3.1302 \(\int \frac{\sqrt{b d+2 c d x}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=143 \[ -\frac{(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac{2 c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac{2 c \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}} \]

[Out]

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

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Rubi [A]  time = 0.097171, antiderivative size = 143, 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 = {687, 694, 329, 298, 203, 206} \[ -\frac{(b d+2 c d x)^{3/2}}{d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac{2 c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac{2 c \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(2*c*(d + e*x)^(m +
1)*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[(2*c*e*(m + 2*p + 3))/(e*(p + 1)*(b^2 - 4*a
*c)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && RationalQ[m] && 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{\sqrt{b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac{(b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac{c \int \frac{\sqrt{b d+2 c d x}}{a+b x+c x^2} \, dx}{b^2-4 a c}\\ &=-\frac{(b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac{\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 )}{2 \left (b^2-4 a c\right ) d}\\ &=-\frac{(b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac{\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 )}{\left (b^2-4 a c\right ) d}\\ &=-\frac{(b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}+\frac{(2 c d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d-x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )}{b^2-4 a c}-\frac{(2 c d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b^2-4 a c} d+x^2} \, dx,x,\sqrt{d (b+2 c x)}\right )}{b^2-4 a c}\\ &=-\frac{(b d+2 c d x)^{3/2}}{\left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )}-\frac{2 c \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac{2 c \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )}{\left (b^2-4 a c\right )^{5/4}}\\ \end{align*}

Mathematica [C]  time = 0.0384849, size = 57, normalized size = 0.4 \[ \frac{16 c (d (b+2 c x))^{3/2} \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.194, size = 344, normalized size = 2.4 \begin{align*} 4\,{\frac{c{d}^{3} \left ( 2\,cdx+bd \right ) ^{3/2}}{ \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) \left ( 4\,{c}^{2}{d}^{2}{x}^{2}+4\,bc{d}^{2}x+4\,ac{d}^{2} \right ) }}+{\frac{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 ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{5}{4}}}}+{c{d}^{3}\sqrt{2}\arctan \left ({\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{5}{4}}}}-{c{d}^{3}\sqrt{2}\arctan \left ( -{\sqrt{2}\sqrt{2\,cdx+bd}{\frac{1}{\sqrt [4]{4\,ac{d}^{2}-{b}^{2}{d}^{2}}}}}+1 \right ) \left ( 4\,ac{d}^{2}-{b}^{2}{d}^{2} \right ) ^{-{\frac{5}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^2,x)

[Out]

4*c*d^3*(2*c*d*x+b*d)^(3/2)/(4*a*c*d^2-b^2*d^2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)+1/2*c*d^3/(4*a*c*d^2-b^2
*d^2)^(5/4)*2^(1/2)*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)))+c*d^3/(4
*a*c*d^2-b^2*d^2)^(5/4)*2^(1/2)*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-c*d^3/(4*a*c*d
^2-b^2*d^2)^(5/4)*2^(1/2)*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)^(1/2)/(c*x^2+b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.19359, size = 2037, normalized size = 14.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")

[Out]

-(4*(c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*
(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*arctan(((b^2*c^3 - 4*a*c^4)*(c^4*d^2/(b^10 - 20*
a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*sqrt(2*c*d*x + b*d)*d -
sqrt(2*c^7*d^3*x + b*c^6*d^3 + (b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*sqrt(c^4*d^2/(b^10 - 20*
a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))*d^2)*(c^4*d^2/(b^10 - 20*a*b^8
*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(b^2 - 4*a*c))/(c^4*d^2)) - (
c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(a*b^
2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*log(sqrt(2*c*d*x + b*d)*c^3*d + (b^8 - 16*a*b^6*c + 9
6*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*(c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3
+ 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/4)) + (c^4*d^2/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3
+ 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)*log(sq
rt(2*c*d*x + b*d)*c^3*d - (b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4)*(c^4*d^2/(b^10 -
20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5))^(3/4)) + sqrt(2*c*d*x + b*
d)*(2*c*x + b))/(a*b^2 - 4*a^2*c + (b^2*c - 4*a*c^2)*x^2 + (b^3 - 4*a*b*c)*x)

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Sympy [B]  time = 98.0098, size = 279, normalized size = 1.95 \begin{align*} \frac{16 c d^{3} \left (b d + 2 c d x\right )^{\frac{3}{2}}}{64 a^{2} c^{2} d^{4} - 32 a b^{2} c d^{4} + 16 a c d^{2} \left (b d + 2 c d x\right )^{2} + 4 b^{4} d^{4} - 4 b^{2} d^{2} \left (b d + 2 c d x\right )^{2}} + 16 c d^{3} \operatorname{RootSum}{\left (t^{4} \left (67108864 a^{5} c^{5} d^{10} - 83886080 a^{4} b^{2} c^{4} d^{10} + 41943040 a^{3} b^{4} c^{3} d^{10} - 10485760 a^{2} b^{6} c^{2} d^{10} + 1310720 a b^{8} c d^{10} - 65536 b^{10} d^{10}\right ) + 1, \left ( t \mapsto t \log{\left (1048576 t^{3} a^{4} c^{4} d^{8} - 1048576 t^{3} a^{3} b^{2} c^{3} d^{8} + 393216 t^{3} a^{2} b^{4} c^{2} d^{8} - 65536 t^{3} a b^{6} c d^{8} + 4096 t^{3} b^{8} d^{8} + \sqrt{b d + 2 c d x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(1/2)/(c*x**2+b*x+a)**2,x)

[Out]

16*c*d**3*(b*d + 2*c*d*x)**(3/2)/(64*a**2*c**2*d**4 - 32*a*b**2*c*d**4 + 16*a*c*d**2*(b*d + 2*c*d*x)**2 + 4*b*
*4*d**4 - 4*b**2*d**2*(b*d + 2*c*d*x)**2) + 16*c*d**3*RootSum(_t**4*(67108864*a**5*c**5*d**10 - 83886080*a**4*
b**2*c**4*d**10 + 41943040*a**3*b**4*c**3*d**10 - 10485760*a**2*b**6*c**2*d**10 + 1310720*a*b**8*c*d**10 - 655
36*b**10*d**10) + 1, Lambda(_t, _t*log(1048576*_t**3*a**4*c**4*d**8 - 1048576*_t**3*a**3*b**2*c**3*d**8 + 3932
16*_t**3*a**2*b**4*c**2*d**8 - 65536*_t**3*a*b**6*c*d**8 + 4096*_t**3*b**8*d**8 + sqrt(b*d + 2*c*d*x))))

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Giac [B]  time = 1.23593, size = 680, normalized size = 4.76 \begin{align*} \frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} + \frac{\sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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^{4} d - 8 \, a b^{2} c d + 16 \, a^{2} c^{2} d} - \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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^{4} d - 8 \, \sqrt{2} a b^{2} c d + 16 \, \sqrt{2} a^{2} c^{2} d} + \frac{{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{3}{4}} c \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^{4} d - 8 \, \sqrt{2} a b^{2} c d + 16 \, \sqrt{2} a^{2} c^{2} d} + \frac{4 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c d}{{\left (b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}\right )}{\left (b^{2} - 4 \, a c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d) + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)
^(3/4)*c*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^4*d - 8*a*b^2*c*d + 16*a^2*c^2*d) - (-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^4*d - 8*sqrt(2)*a*b^
2*c*d + 16*sqrt(2)*a^2*c^2*d) + (-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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^4*d - 8*sqrt(2)*a*b^2*c*d + 16*sqrt(2)*
a^2*c^2*d) + 4*(2*c*d*x + b*d)^(3/2)*c*d/((b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)*(b^2 - 4*a*c))