3.1296 \(\int \frac {(b d+2 c d x)^{13/2}}{(a+b x+c x^2)^2} \, dx\)

Optimal. Leaf size=181 \[ 22 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-22 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+\frac {44}{3} c d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2} \]

[Out]

44/3*c*(-4*a*c+b^2)*d^5*(2*c*d*x+b*d)^(3/2)+44/7*c*d^3*(2*c*d*x+b*d)^(7/2)-d*(2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a
)+22*c*(-4*a*c+b^2)^(7/4)*d^(13/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-22*c*(-4*a*c+b^2)^(7
/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 = 0.16, antiderivative size = 181, 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} \[ \frac {44}{3} c d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}+22 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-22 c d^{13/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(44*c*(b^2 - 4*a*c)*d^5*(b*d + 2*c*d*x)^(3/2))/3 + (44*c*d^3*(b*d + 2*c*d*x)^(7/2))/7 - (d*(b*d + 2*c*d*x)^(11
/2))/(a + b*x + c*x^2) + 22*c*(b^2 - 4*a*c)^(7/4)*d^(13/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqr
t[d])] - 22*c*(b^2 - 4*a*c)^(7/4)*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 )^2} \, dx &=-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\left (11 c d^2\right ) \int \frac {(b d+2 c d x)^{9/2}}{a+b x+c x^2} \, dx\\ &=\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\left (11 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=\frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\left (11 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=\frac {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\frac {1}{2} \left (11 \left (b^2-4 a c\right )^2 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 {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+\left (11 \left (b^2-4 a c\right )^2 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 {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}-\left (22 c \left (b^2-4 a c\right )^2 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 (22 c \left (b^2-4 a c\right )^2 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 {44}{3} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{3/2}+\frac {44}{7} c d^3 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{11/2}}{a+b x+c x^2}+22 c \left (b^2-4 a c\right )^{7/4} d^{13/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-22 c \left (b^2-4 a c\right )^{7/4} 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 = 0.17, size = 171, normalized size = 0.94 \[ \frac {4 d^5 (d (b+2 c x))^{3/2} \left (308 c \left (4 a^2 c+a \left (-b^2+4 b c x+4 c^2 x^2\right )-b^2 x (b+c x)\right ) \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};\frac {(b+2 c x)^2}{b^2-4 a c}\right )+16 c^2 \left (-77 a^2-11 a c x^2+3 c^2 x^4\right )+4 b^2 c \left (143 a+29 c x^2\right )+16 b c^2 x \left (6 c x^2-11 a\right )-63 b^4+68 b^3 c x\right )}{21 (a+x (b+c x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*d^5*(d*(b + 2*c*x))^(3/2)*(-63*b^4 + 68*b^3*c*x + 16*b*c^2*x*(-11*a + 6*c*x^2) + 4*b^2*c*(143*a + 29*c*x^2)
+ 16*c^2*(-77*a^2 - 11*a*c*x^2 + 3*c^2*x^4) + 308*c*(4*a^2*c - b^2*x*(b + c*x) + a*(-b^2 + 4*b*c*x + 4*c^2*x^
2))*Hypergeometric2F1[3/4, 2, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(21*(a + x*(b + c*x)))

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fricas [B]  time = 0.74, size = 1523, normalized size = 8.41 \[ \text {result too large to display} \]

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)^2,x, algorithm="fricas")

[Out]

-1/21*(924*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4
*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4)*(c*x^2 + b*x + a)*arctan(-(((b^14*c^4 - 28*a*b^12*c^5
+ 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*
c^11)*d^26)^(1/4)*(b^10*c^3 - 20*a*b^8*c^4 + 160*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c
^8)*sqrt(2*c*d*x + b*d)*d^19 + sqrt(2*(b^20*c^7 - 40*a*b^18*c^8 + 720*a^2*b^16*c^9 - 7680*a^3*b^14*c^10 + 5376
0*a^4*b^12*c^11 - 258048*a^5*b^10*c^12 + 860160*a^6*b^8*c^13 - 1966080*a^7*b^6*c^14 + 2949120*a^8*b^4*c^15 - 2
621440*a^9*b^2*c^16 + 1048576*a^10*c^17)*d^39*x + (b^21*c^6 - 40*a*b^19*c^7 + 720*a^2*b^17*c^8 - 7680*a^3*b^15
*c^9 + 53760*a^4*b^13*c^10 - 258048*a^5*b^11*c^11 + 860160*a^6*b^9*c^12 - 1966080*a^7*b^7*c^13 + 2949120*a^8*b
^5*c^14 - 2621440*a^9*b^3*c^15 + 1048576*a^10*b*c^16)*d^39 + sqrt((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6
- 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)*(b^14*
c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b
^2*c^10 - 16384*a^7*c^11)*d^26)*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^
6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4))/((b^14*c^4 - 28*a*b^12*c^5 + 336
*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)
*d^26)) - 231*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*
b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4)*(c*x^2 + b*x + a)*log(-1331*(b^10*c^3 - 20*a*b^8*c^
4 + 160*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt(2*c*d*x + b*d)*d^19 + 1331*((b^1
4*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6
*b^2*c^10 - 16384*a^7*c^11)*d^26)^(3/4)) + 231*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^
7 + 8960*a^4*b^6*c^8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(1/4)*(c*x^2 + b*x + a)*
log(-1331*(b^10*c^3 - 20*a*b^8*c^4 + 160*a^2*b^6*c^5 - 640*a^3*b^4*c^6 + 1280*a^4*b^2*c^7 - 1024*a^5*c^8)*sqrt
(2*c*d*x + b*d)*d^19 - 1331*((b^14*c^4 - 28*a*b^12*c^5 + 336*a^2*b^10*c^6 - 2240*a^3*b^8*c^7 + 8960*a^4*b^6*c^
8 - 21504*a^5*b^4*c^9 + 28672*a^6*b^2*c^10 - 16384*a^7*c^11)*d^26)^(3/4)) - (384*c^5*d^6*x^5 + 960*b*c^4*d^6*x
^4 + 32*(41*b^2*c^3 - 44*a*c^4)*d^6*x^3 + 48*(21*b^3*c^2 - 44*a*b*c^3)*d^6*x^2 + 2*(115*b^4*c + 88*a*b^2*c^2 -
1232*a^2*c^3)*d^6*x - (21*b^5 - 440*a*b^3*c + 1232*a^2*b*c^2)*d^6)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)

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giac [B]  time = 0.32, size = 646, normalized size = 3.57 \[ \frac {32}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c d^{5} - \frac {128}{3} \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{2} d^{5} + \frac {16}{7} \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c d^{3} - 11 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c^{2} d^{5}\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 ) - 11 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c^{2} d^{5}\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 {11}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c^{2} d^{5}\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 {11}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c d^{5} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a c^{2} d^{5}\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 {4 \, {\left ({\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{4} c d^{7} - 8 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a b^{2} c^{2} d^{7} + 16 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a^{2} c^{3} d^{7}\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)^(13/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")

[Out]

32/3*(2*c*d*x + b*d)^(3/2)*b^2*c*d^5 - 128/3*(2*c*d*x + b*d)^(3/2)*a*c^2*d^5 + 16/7*(2*c*d*x + b*d)^(7/2)*c*d^
3 - 11*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*c^2*d^5)*arc
tan(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)) -
11*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*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)) + 1
1/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*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)) - 11/2*(s
qrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c*d^5 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*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)) + 4*((2*c*d*x +
b*d)^(3/2)*b^4*c*d^7 - 8*(2*c*d*x + b*d)^(3/2)*a*b^2*c^2*d^7 + 16*(2*c*d*x + b*d)^(3/2)*a^2*c^3*d^7)/(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 = 1090, normalized size = 6.02 \[ -\frac {176 \sqrt {2}\, a^{2} 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 {176 \sqrt {2}\, a^{2} 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 {88 \sqrt {2}\, a^{2} 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 {88 \sqrt {2}\, a \,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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {88 \sqrt {2}\, a \,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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {44 \sqrt {2}\, a \,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 )}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {11 \sqrt {2}\, b^{4} c \,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 {11 \sqrt {2}\, b^{4} c \,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 {11 \sqrt {2}\, b^{4} c \,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 )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}-\frac {64 \left (2 c d x +b d \right )^{\frac {3}{2}} a^{2} c^{3} d^{7}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}}+\frac {32 \left (2 c d x +b d \right )^{\frac {3}{2}} a \,b^{2} c^{2} d^{7}}{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^{4} c \,d^{7}}{4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}}-\frac {128 \left (2 c d x +b d \right )^{\frac {3}{2}} a \,c^{2} d^{5}}{3}+\frac {32 \left (2 c d x +b d \right )^{\frac {3}{2}} b^{2} c \,d^{5}}{3}+\frac {16 \left (2 c d x +b d \right )^{\frac {7}{2}} c \,d^{3}}{7} \]

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)^2,x)

[Out]

16/7*c*d^3*(2*c*d*x+b*d)^(7/2)-128/3*c^2*d^5*(2*c*d*x+b*d)^(3/2)*a+32/3*c*d^5*(2*c*d*x+b*d)^(3/2)*b^2-64*c^3*d
^7*(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^2+32*c^2*d^7*(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*b^2-4*c*d^7*(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^4+88*c^3*d^7
*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a^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)))+176*c^3*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a^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)-176*c^3*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a^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)-44*c^2*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*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)))-88*c^2*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*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)+88*c^2*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*a*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)+11/2*c*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*b^4*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)))+11*c*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^2
)^(1/4)*b^4*arctan(2^(1/2)/(4*a*c*d^2-b^2*d^2)^(1/4)*(2*c*d*x+b*d)^(1/2)+1)-11*c*d^7*2^(1/2)/(4*a*c*d^2-b^2*d^
2)^(1/4)*b^4*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)^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 = 0.60, size = 249, normalized size = 1.38 \[ \frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{7}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (64\,a^2\,c^3\,d^7-32\,a\,b^2\,c^2\,d^7+4\,b^4\,c\,d^7\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}+22\,c\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}-\frac {32\,c\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (4\,a\,c-b^2\right )}{3}+c\,d^{13/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}\,1{}\mathrm {i}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}\,22{}\mathrm {i} \]

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)^2,x)

[Out]

(16*c*d^3*(b*d + 2*c*d*x)^(7/2))/7 - ((b*d + 2*c*d*x)^(3/2)*(4*b^4*c*d^7 + 64*a^2*c^3*d^7 - 32*a*b^2*c^2*d^7))
/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2) + 22*c*d^(13/2)*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) + c*d^(13/2)*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)*22i - (32*c*d^5*(b*d + 2*c*d*x)^(
3/2)*(4*a*c - b^2))/3

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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)**2,x)

[Out]

Timed out

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