 \, \left (\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac{3}{4}}{\left (b^{2} - 4 \, a c\right )}\right ) - 3 \, \left (\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac{1}{4}}{\left (c x^{2} + b x + a\right )} \log \left (27 \, \sqrt{2 \, c d x + b d} c^{3} d^{7} - 27 \, \left (\frac{c^{4} d^{10}}{b^{2} - 4 \, a c}\right )^{\frac{3}{4}}{\left (b^{2} - 4 \, a c\right )}\right ) +{\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt{2 \, c d x + b d}}{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)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-(12*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*arctan(-((c^4*d^10/(b^2 - 4*a*c))^(1/4)*sqrt(2*c*d*x + b
*d)*c^3*d^7 - sqrt(2*c^7*d^15*x + b*c^6*d^15 + sqrt(c^4*d^10/(b^2 - 4*a*c))*(b^2*c^4 - 4*a*c^5)*d^10)*(c^4*d^1
0/(b^2 - 4*a*c))^(1/4))/(c^4*d^10)) + 3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b*x + a)*log(27*sqrt(2*c*d*x +
b*d)*c^3*d^7 + 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) - 3*(c^4*d^10/(b^2 - 4*a*c))^(1/4)*(c*x^2 + b
*x + a)*log(27*sqrt(2*c*d*x + b*d)*c^3*d^7 - 27*(c^4*d^10/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) + (2*c*d^2*x + b
*d^2)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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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)**(5/2)/(c*x**2+b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.26524, size = 593, normalized size = 4.53 \begin{align*} \frac{4 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c d^{3}}{b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}} - \frac{3 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c d \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} + 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{b^{2} - 4 \, a c} - \frac{3 \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c d \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} - 2 \, \sqrt{2 \, c d x + b d}\right )}}{2 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}}}\right )}{b^{2} - 4 \, a c} + \frac{3 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c d \log \left (2 \, c d x + b d + \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt{2} b^{2} - 4 \, \sqrt{2} a c} - \frac{3 \,{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c d \log \left (2 \, c d x + b d - \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} \sqrt{2 \, c d x + b d} + \sqrt{-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt{2} b^{2} - 4 \, \sqrt{2} a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

4*(2*c*d*x + b*d)^(3/2)*c*d^3/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2) - 3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/
4)*c*d*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2
)^(1/4))/(b^2 - 4*a*c) - 3*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4
*a*c*d^2)^(1/4) - 2*sqrt(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4))/(b^2 - 4*a*c) + 3*(-b^2*d^2 + 4*a*c*d^2
)^(3/4)*c*d*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a
*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 3*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d*log(2*c*d*x + b*d - sqrt(2)*(-b^2*
d^2 + 4*a*c*d^2)^(1/4)*sqrt(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c)