3.1309 \(\int \frac {(b d+2 c d x)^{13/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=193 \[ 77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {154}{3} c^2 d^5 (b d+2 c d x)^{3/2} \]

[Out]

154/3*c^2*d^5*(2*c*d*x+b*d)^(3/2)-1/2*d*(2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)^2-11/2*c*d^3*(2*c*d*x+b*d)^(7/2)/(c
*x^2+b*x+a)+77*c^2*(-4*a*c+b^2)^(3/4)*d^(13/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-77*c^2*(
-4*a*c+b^2)^(3/4)*d^(13/2)*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 = 0.15, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ 77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-77 c^2 d^{13/2} \left (b^2-4 a c\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {154}{3} c^2 d^5 (b d+2 c d x)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^3,x]

[Out]

(154*c^2*d^5*(b*d + 2*c*d*x)^(3/2))/3 - (d*(b*d + 2*c*d*x)^(11/2))/(2*(a + b*x + c*x^2)^2) - (11*c*d^3*(b*d +
2*c*d*x)^(7/2))/(2*(a + b*x + c*x^2)) + 77*c^2*(b^2 - 4*a*c)^(3/4)*d^(13/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 -
4*a*c)^(1/4)*Sqrt[d])] - 77*c^2*(b^2 - 4*a*c)^(3/4)*d^(13/2)*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 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)^{13/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (11 c d^2\right ) \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)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (77 c^2 d^4\right ) \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=\frac {154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (77 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=\frac {154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{4} \left (77 c \left (b^2-4 a c\right ) d^5\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 {154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (77 c \left (b^2-4 a c\right ) d^5\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 {154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}-\left (77 c^2 \left (b^2-4 a c\right ) d^7\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 (77 c^2 \left (b^2-4 a c\right ) d^7\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 {154}{3} c^2 d^5 (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {11 c d^3 (b d+2 c d x)^{7/2}}{2 \left (a+b x+c x^2\right )}+77 c^2 \left (b^2-4 a c\right )^{3/4} d^{13/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-77 c^2 \left (b^2-4 a c\right )^{3/4} d^{13/2} \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 = 0.15, size = 145, normalized size = 0.75 \[ -\frac {4 d^5 (d (b+2 c x))^{3/2} \left (-16 c^2 \left (77 a^2+55 a c x^2+5 c^2 x^4\right )+1232 c^2 (a+x (b+c x))^2 \, _2F_1\left (\frac {3}{4},3;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )+4 b^2 c \left (99 a+25 c x^2\right )-80 b c^2 x \left (11 a+2 c x^2\right )-27 b^4+180 b^3 c x\right )}{15 (a+x (b+c x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^3,x]

[Out]

(-4*d^5*(d*(b + 2*c*x))^(3/2)*(-27*b^4 + 180*b^3*c*x - 80*b*c^2*x*(11*a + 2*c*x^2) + 4*b^2*c*(99*a + 25*c*x^2)
- 16*c^2*(77*a^2 + 55*a*c*x^2 + 5*c^2*x^4) + 1232*c^2*(a + x*(b + c*x))^2*Hypergeometric2F1[3/4, 3, 7/4, (b +
2*c*x)^2/(b^2 - 4*a*c)]))/(15*(a + x*(b + c*x))^2)

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fricas [B]  time = 0.84, size = 951, normalized size = 4.93 \[ \frac {924 \, \left ({\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}\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 ({\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}\right )^{\frac {1}{4}} {\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} \sqrt {2 \, c d x + b d} d^{19} - \sqrt {2 \, {\left (b^{8} c^{13} - 16 \, a b^{6} c^{14} + 96 \, a^{2} b^{4} c^{15} - 256 \, a^{3} b^{2} c^{16} + 256 \, a^{4} c^{17}\right )} d^{39} x + {\left (b^{9} c^{12} - 16 \, a b^{7} c^{13} + 96 \, a^{2} b^{5} c^{14} - 256 \, a^{3} b^{3} c^{15} + 256 \, a^{4} b c^{16}\right )} d^{39} + \sqrt {{\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}} {\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}} \left ({\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}\right )^{\frac {1}{4}}}{{\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}}\right ) - 231 \, \left ({\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}\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 (456533 \, {\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} \sqrt {2 \, c d x + b d} d^{19} + 456533 \, \left ({\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}\right )^{\frac {3}{4}}\right ) + 231 \, \left ({\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}\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 (456533 \, {\left (b^{4} c^{6} - 8 \, a b^{2} c^{7} + 16 \, a^{2} c^{8}\right )} \sqrt {2 \, c d x + b d} d^{19} - 456533 \, \left ({\left (b^{6} c^{8} - 12 \, a b^{4} c^{9} + 48 \, a^{2} b^{2} c^{10} - 64 \, a^{3} c^{11}\right )} d^{26}\right )^{\frac {3}{4}}\right ) + {\left (256 \, c^{5} d^{6} x^{5} + 640 \, b c^{4} d^{6} x^{4} + 2 \, {\left (199 \, b^{2} c^{3} + 484 \, a c^{4}\right )} d^{6} x^{3} - {\left (43 \, b^{3} c^{2} - 1452 \, a b c^{3}\right )} d^{6} x^{2} - {\left (63 \, b^{4} c - 418 \, a b^{2} c^{2} - 616 \, a^{2} c^{3}\right )} d^{6} x - {\left (3 \, b^{5} + 33 \, a b^{3} c - 308 \, a^{2} b c^{2}\right )} d^{6}\right )} \sqrt {2 \, c d x + b d}}{6 \, {\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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

1/6*(924*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(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((((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(1/4)*(b^4*
c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*sqrt(2*c*d*x + b*d)*d^19 - sqrt(2*(b^8*c^13 - 16*a*b^6*c^14 + 96*a^2*b^4*c^15
- 256*a^3*b^2*c^16 + 256*a^4*c^17)*d^39*x + (b^9*c^12 - 16*a*b^7*c^13 + 96*a^2*b^5*c^14 - 256*a^3*b^3*c^15 + 2
56*a^4*b*c^16)*d^39 + sqrt((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)*(b^6*c^8 - 12*a*b^4*
c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(1/
4))/((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)) - 231*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b
^2*c^10 - 64*a^3*c^11)*d^26)^(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(456533*(b^4*c
^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*sqrt(2*c*d*x + b*d)*d^19 + 456533*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 -
64*a^3*c^11)*d^26)^(3/4)) + 231*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(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(456533*(b^4*c^6 - 8*a*b^2*c^7 + 16*a^2*c^8)*sqrt(2*c*d*x
+ b*d)*d^19 - 456533*((b^6*c^8 - 12*a*b^4*c^9 + 48*a^2*b^2*c^10 - 64*a^3*c^11)*d^26)^(3/4)) + (256*c^5*d^6*x^
5 + 640*b*c^4*d^6*x^4 + 2*(199*b^2*c^3 + 484*a*c^4)*d^6*x^3 - (43*b^3*c^2 - 1452*a*b*c^3)*d^6*x^2 - (63*b^4*c
- 418*a*b^2*c^2 - 616*a^2*c^3)*d^6*x - (3*b^5 + 33*a*b^3*c - 308*a^2*b*c^2)*d^6)*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)

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giac [B]  time = 0.34, size = 521, normalized size = 2.70 \[ -\frac {77}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{5} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {77}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{5} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) + \frac {77}{4} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{5} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) - \frac {77}{4} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} c^{2} d^{5} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {64}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} c^{2} d^{5} + \frac {2 \, {\left (15 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{4} c^{2} d^{9} - 120 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a b^{2} c^{3} d^{9} + 240 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a^{2} c^{4} d^{9} - 19 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b^{2} c^{2} d^{7} + 76 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} a c^{3} d^{7}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-77/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*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)) - 77/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*arcta
n(-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)) +
77/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*log(2*c*d*x + b*d + sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*sqr
t(2*c*d*x + b*d) + sqrt(-b^2*d^2 + 4*a*c*d^2)) - 77/4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*c^2*d^5*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)) + 64/3*(2*c*d*
x + b*d)^(3/2)*c^2*d^5 + 2*(15*(2*c*d*x + b*d)^(3/2)*b^4*c^2*d^9 - 120*(2*c*d*x + b*d)^(3/2)*a*b^2*c^3*d^9 + 2
40*(2*c*d*x + b*d)^(3/2)*a^2*c^4*d^9 - 19*(2*c*d*x + b*d)^(7/2)*b^2*c^2*d^7 + 76*(2*c*d*x + b*d)^(7/2)*a*c^3*d
^7)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2

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maple [B]  time = 0.07, size = 857, normalized size = 4.44 \[ \frac {480 \left (2 c d x +b d \right )^{\frac {3}{2}} a^{2} c^{4} d^{9}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}-\frac {240 \left (2 c d x +b d \right )^{\frac {3}{2}} a \,b^{2} c^{3} d^{9}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {30 \left (2 c d x +b d \right )^{\frac {3}{2}} b^{4} c^{2} d^{9}}{\left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {154 \sqrt {2}\, a \,c^{3} d^{7} \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 {154 \sqrt {2}\, a \,c^{3} d^{7} \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 {77 \sqrt {2}\, a \,c^{3} d^{7} \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 {77 \sqrt {2}\, b^{2} c^{2} d^{7} \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 {77 \sqrt {2}\, b^{2} c^{2} d^{7} \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 {77 \sqrt {2}\, b^{2} c^{2} d^{7} \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 {152 \left (2 c d x +b d \right )^{\frac {7}{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 {38 \left (2 c d x +b d \right )^{\frac {7}{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 {64 \left (2 c d x +b d \right )^{\frac {3}{2}} c^{2} d^{5}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a)^3,x)

[Out]

64/3*c^2*d^5*(2*c*d*x+b*d)^(3/2)+152*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)^(7/2)*a-38*
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)^(7/2)*b^2+480*c^4*d^9/(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^2-240*c^3*d^9/(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*b^2+30*c^2*d^9/(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^4-77*c^3*d^7*2^(1/2)/(4*a*c*d^2
-b^2*d^2)^(1/4)*a*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)))-154*c^3*d^
7*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)+154*c^3*
d^7*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)+77/4*
c^2*d^7*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)))+77/2*c^2*d^7*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)-77/2*c^2*d^7*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)^(13/2)/(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 = 0.63, size = 264, normalized size = 1.37 \[ \frac {{\left (b\,d+2\,c\,d\,x\right )}^{7/2}\,\left (152\,a\,c^3\,d^7-38\,b^2\,c^2\,d^7\right )+{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (480\,a^2\,c^4\,d^9-240\,a\,b^2\,c^3\,d^9+30\,b^4\,c^2\,d^9\right )}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}+\frac {64\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}}{3}+77\,c^2\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}+c^2\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}\,77{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(13/2)/(a + b*x + c*x^2)^3,x)

[Out]

((b*d + 2*c*d*x)^(7/2)*(152*a*c^3*d^7 - 38*b^2*c^2*d^7) + (b*d + 2*c*d*x)^(3/2)*(480*a^2*c^4*d^9 + 30*b^4*c^2*
d^9 - 240*a*b^2*c^3*d^9))/((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) + (64*c^2*d^5*(b*d + 2*c*d*x)^(3/2))/3 + 77*c^2*d^(13/2)*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^2*d^(13/2)*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)*77i

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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)**(13/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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