3.1298 \(\int \frac {(b d+2 c d x)^{9/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=152 \[ 14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac {28}{3} c d^3 (b d+2 c d x)^{3/2} \]

[Out]

28/3*c*d^3*(2*c*d*x+b*d)^(3/2)-d*(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)+14*c*(-4*a*c+b^2)^(3/4)*d^(9/2)*arctan((d*(
2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-14*c*(-4*a*c+b^2)^(3/4)*d^(9/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*
c+b^2)^(1/4)/d^(1/2))

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Rubi [A]  time = 0.12, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {686, 692, 694, 329, 298, 203, 206} \[ 14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-14 c d^{9/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac {28}{3} c d^3 (b d+2 c d x)^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(28*c*d^3*(b*d + 2*c*d*x)^(3/2))/3 - (d*(b*d + 2*c*d*x)^(7/2))/(a + b*x + c*x^2) + 14*c*(b^2 - 4*a*c)^(3/4)*d^
(9/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 14*c*(b^2 - 4*a*c)^(3/4)*d^(9/2)*ArcTanh[Sqr
t[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*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 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 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]

Rubi steps

\begin {align*} \int \frac {(b d+2 c d x)^{9/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 c d^2\right ) \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=\frac {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=\frac {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\frac {1}{2} \left (7 \left (b^2-4 a c\right ) 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 {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+\left (7 \left (b^2-4 a c\right ) 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 {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}-\left (14 c \left (b^2-4 a c\right ) 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 (14 c \left (b^2-4 a c\right ) 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 {28}{3} c d^3 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{7/2}}{a+b x+c x^2}+14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-14 c \left (b^2-4 a c\right )^{3/4} d^{9/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 92, normalized size = 0.61 \[ -\frac {8 d^3 (d (b+2 c x))^{3/2} \left (14 c (a+x (b+c x)) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )-2 c \left (7 a+c x^2\right )+3 b^2-2 b c x\right )}{3 (a+x (b+c x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*d^3*(d*(b + 2*c*x))^(3/2)*(3*b^2 - 2*b*c*x - 2*c*(7*a + c*x^2) + 14*c*(a + x*(b + c*x))*Hypergeometric2F1[
3/4, 2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(3*(a + x*(b + c*x)))

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fricas [B]  time = 0.94, size = 782, normalized size = 5.14 \[ \frac {84 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \arctan \left (\frac {\left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} - \sqrt {2 \, {\left (b^{8} c^{7} - 16 \, a b^{6} c^{8} + 96 \, a^{2} b^{4} c^{9} - 256 \, a^{3} b^{2} c^{10} + 256 \, a^{4} c^{11}\right )} d^{27} x + {\left (b^{9} c^{6} - 16 \, a b^{7} c^{7} + 96 \, a^{2} b^{5} c^{8} - 256 \, a^{3} b^{3} c^{9} + 256 \, a^{4} b c^{10}\right )} d^{27} + \sqrt {{\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}} {\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}} \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}}}{{\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}}\right ) - 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} + 343 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) + 21 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (343 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {2 \, c d x + b d} d^{13} - 343 \, \left ({\left (b^{6} c^{4} - 12 \, a b^{4} c^{5} + 48 \, a^{2} b^{2} c^{6} - 64 \, a^{3} c^{7}\right )} d^{18}\right )^{\frac {3}{4}}\right ) + {\left (32 \, c^{3} d^{4} x^{3} + 48 \, b c^{2} d^{4} x^{2} + 2 \, {\left (5 \, b^{2} c + 28 \, a c^{2}\right )} d^{4} x - {\left (3 \, b^{3} - 28 \, a b c\right )} d^{4}\right )} \sqrt {2 \, c d x + b d}}{3 \, {\left (c x^{2} + b x + a\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

1/3*(84*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x + a)*arctan((((b^6*c^
4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x
+ b*d)*d^13 - sqrt(2*(b^8*c^7 - 16*a*b^6*c^8 + 96*a^2*b^4*c^9 - 256*a^3*b^2*c^10 + 256*a^4*c^11)*d^27*x + (b^9
*c^6 - 16*a*b^7*c^7 + 96*a^2*b^5*c^8 - 256*a^3*b^3*c^9 + 256*a^4*b*c^10)*d^27 + sqrt((b^6*c^4 - 12*a*b^4*c^5 +
48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)*(b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)*((b^6*c^4 - 1
2*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4))/((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7
)*d^18)) - 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x + a)*log(343*(b
^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 + 343*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 -
64*a^3*c^7)*d^18)^(3/4)) + 21*((b^6*c^4 - 12*a*b^4*c^5 + 48*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(1/4)*(c*x^2 + b*x
+ a)*log(343*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(2*c*d*x + b*d)*d^13 - 343*((b^6*c^4 - 12*a*b^4*c^5 + 4
8*a^2*b^2*c^6 - 64*a^3*c^7)*d^18)^(3/4)) + (32*c^3*d^4*x^3 + 48*b*c^2*d^4*x^2 + 2*(5*b^2*c + 28*a*c^2)*d^4*x -
(3*b^3 - 28*a*b*c)*d^4)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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giac [B]  time = 0.28, size = 441, normalized size = 2.90 \[ -7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - 7 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) + \frac {7}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) - \frac {7}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {16}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c d^{3} + \frac {4 \, {\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c d^{5} - 4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \]

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)^2,x, algorithm="giac")

[Out]

-7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*arctan(1/2*sqrt(2)*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4) + 2*sqr
t(2*c*d*x + b*d))/(-b^2*d^2 + 4*a*c*d^2)^(1/4)) - 7*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*d^3*arctan(-1/2*sqr
t(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)) + 7/2*sqrt(2
)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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)) - 7/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c*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)) + 16/3*(2*c*d*x + b*d)^(3/2)*c
*d^3 + 4*((2*c*d*x + b*d)^(3/2)*b^2*c*d^5 - 4*(2*c*d*x + b*d)^(3/2)*a*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x
+ b*d)^2)

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maple [B]  time = 0.06, size = 693, normalized size = 4.56 \[ \frac {28 \sqrt {2}\, a \,c^{2} d^{5} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {28 \sqrt {2}\, a \,c^{2} d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {14 \sqrt {2}\, a \,c^{2} d^{5} \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {7 \sqrt {2}\, b^{2} c \,d^{5} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {7 \sqrt {2}\, b^{2} c \,d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {7 \sqrt {2}\, b^{2} c \,d^{5} \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {16 \left (2 c d x +b d \right )^{\frac {3}{2}} a \,c^{2} d^{5}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}}-\frac {4 \left (2 c d x +b d \right )^{\frac {3}{2}} b^{2} c \,d^{5}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}}+\frac {16 \left (2 c d x +b d \right )^{\frac {3}{2}} c \,d^{3}}{3} \]

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)^2,x)

[Out]

16/3*c*d^3*(2*c*d*x+b*d)^(3/2)+16*c^2*d^5*(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)*a-4*c*d^5*
(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)*b^2-14*c^2*d^5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*l
n((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)))-28*c^2*d^5*2^(1/2)/(4*a*c*d^2
-b^2*d^2)^(1/4)*a*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)+28*c^2*d^5*2^(1/2)/(4*a*c*d^
2-b^2*d^2)^(1/4)*a*arctan(-2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)+7/2*c*d^5*2^(1/2)/(4*a*c*d
^2-b^2*d^2)^(1/4)*b^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)))+7*c*d^
5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*b^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)-7*c*d^
5*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*b^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)^(9/2)/(c*x^2+b*x+a)^2,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 = 0.16, size = 165, normalized size = 1.09 \[ \frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{3}+\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (16\,a\,c^2\,d^5-4\,b^2\,c\,d^5\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}+14\,c\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}+c\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}\,14{}\mathrm {i} \]

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)^2,x)

[Out]

(16*c*d^3*(b*d + 2*c*d*x)^(3/2))/3 + ((b*d + 2*c*d*x)^(3/2)*(16*a*c^2*d^5 - 4*b^2*c*d^5))/((b*d + 2*c*d*x)^2 -
b^2*d^2 + 4*a*c*d^2) + 14*c*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)^(
3/4) + c*d^(9/2)*atan(((b*d + 2*c*d*x)^(1/2)*1i)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(3/4)*14i

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

[Out]

Timed out

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