3.1315 \(\int \frac {\sqrt {b d+2 c d x}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=192 \[ \frac {5 c^2 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac {5 c^2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}+\frac {5 c (b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {(b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

[Out]

-1/2*(2*c*d*x+b*d)^(3/2)/(-4*a*c+b^2)/d/(c*x^2+b*x+a)^2+5/2*c*(2*c*d*x+b*d)^(3/2)/(-4*a*c+b^2)^2/d/(c*x^2+b*x+
a)+5*c^2*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))*d^(1/2)/(-4*a*c+b^2)^(9/4)-5*c^2*arctanh((d*(2
*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))*d^(1/2)/(-4*a*c+b^2)^(9/4)

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Rubi [A]  time = 0.13, 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, 298, 203, 206} \[ \frac {5 c^2 \sqrt {d} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}-\frac {5 c^2 \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{9/4}}+\frac {5 c (b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {(b d+2 c d x)^{3/2}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}-\frac {(5 c) \int \frac {\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 {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {\left (5 c^2\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx}{2 \left (b^2-4 a c\right )^2}\\ &=-\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(5 c) \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 )}{4 \left (b^2-4 a c\right )^2 d}\\ &=-\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(5 c) \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 )}{2 \left (b^2-4 a c\right )^2 d}\\ &=-\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {\left (5 c^2 d\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 )^2}+\frac {\left (5 c^2 d\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 )^2}\\ &=-\frac {(b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {5 c (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {5 c^2 \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 )^{9/4}}-\frac {5 c^2 \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 )^{9/4}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 59, normalized size = 0.31 \[ -\frac {64 c^2 (d (b+2 c x))^{3/2} \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 d \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*c^2*(d*(b + 2*c*x))^(3/2)*Hypergeometric2F1[3/4, 3, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(3*(b^2 - 4*a*c)^3
*d)

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fricas [B]  time = 1.05, size = 1796, normalized size = 9.35 \[ \text {result too large to display} \]

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)^3,x, algorithm="fricas")

[Out]

-1/2*(20*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5
*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(1/4)*(a^2*b^4 - 8*
a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3
+ (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*arctan(-((b^4*c^6 - 8*a*b^2*
c^7 + 16*a^2*c^8)*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 1
29024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(1/4)*sqrt
(2*c*d*x + b*d)*d - sqrt(2*c^13*d^3*x + b*c^12*d^3 + (b^10*c^8 - 20*a*b^8*c^9 + 160*a^2*b^6*c^10 - 640*a^3*b^4
*c^11 + 1280*a^4*b^2*c^12 - 1024*a^5*c^13)*sqrt(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12
*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8
- 262144*a^9*c^9))*d^2)*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c
^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(1/4
)*(b^4 - 8*a*b^2*c + 16*a^2*c^2))/(c^8*d^2)) + 5*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^
12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^
8 - 262144*a^9*c^9))^(1/4)*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*
(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^
3*b*c^2)*x)*log(125*sqrt(2*c*d*x + b*d)*c^6*d + 125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^3*b^8*c^3
+ 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*(c^8*d^2/(b^18 - 36*a*b^16*c + 576
*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*
b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(3/4)) - 5*(c^8*d^2/(b^18 - 36*a*b^16*c + 576*a^2*b^14*c^2 - 5
376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4*c^7 + 589824*
a^8*b^2*c^8 - 262144*a^9*c^9))^(1/4)*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4
)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3
*c + 16*a^3*b*c^2)*x)*log(125*sqrt(2*c*d*x + b*d)*c^6*d - 125*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^
3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*(c^8*d^2/(b^18 - 36*a*b^
16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 5
89824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9))^(3/4)) - (10*c^3*x^3 + 15*b*c^2*x^2 - b^3 + 9*a*b*c
+ 3*(b^2*c + 6*a*c^2)*x)*sqrt(2*c*d*x + b*d))/(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 1
6*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 -
8*a^2*b^3*c + 16*a^3*b*c^2)*x)

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giac [B]  time = 0.28, size = 645, normalized size = 3.36 \[ -\frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{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 (9 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{3} - 36 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{3} - 5 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{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((2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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*sq
rt(2)*a^3*c^3*d) - 5*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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) + 5/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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) - 5/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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*sqr
t(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*(9*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^3 - 3
6*(2*c*d*x + b*d)^(3/2)*a*c^3*d^3 - 5*(2*c*d*x + b*d)^(7/2)*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 \[ -\frac {5 \sqrt {2}\, c^{2} d^{5} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {9}{4}}}+\frac {5 \sqrt {2}\, c^{2} d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {9}{4}}}+\frac {5 \sqrt {2}\, c^{2} d^{5} \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{4 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {9}{4}}}+\frac {8 \left (2 c d x +b d \right )^{\frac {3}{2}} c^{2} d^{5}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {10 \left (2 c d x +b d \right )^{\frac {3}{2}} c^{2} d^{5}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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)^(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)^2+10*c^2*d^5/(4*a*c*d^
2-b^2*d^2)^2*(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)+5/4*c^2*d^5/(4*a*c*d^2-b^2*d^2)^(9/4)*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)))+5/2*c^2*d^5/(4*a*c*d^
2-b^2*d^2)^(9/4)*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)-5/2*c^2*d^5/(4*a*c*d^
2-b^2*d^2)^(9/4)*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((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 = 0.69, size = 320, normalized size = 1.67 \[ \frac {\frac {10\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {18\,c^2\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{4\,a\,c-b^2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}+\frac {5\,c^2\,\sqrt {d}\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{9/4}}+\frac {c^2\,\sqrt {d}\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}+a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,16{}\mathrm {i}-a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}\,8{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,5{}\mathrm {i}}{{\left (b^2-4\,a\,c\right )}^{9/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((10*c^2*d*(b*d + 2*c*d*x)^(7/2))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (18*c^2*d^3*(b*d + 2*c*d*x)^(3/2))/(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) + (5*c^2*d^(1/2)*atan((b^4*(b*d + 2*c*d*x)^(1/2) + 16*a^2*c^2*(b*d + 2*c*d*x)^(1/2) - 8*a*b^2*c*(b*d + 2*c
*d*x)^(1/2))/(d^(1/2)*(b^2 - 4*a*c)^(9/4))))/(b^2 - 4*a*c)^(9/4) + (c^2*d^(1/2)*atan((b^4*(b*d + 2*c*d*x)^(1/2
)*1i + a^2*c^2*(b*d + 2*c*d*x)^(1/2)*16i - a*b^2*c*(b*d + 2*c*d*x)^(1/2)*8i)/(d^(1/2)*(b^2 - 4*a*c)^(9/4)))*5i
)/(b^2 - 4*a*c)^(9/4)

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

[Out]

Timed out

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