3.1310 \(\int \frac{(b d+2 c d x)^{11/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=191 \[ -45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}+90 c^2 d^5 \sqrt{b d+2 c d x} \]

[Out]

90*c^2*d^5*Sqrt[b*d + 2*c*d*x] - (d*(b*d + 2*c*d*x)^(9/2))/(2*(a + b*x + c*x^2)^2) - (9*c*d^3*(b*d + 2*c*d*x)^
(5/2))/(2*(a + b*x + c*x^2)) - 45*c^2*(b^2 - 4*a*c)^(1/4)*d^(11/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(
1/4)*Sqrt[d])] - 45*c^2*(b^2 - 4*a*c)^(1/4)*d^(11/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])
]

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Rubi [A]  time = 0.147768, antiderivative size = 191, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {686, 692, 694, 329, 212, 206, 203} \[ -45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-45 c^2 d^{11/2} \sqrt [4]{b^2-4 a c} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right )-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}+90 c^2 d^5 \sqrt{b d+2 c d x} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^3,x]

[Out]

90*c^2*d^5*Sqrt[b*d + 2*c*d*x] - (d*(b*d + 2*c*d*x)^(9/2))/(2*(a + b*x + c*x^2)^2) - (9*c*d^3*(b*d + 2*c*d*x)^
(5/2))/(2*(a + b*x + c*x^2)) - 45*c^2*(b^2 - 4*a*c)^(1/4)*d^(11/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(
1/4)*Sqrt[d])] - 45*c^2*(b^2 - 4*a*c)^(1/4)*d^(11/2)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])
]

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 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 212

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

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])

Rubi steps

\begin{align*} \int \frac{(b d+2 c d x)^{11/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}+\frac{1}{2} \left (9 c d^2\right ) \int \frac{(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (45 c^2 d^4\right ) \int \frac{(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=90 c^2 d^5 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (45 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac{1}{\sqrt{b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=90 c^2 d^5 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{4} \left (45 c \left (b^2-4 a c\right ) d^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (a-\frac{b^2}{4 c}+\frac{x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )\\ &=90 c^2 d^5 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}+\frac{1}{2} \left (45 c \left (b^2-4 a c\right ) d^5\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b^2}{4 c}+\frac{x^4}{4 c d^2}} \, dx,x,\sqrt{d (b+2 c x)}\right )\\ &=90 c^2 d^5 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-\left (45 c^2 \sqrt{b^2-4 a c} d^6\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 (45 c^2 \sqrt{b^2-4 a c} d^6\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 )\\ &=90 c^2 d^5 \sqrt{b d+2 c d x}-\frac{d (b d+2 c d x)^{9/2}}{2 \left (a+b x+c x^2\right )^2}-\frac{9 c d^3 (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )}-45 c^2 \sqrt [4]{b^2-4 a c} d^{11/2} \tan ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )-45 c^2 \sqrt [4]{b^2-4 a c} d^{11/2} \tanh ^{-1}\left (\frac{\sqrt{d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\\ \end{align*}

Mathematica [A]  time = 0.349355, size = 226, normalized size = 1.18 \[ -\frac{d^5 \sqrt{d (b+2 c x)} \left (\sqrt{b+2 c x} \left (-4 c^2 \left (45 a^2+81 a c x^2+32 c^2 x^4\right )+3 b^2 c \left (3 a-37 c x^2\right )-4 b c^2 x \left (81 a+64 c x^2\right )+17 b^3 c x+b^4\right )+90 c^2 \sqrt [4]{b^2-4 a c} (a+x (b+c x))^2 \tan ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+90 c^2 \sqrt [4]{b^2-4 a c} (a+x (b+c x))^2 \tanh ^{-1}\left (\frac{\sqrt{b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{2 \sqrt{b+2 c x} (a+x (b+c x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(11/2)/(a + b*x + c*x^2)^3,x]

[Out]

-(d^5*Sqrt[d*(b + 2*c*x)]*(Sqrt[b + 2*c*x]*(b^4 + 17*b^3*c*x + 3*b^2*c*(3*a - 37*c*x^2) - 4*b*c^2*x*(81*a + 64
*c*x^2) - 4*c^2*(45*a^2 + 81*a*c*x^2 + 32*c^2*x^4)) + 90*c^2*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))^2*ArcTan[Sq
rt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + 90*c^2*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))^2*ArcTanh[Sqrt[b + 2*c*x]/(b
^2 - 4*a*c)^(1/4)]))/(2*Sqrt[b + 2*c*x]*(a + x*(b + c*x))^2)

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Maple [B]  time = 0.199, size = 857, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^3,x)

[Out]

64*c^2*d^5*(2*c*d*x+b*d)^(1/2)+136*c^3*d^7/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(5/2)*a-34*c^
2*d^7/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(5/2)*b^2+416*c^4*d^9/(4*c^2*d^2*x^2+4*b*c*d^2*x+4
*a*c*d^2)^2*(2*c*d*x+b*d)^(1/2)*a^2-208*c^3*d^9/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(1/2)*a*
b^2+26*c^2*d^9/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2*(2*c*d*x+b*d)^(1/2)*b^4-90*c^3*d^7/(4*a*c*d^2-b^2*d^2)^
(3/4)*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)*a+45/2*c^2*d^7/(4*a*c*d^2-b^2*d^
2)^(3/4)*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)*b^2+90*c^3*d^7/(4*a*c*d^2-b^2
*d^2)^(3/4)*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)*a-45/2*c^2*d^7/(4*a*c*d^2
-b^2*d^2)^(3/4)*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)*b^2-45*c^3*d^7/(4*a*c
*d^2-b^2*d^2)^(3/4)*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)))*
a+45/4*c^2*d^7/(4*a*c*d^2-b^2*d^2)^(3/4)*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)))*b^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)^(11/2)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.7876, size = 1170, normalized size = 6.13 \begin{align*} -\frac{180 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac{1}{4}}{\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 )} \arctan \left (\frac{\left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac{3}{4}} \sqrt{2 \, c d x + b d} c^{2} d^{5} - \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac{3}{4}} \sqrt{2 \, c^{5} d^{11} x + b c^{4} d^{11} + \sqrt{{\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}}}}{{\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}}\right ) + 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac{1}{4}}{\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 )} \log \left (45 \, \sqrt{2 \, c d x + b d} c^{2} d^{5} + 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac{1}{4}}\right ) - 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac{1}{4}}{\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 )} \log \left (45 \, \sqrt{2 \, c d x + b d} c^{2} d^{5} - 45 \, \left ({\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{22}\right )^{\frac{1}{4}}\right ) -{\left (128 \, c^{4} d^{5} x^{4} + 256 \, b c^{3} d^{5} x^{3} + 3 \,{\left (37 \, b^{2} c^{2} + 108 \, a c^{3}\right )} d^{5} x^{2} -{\left (17 \, b^{3} c - 324 \, a b c^{2}\right )} d^{5} x -{\left (b^{4} + 9 \, a b^{2} c - 180 \, a^{2} c^{2}\right )} d^{5}\right )} \sqrt{2 \, c d x + b d}}{2 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(180*((b^2*c^8 - 4*a*c^9)*d^22)^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*arctan(((
(b^2*c^8 - 4*a*c^9)*d^22)^(3/4)*sqrt(2*c*d*x + b*d)*c^2*d^5 - ((b^2*c^8 - 4*a*c^9)*d^22)^(3/4)*sqrt(2*c^5*d^11
*x + b*c^4*d^11 + sqrt((b^2*c^8 - 4*a*c^9)*d^22)))/((b^2*c^8 - 4*a*c^9)*d^22)) + 45*((b^2*c^8 - 4*a*c^9)*d^22)
^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*log(45*sqrt(2*c*d*x + b*d)*c^2*d^5 + 45*((b^2
*c^8 - 4*a*c^9)*d^22)^(1/4)) - 45*((b^2*c^8 - 4*a*c^9)*d^22)^(1/4)*(c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a
*c)*x^2 + a^2)*log(45*sqrt(2*c*d*x + b*d)*c^2*d^5 - 45*((b^2*c^8 - 4*a*c^9)*d^22)^(1/4)) - (128*c^4*d^5*x^4 +
256*b*c^3*d^5*x^3 + 3*(37*b^2*c^2 + 108*a*c^3)*d^5*x^2 - (17*b^3*c - 324*a*b*c^2)*d^5*x - (b^4 + 9*a*b^2*c - 1
80*a^2*c^2)*d^5)*sqrt(2*c*d*x + b*d))/(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)**(11/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.25903, size = 703, normalized size = 3.68 \begin{align*} -\frac{45}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c^{2} d^{5} \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 ) - \frac{45}{2} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c^{2} d^{5} \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 ) - \frac{45}{4} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c^{2} d^{5} \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 ) + \frac{45}{4} \, \sqrt{2}{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac{1}{4}} c^{2} d^{5} \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 ) + 64 \, \sqrt{2 \, c d x + b d} c^{2} d^{5} + \frac{2 \,{\left (13 \, \sqrt{2 \, c d x + b d} b^{4} c^{2} d^{9} - 104 \, \sqrt{2 \, c d x + b d} a b^{2} c^{3} d^{9} + 208 \, \sqrt{2 \, c d x + b d} a^{2} c^{4} d^{9} - 17 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} b^{2} c^{2} d^{7} + 68 \,{\left (2 \, c d x + b d\right )}^{\frac{5}{2}} a c^{3} d^{7}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} -{\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-45/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*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)) - 45/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*arcta
n(-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)) -
45/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqr
t(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2)) + 45/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^5*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)) + 64*sqrt(2*c*
d*x + b*d)*c^2*d^5 + 2*(13*sqrt(2*c*d*x + b*d)*b^4*c^2*d^9 - 104*sqrt(2*c*d*x + b*d)*a*b^2*c^3*d^9 + 208*sqrt(
2*c*d*x + b*d)*a^2*c^4*d^9 - 17*(2*c*d*x + b*d)^(5/2)*b^2*c^2*d^7 + 68*(2*c*d*x + b*d)^(5/2)*a*c^3*d^7)/(b^2*d
^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2