3.12.25 \(\int \frac {(b d+2 c d x)^{5/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=178 \[ -\frac {3 c^2 d^{5/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}+\frac {3 c^2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/4}}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2} \]

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Rubi [A]  time = 0.13, antiderivative size = 178, 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, 687, 694, 329, 298, 203, 206} \begin {gather*} -\frac {3 c^2 d^{5/2} \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 )^{5/4}}+\frac {3 c^2 d^{5/2} \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 )^{5/4}}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2)^3,x]

[Out]

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

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 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 {(b d+2 c d x)^{5/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (3 c d^2\right ) \int \frac {\sqrt {b d+2 c d x}}{\left (a+b x+c x^2\right )^2} \, dx\\ &=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {\left (3 c^2 d^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 )}\\ &=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(3 c d) \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 )}\\ &=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {(3 c d) \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 )}\\ &=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}+\frac {\left (3 c^2 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 )}{b^2-4 a c}-\frac {\left (3 c^2 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 )}{b^2-4 a c}\\ &=-\frac {d (b d+2 c d x)^{3/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {3 c d (b d+2 c d x)^{3/2}}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )}-\frac {3 c^2 d^{5/2} \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 )^{5/4}}+\frac {3 c^2 d^{5/2} \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 )^{5/4}}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 77, normalized size = 0.43 \begin {gather*} \frac {64}{5} c^2 d (d (b+2 c x))^{3/2} \left (\frac {\, _2F_1\left (\frac {3}{4},3;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right )^2}-\frac {1}{16 c^2 (a+x (b+c x))^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2)^3,x]

[Out]

(64*c^2*d*(d*(b + 2*c*x))^(3/2)*(-1/16*1/(c^2*(a + x*(b + c*x))^2) + Hypergeometric2F1[3/4, 3, 7/4, (b + 2*c*x
)^2/(b^2 - 4*a*c)]/(b^2 - 4*a*c)^2))/5

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IntegrateAlgebraic [C]  time = 2.21, size = 306, normalized size = 1.72 \begin {gather*} \frac {\left (\frac {3}{2}+\frac {3 i}{2}\right ) c^2 d^{5/2} \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 )}{\left (b^2-4 a c\right )^{5/4}}+\frac {\left (\frac {3}{2}+\frac {3 i}{2}\right ) c^2 d^{5/2} \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 )}{\left (b^2-4 a c\right )^{5/4}}+\frac {\sqrt {b d+2 c d x} \left (a b c d^2+2 a c^2 d^2 x+b^3 \left (-d^2\right )-5 b^2 c d^2 x-9 b c^2 d^2 x^2-6 c^3 d^2 x^3\right )}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*d + 2*c*d*x)^(5/2)/(a + b*x + c*x^2)^3,x]

[Out]

(Sqrt[b*d + 2*c*d*x]*(-(b^3*d^2) + a*b*c*d^2 - 5*b^2*c*d^2*x + 2*a*c^2*d^2*x - 9*b*c^2*d^2*x^2 - 6*c^3*d^2*x^3
))/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + ((3/2 + (3*I)/2)*c^2*d^(5/2)*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)^(5/4) + ((3/2 + (3*I)/2)*c^2*d^(5/2)*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)^(5/4)

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fricas [B]  time = 0.45, size = 1174, normalized size = 6.60

result too large to display

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)^3,x, algorithm="fricas")

[Out]

-1/2*(12*(c^8*d^10/(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))^
(1/4)*((b^2*c^2 - 4*a*c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)
*x^2 + 2*(a*b^3 - 4*a^2*b*c)*x)*arctan(((c^8*d^10/(b^10 - 20*a*b^8*c + 160*a^2*b^6*c^2 - 640*a^3*b^4*c^3 + 128
0*a^4*b^2*c^4 - 1024*a^5*c^5))^(1/4)*(b^2*c^6 - 4*a*c^7)*sqrt(2*c*d*x + b*d)*d^7 - sqrt(2*c^13*d^15*x + b*c^12
*d^15 + sqrt(c^8*d^10/(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
))*(b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^10)*(c^8*d^10/(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))^(1/4)*(b^2 - 4*a*c))/(c^8*d^10)) - 3*(c^8*d^10/(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))^(1/4)*((b^2*c^2 - 4*a*c^3
)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*x^2 + 2*(a*b^3 - 4*a^2*b
*c)*x)*log(27*sqrt(2*c*d*x + b*d)*c^6*d^7 + 27*(c^8*d^10/(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))^(3/4)*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^3*b^2*c^3 + 256*a^4*c^4
)) + 3*(c^8*d^10/(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))^(1
/4)*((b^2*c^2 - 4*a*c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*a*b^2*c - 8*a^2*c^2)*x
^2 + 2*(a*b^3 - 4*a^2*b*c)*x)*log(27*sqrt(2*c*d*x + b*d)*c^6*d^7 - 27*(c^8*d^10/(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))^(3/4)*(b^8 - 16*a*b^6*c + 96*a^2*b^4*c^2 - 256*a^
3*b^2*c^3 + 256*a^4*c^4)) + (6*c^3*d^2*x^3 + 9*b*c^2*d^2*x^2 + (5*b^2*c - 2*a*c^2)*d^2*x + (b^3 - a*b*c)*d^2)*
sqrt(2*c*d*x + b*d))/((b^2*c^2 - 4*a*c^3)*x^4 + a^2*b^2 - 4*a^3*c + 2*(b^3*c - 4*a*b*c^2)*x^3 + (b^4 - 2*a*b^2
*c - 8*a^2*c^2)*x^2 + 2*(a*b^3 - 4*a^2*b*c)*x)

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giac [B]  time = 0.32, size = 567, normalized size = 3.19 \begin {gather*} \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} 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} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} 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} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}} - \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} 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 )}{2 \, {\left (\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}\right )}} + \frac {3 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} 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 )}{2 \, {\left (\sqrt {2} b^{4} - 8 \, \sqrt {2} a b^{2} c + 16 \, \sqrt {2} a^{2} c^{2}\right )}} - \frac {2 \, {\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{5} - 4 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{5} + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2} d^{3}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2} {\left (b^{2} - 4 \, a c\right )}} \end {gather*}

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)^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 431, normalized size = 2.42 \begin {gather*} -\frac {2 \left (2 c d x +b d \right )^{\frac {3}{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}}-\frac {3 \sqrt {2}\, c^{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 )}{2 \left (4 a c -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {2}\, c^{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 )}{2 \left (4 a c -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {3 \sqrt {2}\, c^{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 )}{4 \left (4 a c -b^{2}\right ) \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+\frac {6 \left (2 c d x +b d \right )^{\frac {7}{2}} c^{2} d^{3}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2} \left (4 a c -b^{2}\right )} \end {gather*}

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)^3,x)

[Out]

6*c^2*d^3/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)^2/(4*a*c-b^2)*(2*c*d*x+b*d)^(7/2)-2*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)^(3/2)+3/4*c^2*d^3/(4*a*c-b^2)/(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)))+3/2*c^2*d^3/(4*a*c-b^2)/(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)-3/2*c^2*d^3/(4*a*c-b^2)/(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)

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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)^(5/2)/(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 = 0.14, size = 251, normalized size = 1.41 \begin {gather*} \frac {3\,c^2\,d^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {3\,c^2\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{5/4}}-\frac {2\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}-\frac {6\,c^2\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{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} \end {gather*}

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)^3,x)

[Out]

(3*c^2*d^(5/2)*atanh(((b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(d^(1/2)*(b^2 - 4*a*c)^(9/4))))/(b
^2 - 4*a*c)^(5/4) - (3*c^2*d^(5/2)*atan(((b*d + 2*c*d*x)^(1/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(d^(1/2)*(b^2 -
4*a*c)^(9/4))))/(b^2 - 4*a*c)^(5/4) - (2*c^2*d^5*(b*d + 2*c*d*x)^(3/2) - (6*c^2*d^3*(b*d + 2*c*d*x)^(7/2))/(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)

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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)**(5/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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