3.12.23 \(\int \frac {(b d+2 c d x)^{9/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=170 \[ \frac {21 c^2 d^{9/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {21 c^2 d^{9/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 170, 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 = {686, 694, 329, 298, 203, 206} \begin {gather*} \frac {21 c^2 d^{9/2} \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 {21 c^2 d^{9/2} \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 {7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^3,x]

[Out]

-(d*(b*d + 2*c*d*x)^(7/2))/(2*(a + b*x + c*x^2)^2) - (7*c*d^3*(b*d + 2*c*d*x)^(3/2))/(2*(a + b*x + c*x^2)) + (
21*c^2*d^(9/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4) - (21*c^2*d^(9/2
)*ArcTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(1/4)

Rule 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 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]

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {d (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (7 c d^2\right ) \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)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (21 c^2 d^4\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)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{4} \left (21 c d^3\right ) \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)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (21 c d^3\right ) \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)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}-\left (21 c^2 d^5\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 (21 c^2 d^5\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)^{7/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {7 c d^3 (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )}+\frac {21 c^2 d^{9/2} \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 {21 c^2 d^{9/2} \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 = 0.14, size = 119, normalized size = 0.70 \begin {gather*} \frac {4 (d (b+2 c x))^{9/2} \left (-112 c^2 (a+x (b+c x))^2 \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )-5 \left (b^2-4 a c\right ) (b+2 c x)^2+7 \left (b^2-4 a c\right )^2\right )}{5 \left (b^2-4 a c\right ) (b+2 c x)^3 (a+x (b+c x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^3,x]

[Out]

(4*(d*(b + 2*c*x))^(9/2)*(7*(b^2 - 4*a*c)^2 - 5*(b^2 - 4*a*c)*(b + 2*c*x)^2 - 112*c^2*(a + x*(b + c*x))^2*Hype
rgeometric2F1[3/4, 3, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(5*(b^2 - 4*a*c)*(b + 2*c*x)^3*(a + x*(b + c*x))^2)

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 1.98, size = 297, normalized size = 1.75 \begin {gather*} -\frac {\left (\frac {21}{2}+\frac {21 i}{2}\right ) c^2 d^{9/2} \tan ^{-1}\left (\frac {-\frac {(1+i) c \sqrt {d} x}{\sqrt [4]{b^2-4 a c}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {d}}{\sqrt [4]{b^2-4 a c}}+\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {d} \sqrt [4]{b^2-4 a c}}{\sqrt {b d+2 c d x}}\right )}{\sqrt [4]{b^2-4 a c}}-\frac {\left (\frac {21}{2}+\frac {21 i}{2}\right ) c^2 d^{9/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b d+2 c d x}}{\sqrt {d} \left (\sqrt {b^2-4 a c}+i b+2 i c x\right )}\right )}{\sqrt [4]{b^2-4 a c}}+\frac {\sqrt {b d+2 c d x} \left (-7 a b c d^4-14 a c^2 d^4 x-b^3 d^4-13 b^2 c d^4 x-33 b c^2 d^4 x^2-22 c^3 d^4 x^3\right )}{2 \left (a+b x+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^3,x]

[Out]

(Sqrt[b*d + 2*c*d*x]*(-(b^3*d^4) - 7*a*b*c*d^4 - 13*b^2*c*d^4*x - 14*a*c^2*d^4*x - 33*b*c^2*d^4*x^2 - 22*c^3*d
^4*x^3))/(2*(a + b*x + c*x^2)^2) - ((21/2 + (21*I)/2)*c^2*d^(9/2)*ArcTan[(((-1/2 - I/2)*b*Sqrt[d])/(b^2 - 4*a*
c)^(1/4) + (1/2 - I/2)*(b^2 - 4*a*c)^(1/4)*Sqrt[d] - ((1 + I)*c*Sqrt[d]*x)/(b^2 - 4*a*c)^(1/4))/Sqrt[b*d + 2*c
*d*x]])/(b^2 - 4*a*c)^(1/4) - ((21/2 + (21*I)/2)*c^2*d^(9/2)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b*d + 2
*c*d*x])/(Sqrt[d]*(I*b + Sqrt[b^2 - 4*a*c] + (2*I)*c*x))])/(b^2 - 4*a*c)^(1/4)

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 500, normalized size = 2.94 \begin {gather*} -\frac {84 \, \left (\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}\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 (\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}} \sqrt {2 \, c d x + b d} c^{6} d^{13} - \sqrt {2 \, c^{13} d^{27} x + b c^{12} d^{27} + \sqrt {\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}} {\left (b^{2} c^{8} - 4 \, a c^{9}\right )} d^{18}} \left (\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac {1}{4}}}{c^{8} d^{18}}\right ) + 21 \, \left (\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}\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 (9261 \, \sqrt {2 \, c d x + b d} c^{6} d^{13} + 9261 \, \left (\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - 21 \, \left (\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}\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 (9261 \, \sqrt {2 \, c d x + b d} c^{6} d^{13} - 9261 \, \left (\frac {c^{8} d^{18}}{b^{2} - 4 \, a c}\right )^{\frac {3}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) + {\left (22 \, c^{3} d^{4} x^{3} + 33 \, b c^{2} d^{4} x^{2} + {\left (13 \, b^{2} c + 14 \, a c^{2}\right )} d^{4} x + {\left (b^{3} + 7 \, a b c\right )} d^{4}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(84*(c^8*d^18/(b^2 - 4*a*c))^(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(-((c^
8*d^18/(b^2 - 4*a*c))^(1/4)*sqrt(2*c*d*x + b*d)*c^6*d^13 - sqrt(2*c^13*d^27*x + b*c^12*d^27 + sqrt(c^8*d^18/(b
^2 - 4*a*c))*(b^2*c^8 - 4*a*c^9)*d^18)*(c^8*d^18/(b^2 - 4*a*c))^(1/4))/(c^8*d^18)) + 21*(c^8*d^18/(b^2 - 4*a*c
))^(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(9261*sqrt(2*c*d*x + b*d)*c^6*d^13 + 926
1*(c^8*d^18/(b^2 - 4*a*c))^(3/4)*(b^2 - 4*a*c)) - 21*(c^8*d^18/(b^2 - 4*a*c))^(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(9261*sqrt(2*c*d*x + b*d)*c^6*d^13 - 9261*(c^8*d^18/(b^2 - 4*a*c))^(3/4)*(b
^2 - 4*a*c)) + (22*c^3*d^4*x^3 + 33*b*c^2*d^4*x^2 + (13*b^2*c + 14*a*c^2)*d^4*x + (b^3 + 7*a*b*c)*d^4)*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)

________________________________________________________________________________________

giac [B]  time = 0.32, size = 510, normalized size = 3.00 \begin {gather*} -\frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{3} \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^{2} - 4 \, \sqrt {2} a c} - \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{3} \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^{2} - 4 \, \sqrt {2} a c} + \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{3} \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^{2} - 4 \, \sqrt {2} a c\right )}} - \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{3} \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^{2} - 4 \, \sqrt {2} a c\right )}} + \frac {2 \, {\left (7 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{7} - 28 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{7} - 11 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2} d^{5}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-21*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^3*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^2 - 4*sqrt(2)*a*c) - 21*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2
*d^3*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^2 - 4*sqrt(2)*a*c) + 21/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^3*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) -
21/2*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^3*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) + 2*(7*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^7
- 28*(2*c*d*x + b*d)^(3/2)*a*c^3*d^7 - 11*(2*c*d*x + b*d)^(7/2)*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*
d)^2)^2

________________________________________________________________________________________

maple [B]  time = 0.07, size = 435, normalized size = 2.56 \begin {gather*} -\frac {56 \left (2 c d x +b d \right )^{\frac {3}{2}} a \,c^{3} d^{7}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {14 \left (2 c d x +b d \right )^{\frac {3}{2}} b^{2} c^{2} d^{7}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}-\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 {1}{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 {1}{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 {1}{4}}}-\frac {22 \left (2 c d x +b d \right )^{\frac {7}{2}} c^{2} d^{5}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(9/2)/(c*x^2+b*x+a)^3,x)

[Out]

-22*c^2*d^5/(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)^(7/2)-56*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)^(3/2)*a+14*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)^(3/2)*b^2+
21/4*c^2*d^5*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)))+21/2*c^2*d^5*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)-21/2*c^2*d^5*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)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(9/2)/(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?

________________________________________________________________________________________

mupad [B]  time = 0.59, size = 215, normalized size = 1.26 \begin {gather*} \frac {21\,c^2\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (56\,a\,c^3\,d^7-14\,b^2\,c^2\,d^7\right )+22\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{{\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\,d^{9/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(9/2)/(a + b*x + c*x^2)^3,x)

[Out]

(21*c^2*d^(9/2)*atan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4))))/(b^2 - 4*a*c)^(1/4) - ((b*d + 2*c*d
*x)^(3/2)*(56*a*c^3*d^7 - 14*b^2*c^2*d^7) + 22*c^2*d^5*(b*d + 2*c*d*x)^(7/2))/((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*d^(9/2)*atanh((b*d + 2*c*
d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4))))/(b^2 - 4*a*c)^(1/4)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(9/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

________________________________________________________________________________________