3.1288 \(\int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=119 \[ 2 d^{5/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{5/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {4}{3} d (b d+2 c d x)^{3/2} \]

[Out]

4/3*d*(2*c*d*x+b*d)^(3/2)+2*(-4*a*c+b^2)^(3/4)*d^(5/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-
2*(-4*a*c+b^2)^(3/4)*d^(5/2)*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 = 0.10, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ 2 d^{5/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-2 d^{5/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {4}{3} d (b d+2 c d x)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2),x]

[Out]

(4*d*(b*d + 2*c*d*x)^(3/2))/3 + 2*(b^2 - 4*a*c)^(3/4)*d^(5/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*
Sqrt[d])] - 2*(b^2 - 4*a*c)^(3/4)*d^(5/2)*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)^{5/2}}{a+b x+c x^2} \, dx &=\frac {4}{3} d (b d+2 c d x)^{3/2}+\left (\left (b^2-4 a c\right ) d^2\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=\frac {4}{3} d (b d+2 c d x)^{3/2}+\frac {\left (\left (b^2-4 a c\right ) d\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} d (b d+2 c d x)^{3/2}+\frac {\left (\left (b^2-4 a c\right ) d\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} d (b d+2 c d x)^{3/2}-\left (2 \left (b^2-4 a c\right ) d^3\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 ) d^3\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} d (b d+2 c d x)^{3/2}+2 \left (b^2-4 a c\right )^{3/4} d^{5/2} \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 )^{3/4} d^{5/2} \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 = 0.12, size = 104, normalized size = 0.87 \[ \frac {2 (d (b+2 c x))^{5/2} \left (3 \left (b^2-4 a c\right )^{3/4} \left (\tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-\tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )+2 (b+2 c x)^{3/2}\right )}{3 (b+2 c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2),x]

[Out]

(2*(d*(b + 2*c*x))^(5/2)*(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)])))/(3*(b + 2*c*x)^(5/2))

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fricas [B]  time = 1.04, size = 602, normalized size = 5.06 \[ \frac {4}{3} \, {\left (2 \, c d^{2} x + b d^{2}\right )} \sqrt {2 \, c d x + b d} + 4 \, \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \arctan \left (\frac {\left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} - \sqrt {2 \, {\left (b^{8} c - 16 \, a b^{6} c^{2} + 96 \, a^{2} b^{4} c^{3} - 256 \, a^{3} b^{2} c^{4} + 256 \, a^{4} c^{5}\right )} d^{15} x + {\left (b^{9} - 16 \, a b^{7} c + 96 \, a^{2} b^{5} c^{2} - 256 \, a^{3} b^{3} c^{3} + 256 \, a^{4} b c^{4}\right )} d^{15} + \sqrt {{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}} {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}} \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}}}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}}\right ) - \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} + \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) + \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {1}{4}} \log \left ({\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {2 \, c d x + b d} d^{7} - \left ({\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{10}\right )^{\frac {3}{4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

4/3*(2*c*d^2*x + b*d^2)*sqrt(2*c*d*x + b*d) + 4*((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(1/4)*
arctan((((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(1/4)*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*
d*x + b*d)*d^7 - sqrt(2*(b^8*c - 16*a*b^6*c^2 + 96*a^2*b^4*c^3 - 256*a^3*b^2*c^4 + 256*a^4*c^5)*d^15*x + (b^9
- 16*a*b^7*c + 96*a^2*b^5*c^2 - 256*a^3*b^3*c^3 + 256*a^4*b*c^4)*d^15 + sqrt((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^
2 - 64*a^3*c^3)*d^10)*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)*((b^6 - 12*a*b^4*c + 48*a^2*b^2*c
^2 - 64*a^3*c^3)*d^10)^(1/4))/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)) - ((b^6 - 12*a*b^4*c +
48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*d*x + b*d)*d^7 + ((b^6 -
12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^10)^(3/4)) + ((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^1
0)^(1/4)*log((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(2*c*d*x + b*d)*d^7 - ((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64
*a^3*c^3)*d^10)^(3/4))

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giac [B]  time = 0.26, size = 354, normalized size = 2.97 \[ -\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) - \frac {1}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} d \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {4}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

-sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*d*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*d^2 + 4*a*c*d^2)^(3/4)*d*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)*d*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)*d*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)) + 4/3*(2*c*d*x + b*d)^(3/2)*d

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maple [B]  time = 0.06, size = 582, normalized size = 4.89 \[ \frac {4 \sqrt {2}\, a c \,d^{3} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {4 \sqrt {2}\, a c \,d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {2 \sqrt {2}\, a c \,d^{3} \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {\sqrt {2}\, b^{2} d^{3} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, b^{2} d^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {\sqrt {2}\, b^{2} d^{3} \ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {4 \left (2 c d x +b d \right )^{\frac {3}{2}} d}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a),x)

[Out]

4/3*d*(2*c*d*x+b*d)^(3/2)-4*d^3/(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*c+d^3/(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^2+4*d^3/(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*c-d^3/(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^2-2*d^3/(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*c+1/2*d^3/(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^2

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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)^(5/2)/(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 = 0.13, size = 97, normalized size = 0.82 \[ \frac {4\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{3}+2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}-2\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2),x)

[Out]

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

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sympy [B]  time = 97.72, size = 700, normalized size = 5.88 \[ 32 a b c d^{4} \operatorname {RootSum} {\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left (t \mapsto t \log {\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} - 32 a b c d^{4} \operatorname {RootSum} {\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left (t \mapsto t \log {\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} - 16 a c d^{3} \operatorname {RootSum} {\left (t^{4} \left (1024 a c d^{2} - 256 b^{2} d^{2}\right ) + 1, \left (t \mapsto t \log {\left (256 t^{3} a c d^{2} - 64 t^{3} b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} - 8 b^{3} d^{4} \operatorname {RootSum} {\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left (t \mapsto t \log {\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} + 8 b^{3} d^{4} \operatorname {RootSum} {\left (t^{4} \left (16384 a^{3} c^{3} d^{6} - 12288 a^{2} b^{2} c^{2} d^{6} + 3072 a b^{4} c d^{6} - 256 b^{6} d^{6}\right ) + 1, \left (t \mapsto t \log {\left (16 t a c d^{2} - 4 t b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} + 8 b^{2} d^{3} \operatorname {RootSum} {\left (t^{4} \left (1024 a c d^{2} - 256 b^{2} d^{2}\right ) + 1, \left (t \mapsto t \log {\left (256 t^{3} a c d^{2} - 64 t^{3} b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} - 8 b^{2} d^{3} \operatorname {RootSum} {\left (t^{4} \left (1024 a c d^{2} - 256 b^{2} d^{2}\right ) + 1, \left (t \mapsto t \log {\left (256 t^{3} a c d^{2} - 64 t^{3} b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} + 4 b^{2} d^{3} \operatorname {RootSum} {\left (t^{4} \left (1024 a c d^{2} - 256 b^{2} d^{2}\right ) + 1, \left (t \mapsto t \log {\left (256 t^{3} a c d^{2} - 64 t^{3} b^{2} d^{2} + \sqrt {b d + 2 c d x} \right )} \right )\right )} + \frac {4 d \left (b d + 2 c d x\right )^{\frac {3}{2}}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)**(5/2)/(c*x**2+b*x+a),x)

[Out]

32*a*b*c*d**4*RootSum(_t**4*(16384*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*
d**6) + 1, Lambda(_t, _t*log(16*_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) - 32*a*b*c*d**4*RootSum(
_t**4*(16384*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*d**6) + 1, Lambda(_t,
_t*log(16*_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) - 16*a*c*d**3*RootSum(_t**4*(1024*a*c*d**2 - 2
56*b**2*d**2) + 1, Lambda(_t, _t*log(256*_t**3*a*c*d**2 - 64*_t**3*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) - 8*b**3
*d**4*RootSum(_t**4*(16384*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*d**6) +
1, Lambda(_t, _t*log(16*_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) + 8*b**3*d**4*RootSum(_t**4*(163
84*a**3*c**3*d**6 - 12288*a**2*b**2*c**2*d**6 + 3072*a*b**4*c*d**6 - 256*b**6*d**6) + 1, Lambda(_t, _t*log(16*
_t*a*c*d**2 - 4*_t*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) + 8*b**2*d**3*RootSum(_t**4*(1024*a*c*d**2 - 256*b**2*d*
*2) + 1, Lambda(_t, _t*log(256*_t**3*a*c*d**2 - 64*_t**3*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) - 8*b**2*d**3*Root
Sum(_t**4*(1024*a*c*d**2 - 256*b**2*d**2) + 1, Lambda(_t, _t*log(256*_t**3*a*c*d**2 - 64*_t**3*b**2*d**2 + sqr
t(b*d + 2*c*d*x)))) + 4*b**2*d**3*RootSum(_t**4*(1024*a*c*d**2 - 256*b**2*d**2) + 1, Lambda(_t, _t*log(256*_t*
*3*a*c*d**2 - 64*_t**3*b**2*d**2 + sqrt(b*d + 2*c*d*x)))) + 4*d*(b*d + 2*c*d*x)**(3/2)/3

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