3.1316 \(\int \frac {1}{\sqrt {b d+2 c d x} (a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=192 \[ -\frac {21 c^2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}-\frac {21 c^2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}+\frac {7 c \sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {\sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

[Out]

-21*c^2*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(11/4)/d^(1/2)-21*c^2*arctanh((d*(
2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(11/4)/d^(1/2)-1/2*(2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)/d
/(c*x^2+b*x+a)^2+7/2*c*(2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^2/d/(c*x^2+b*x+a)

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Rubi [A]  time = 0.14, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {687, 694, 329, 212, 206, 203} \[ -\frac {21 c^2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}-\frac {21 c^2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}+\frac {7 c \sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {\sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^3),x]

[Out]

-Sqrt[b*d + 2*c*d*x]/(2*(b^2 - 4*a*c)*d*(a + b*x + c*x^2)^2) + (7*c*Sqrt[b*d + 2*c*d*x])/(2*(b^2 - 4*a*c)^2*d*
(a + b*x + c*x^2)) - (21*c^2*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*
Sqrt[d]) - (21*c^2*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/((b^2 - 4*a*c)^(11/4)*Sqrt[d])

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])

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3} \, dx &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}-\frac {(7 c) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {\left (21 c^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{2 \left (b^2-4 a c\right )^2}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(21 c) \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 )}{4 \left (b^2-4 a c\right )^2 d}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(21 c) \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 )}{2 \left (b^2-4 a c\right )^2 d}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {\left (21 c^2\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 (b^2-4 a c\right )^{5/2}}-\frac {\left (21 c^2\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 (b^2-4 a c\right )^{5/2}}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {21 c^2 \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 )^{11/4} \sqrt {d}}-\frac {21 c^2 \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 )^{11/4} \sqrt {d}}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 179, normalized size = 0.93 \[ \frac {c^2 \left (\frac {\left (b^2-4 a c\right ) (b+2 c x) \left (-c \left (11 a+7 c x^2\right )+b^2-7 b c x\right )}{2 c^2 (a+x (b+c x))^2}+21 \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x} \tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+21 \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x} \tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{\left (4 a c-b^2\right )^3 \sqrt {d (b+2 c x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[b*d + 2*c*d*x]*(a + b*x + c*x^2)^3),x]

[Out]

(c^2*(((b^2 - 4*a*c)*(b + 2*c*x)*(b^2 - 7*b*c*x - c*(11*a + 7*c*x^2)))/(2*c^2*(a + x*(b + c*x))^2) + 21*(b^2 -
4*a*c)^(1/4)*Sqrt[b + 2*c*x]*ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] + 21*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*
c*x]*ArcTanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)]))/((-b^2 + 4*a*c)^3*Sqrt[d*(b + 2*c*x)])

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fricas [B]  time = 0.78, size = 2021, normalized size = 10.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

1/2*(84*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*
a*b^4*c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2
)*d)*(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*
c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10
*b^2*c^10 - 4194304*a^11*c^11)*d^2))^(1/4)*arctan(((b^16 - 32*a*b^14*c + 448*a^2*b^12*c^2 - 3584*a^3*b^10*c^3
+ 17920*a^4*b^8*c^4 - 57344*a^5*b^6*c^5 + 114688*a^6*b^4*c^6 - 131072*a^7*b^2*c^7 + 65536*a^8*c^8)*sqrt(2*c^5*
d*x + b*c^4*d + (b^12 - 24*a*b^10*c + 240*a^2*b^8*c^2 - 1280*a^3*b^6*c^3 + 3840*a^4*b^4*c^4 - 6144*a^5*b^2*c^5
+ 4096*a^6*c^6)*sqrt(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 -
473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9
+ 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^2))*d^2)*(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560
*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 108134
40*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^2))^(3/4)*d - (b^16*c^2
- 32*a*b^14*c^3 + 448*a^2*b^12*c^4 - 3584*a^3*b^10*c^5 + 17920*a^4*b^8*c^6 - 57344*a^5*b^6*c^7 + 114688*a^6*b^
4*c^8 - 131072*a^7*b^2*c^9 + 65536*a^8*c^10)*(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3
+ 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^
8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^2))^(3/4)*sqrt(2*c*d*x + b*d)*d)/c^8)
- 21*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*
b^4*c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*
d)*(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^
5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b
^2*c^10 - 4194304*a^11*c^11)*d^2))^(1/4)*log(21*sqrt(2*c*d*x + b*d)*c^2 + 21*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^
2 - 64*a^3*c^3)*(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 47308
8*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11
534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^2))^(1/4)*d) + 21*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d*x^4 + 2*
(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 1
6*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d)*(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^18*c^2 - 1056
0*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^8*c^7 + 10813
440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^2))^(1/4)*log(21*sqrt(2
*c*d*x + b*d)*c^2 - 21*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*(c^8/((b^22 - 44*a*b^20*c + 880*a^2*b^
18*c^2 - 10560*a^3*b^16*c^3 + 84480*a^4*b^14*c^4 - 473088*a^5*b^12*c^5 + 1892352*a^6*b^10*c^6 - 5406720*a^7*b^
8*c^7 + 10813440*a^8*b^6*c^8 - 14417920*a^9*b^4*c^9 + 11534336*a^10*b^2*c^10 - 4194304*a^11*c^11)*d^2))^(1/4)*
d) + (7*c^2*x^2 + 7*b*c*x - b^2 + 11*a*c)*sqrt(2*c*d*x + b*d))/((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*d*x^4 + 2
*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*d*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*d*x^2 + 2*(a*b^5 - 8*a^2*b^3*c +
16*a^3*b*c^2)*d*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*d)

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giac [B]  time = 0.27, size = 645, normalized size = 3.36 \[ -\frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \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 )}{\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d} - \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \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 )}{\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d} - \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \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 )}{2 \, {\left (\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d\right )}} + \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \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 )}{2 \, {\left (\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d\right )}} - \frac {2 \, {\left (11 \, \sqrt {2 \, c d x + b d} b^{2} c^{2} d^{3} - 44 \, \sqrt {2 \, c d x + b d} a c^{3} d^{3} - 7 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d\right )}}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-21*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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))/(sqrt(2)*b^6*d - 12*sqrt(2)*a*b^4*c*d + 48*sqrt(2)*a^2*b^2*c^2*d - 64*s
qrt(2)*a^3*c^3*d) - 21*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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))/(sqrt(2)*b^6*d - 12*sqrt(2)*a*b^4*c*d + 48*sqrt(2)*
a^2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) - 21/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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^6*d - 12*sqrt(2)*a*b^4*c
*d + 48*sqrt(2)*a^2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) + 21/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*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^6*d - 12
*sqrt(2)*a*b^4*c*d + 48*sqrt(2)*a^2*b^2*c^2*d - 64*sqrt(2)*a^3*c^3*d) - 2*(11*sqrt(2*c*d*x + b*d)*b^2*c^2*d^3
- 44*sqrt(2*c*d*x + b*d)*a*c^3*d^3 - 7*(2*c*d*x + b*d)^(5/2)*c^2*d)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(b^2*d^2 -
4*a*c*d^2 - (2*c*d*x + b*d)^2)^2)

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maple [B]  time = 0.06, size = 419, normalized size = 2.18 \[ -\frac {21 \sqrt {2}\, c^{2} d^{5} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {11}{4}}}+\frac {21 \sqrt {2}\, c^{2} d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {11}{4}}}+\frac {21 \sqrt {2}\, c^{2} d^{5} \ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{4 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {11}{4}}}+\frac {8 \sqrt {2 c d x +b d}\, c^{2} d^{5}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {14 \sqrt {2 c d x +b d}\, c^{2} d^{5}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x)

[Out]

8*c^2*d^5*(2*c*d*x+b*d)^(1/2)/(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)^2+14*c^2*d^5/(4*a*c*d^
2-b^2*d^2)^2*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)+21/4*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(11/4)
*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)))+21/2*c^2*d^5/(4*a*c
*d^2-b^2*d^2)^(11/4)*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)-21/2*c^2*d^5/(4*a
*c*d^2-b^2*d^2)^(11/4)*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 0.18, size = 317, normalized size = 1.65 \[ \frac {\frac {14\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {22\,c^2\,d^3\,\sqrt {b\,d+2\,c\,d\,x}}{4\,a\,c-b^2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}-\frac {21\,c^2\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{15/4}}{\sqrt {d}\,\left (256\,a^4\,c^4-256\,a^3\,b^2\,c^3+96\,a^2\,b^4\,c^2-16\,a\,b^6\,c+b^8\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{11/4}}-\frac {21\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{15/4}}{\sqrt {d}\,\left (256\,a^4\,c^4-256\,a^3\,b^2\,c^3+96\,a^2\,b^4\,c^2-16\,a\,b^6\,c+b^8\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{11/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*d + 2*c*d*x)^(1/2)*(a + b*x + c*x^2)^3),x)

[Out]

((14*c^2*d*(b*d + 2*c*d*x)^(5/2))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (22*c^2*d^3*(b*d + 2*c*d*x)^(1/2))/(4*a*c -
b^2))/((b*d + 2*c*d*x)^4 - (b*d + 2*c*d*x)^2*(2*b^2*d^2 - 8*a*c*d^2) + b^4*d^4 + 16*a^2*c^2*d^4 - 8*a*b^2*c*d
^4) - (21*c^2*atan(((b*d + 2*c*d*x)^(1/2)*(b^2 - 4*a*c)^(15/4))/(d^(1/2)*(b^8 + 256*a^4*c^4 + 96*a^2*b^4*c^2 -
256*a^3*b^2*c^3 - 16*a*b^6*c))))/(d^(1/2)*(b^2 - 4*a*c)^(11/4)) - (21*c^2*atanh(((b*d + 2*c*d*x)^(1/2)*(b^2 -
4*a*c)^(15/4))/(d^(1/2)*(b^8 + 256*a^4*c^4 + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 - 16*a*b^6*c))))/(d^(1/2)*(b^2
- 4*a*c)^(11/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2*c*d*x+b*d)**(1/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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