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

Optimal. Leaf size=147 \[ 2 d^{9/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{9/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {4}{3} d^3 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}+\frac {4}{7} d (b d+2 c d x)^{7/2} \]

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Rubi [A]  time = 0.14, antiderivative size = 147, 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 = {692, 694, 329, 298, 203, 206} \begin {gather*} \frac {4}{3} d^3 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}+2 d^{9/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{9/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {4}{7} d (b d+2 c d x)^{7/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(b^2 - 4*a*c)*d^3*(b*d + 2*c*d*x)^(3/2))/3 + (4*d*(b*d + 2*c*d*x)^(7/2))/7 + 2*(b^2 - 4*a*c)^(7/4)*d^(9/2)*
ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 2*(b^2 - 4*a*c)^(7/4)*d^(9/2)*ArcTanh[Sqrt[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 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}}{a+b x+c x^2} \, dx &=\frac {4}{7} d (b d+2 c d x)^{7/2}+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=\frac {4}{3} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{3/2}+\frac {4}{7} d (b d+2 c d x)^{7/2}+\left (\left (b^2-4 a c\right )^2 d^4\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=\frac {4}{3} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{3/2}+\frac {4}{7} d (b d+2 c d x)^{7/2}+\frac {\left (\left (b^2-4 a c\right )^2 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 )}{2 c}\\ &=\frac {4}{3} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{3/2}+\frac {4}{7} d (b d+2 c d x)^{7/2}+\frac {\left (\left (b^2-4 a c\right )^2 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 )}{c}\\ &=\frac {4}{3} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{3/2}+\frac {4}{7} d (b d+2 c d x)^{7/2}-\left (2 \left (b^2-4 a c\right )^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 (2 \left (b^2-4 a c\right )^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 {4}{3} \left (b^2-4 a c\right ) d^3 (b d+2 c d x)^{3/2}+\frac {4}{7} d (b d+2 c d x)^{7/2}+2 \left (b^2-4 a c\right )^{7/4} d^{9/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-2 \left (b^2-4 a c\right )^{7/4} 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 [A]  time = 0.14, size = 131, normalized size = 0.89 \begin {gather*} \frac {4 (d (b+2 c x))^{9/2} \left (\frac {1}{7} (b+2 c x)^{7/2}-\frac {1}{6} \left (4 a c-b^2\right ) \left (3 \left (b^2-4 a c\right )^{3/4} \left (\tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )+2 (b+2 c x)^{3/2}\right )\right )}{(b+2 c x)^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(d*(b + 2*c*x))^(9/2)*((b + 2*c*x)^(7/2)/7 - ((-b^2 + 4*a*c)*(2*(b + 2*c*x)^(3/2) + 3*(b^2 - 4*a*c)^(3/4)*(
ArcTan[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)] - ArcTanh[Sqrt[b + 2*c*x]/(b^2 - 4*a*c)^(1/4)])))/6))/(b + 2*c*x)^
(9/2)

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IntegrateAlgebraic [C]  time = 0.36, size = 271, normalized size = 1.84 \begin {gather*} (-1-i) d^{9/2} \left (b^2-4 a c\right )^{7/4} \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 )-(1+i) d^{9/2} \left (b^2-4 a c\right )^{7/4} \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 )+\frac {8}{21} \sqrt {b d+2 c d x} \left (-14 a b c d^4-28 a c^2 d^4 x+5 b^3 d^4+16 b^2 c d^4 x+18 b c^2 d^4 x^2+12 c^3 d^4 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[b*d + 2*c*d*x]*(5*b^3*d^4 - 14*a*b*c*d^4 + 16*b^2*c*d^4*x - 28*a*c^2*d^4*x + 18*b*c^2*d^4*x^2 + 12*c^3
*d^4*x^3))/21 - (1 + I)*(b^2 - 4*a*c)^(7/4)*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]] - (1 +
I)*(b^2 - 4*a*c)^(7/4)*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))]

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fricas [B]  time = 0.48, size = 1320, normalized size = 8.98

result too large to display

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),x, algorithm="fricas")

[Out]

8/21*(12*c^3*d^4*x^3 + 18*b*c^2*d^4*x^2 + 4*(4*b^2*c - 7*a*c^2)*d^4*x + (5*b^3 - 14*a*b*c)*d^4)*sqrt(2*c*d*x +
b*d) - 4*((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)*d^18)^(1/4)*arctan(-(((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)*d^18)^(1/4)*(b^10 - 20*a*b^8*
c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*sqrt(2*c*d*x + b*d)*d^13 + sqrt(2*(b^
20*c - 40*a*b^18*c^2 + 720*a^2*b^16*c^3 - 7680*a^3*b^14*c^4 + 53760*a^4*b^12*c^5 - 258048*a^5*b^10*c^6 + 86016
0*a^6*b^8*c^7 - 1966080*a^7*b^6*c^8 + 2949120*a^8*b^4*c^9 - 2621440*a^9*b^2*c^10 + 1048576*a^10*c^11)*d^27*x +
(b^21 - 40*a*b^19*c + 720*a^2*b^17*c^2 - 7680*a^3*b^15*c^3 + 53760*a^4*b^13*c^4 - 258048*a^5*b^11*c^5 + 86016
0*a^6*b^9*c^6 - 1966080*a^7*b^7*c^7 + 2949120*a^8*b^5*c^8 - 2621440*a^9*b^3*c^9 + 1048576*a^10*b*c^10)*d^27 +
sqrt((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)*d^18)*(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)*d^18)*((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 224
0*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)*d^18)^(1/4))/((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)*d^18)) + ((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 - 215
04*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^18)^(1/4)*log(-(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 64
0*a^3*b^4*c^3 + 1280*a^4*b^2*c^4 - 1024*a^5*c^5)*sqrt(2*c*d*x + b*d)*d^13 + ((b^14 - 28*a*b^12*c + 336*a^2*b^1
0*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)*d^18)^(3/
4)) - ((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 + 2867
2*a^6*b^2*c^6 - 16384*a^7*c^7)*d^18)^(1/4)*log(-(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 1280*
a^4*b^2*c^4 - 1024*a^5*c^5)*sqrt(2*c*d*x + b*d)*d^13 - ((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)*d^18)^(3/4))

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giac [B]  time = 0.31, size = 531, normalized size = 3.61 \begin {gather*} \frac {4}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} d^{3} - \frac {16}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c d^{3} + \frac {4}{7} \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} d - {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c d^{3}\right )} \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 ) - {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c d^{3}\right )} \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 {1}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c d^{3}\right )} \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 {1}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} d^{3} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c d^{3}\right )} \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 ) \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),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 922, normalized size = 6.27 \begin {gather*} -\frac {16 \sqrt {2}\, a^{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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {16 \sqrt {2}\, a^{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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {8 \sqrt {2}\, a^{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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {8 \sqrt {2}\, a \,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 {8 \sqrt {2}\, a \,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 {4 \sqrt {2}\, a \,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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, b^{4} 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 {\sqrt {2}\, b^{4} 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 {\sqrt {2}\, b^{4} 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 \,d^{3}}{3}+\frac {4 \left (2 c d x +b d \right )^{\frac {3}{2}} b^{2} d^{3}}{3}+\frac {4 \left (2 c d x +b d \right )^{\frac {7}{2}} d}{7} \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),x)

[Out]

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

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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),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 = 0.58, size = 168, normalized size = 1.14 \begin {gather*} \frac {4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{7}-\frac {4\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (4\,a\,c-b^2\right )}{3}+2\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}+d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}\,1{}\mathrm {i}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}\,2{}\mathrm {i} \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),x)

[Out]

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

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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),x)

[Out]

Timed out

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