3.1295 \(\int \frac {(b d+2 c d x)^{15/2}}{(a+b x+c x^2)^2} \, dx\)
Optimal. Leaf size=210 \[ -26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+52 c d^7 \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}+\frac {52}{5} c d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2} \]
[Out]
52/5*c*(-4*a*c+b^2)*d^5*(2*c*d*x+b*d)^(5/2)+52/9*c*d^3*(2*c*d*x+b*d)^(9/2)-d*(2*c*d*x+b*d)^(13/2)/(c*x^2+b*x+a
)-26*c*(-4*a*c+b^2)^(9/4)*d^(15/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-26*c*(-4*a*c+b^2)^(9
/4)*d^(15/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))+52*c*(-4*a*c+b^2)^2*d^7*(2*c*d*x+b*d)^(1/
2)
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Rubi [A] time = 0.18, antiderivative size = 210, normalized size of antiderivative = 1.00,
number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used =
{686, 692, 694, 329, 212, 206, 203} \[ 52 c d^7 \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}+\frac {52}{5} c d^5 \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}-26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-26 c d^{15/2} \left (b^2-4 a c\right )^{9/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2} \]
Antiderivative was successfully verified.
[In]
Int[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x]
[Out]
52*c*(b^2 - 4*a*c)^2*d^7*Sqrt[b*d + 2*c*d*x] + (52*c*(b^2 - 4*a*c)*d^5*(b*d + 2*c*d*x)^(5/2))/5 + (52*c*d^3*(b
*d + 2*c*d*x)^(9/2))/9 - (d*(b*d + 2*c*d*x)^(13/2))/(a + b*x + c*x^2) - 26*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*ArcT
an[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 26*c*(b^2 - 4*a*c)^(9/4)*d^(15/2)*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 212
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), 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)^{15/2}}{\left (a+b x+c x^2\right )^2} \, dx &=-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c d^2\right ) \int \frac {(b d+2 c d x)^{11/2}}{a+b x+c x^2} \, dx\\ &=\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c \left (b^2-4 a c\right ) d^4\right ) \int \frac {(b d+2 c d x)^{7/2}}{a+b x+c x^2} \, dx\\ &=\frac {52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c \left (b^2-4 a c\right )^2 d^6\right ) \int \frac {(b d+2 c d x)^{3/2}}{a+b x+c x^2} \, dx\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt {b d+2 c d x}+\frac {52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 c \left (b^2-4 a c\right )^3 d^8\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt {b d+2 c d x}+\frac {52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\frac {1}{2} \left (13 \left (b^2-4 a c\right )^3 d^7\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt {b d+2 c d x}+\frac {52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}+\left (13 \left (b^2-4 a c\right )^3 d^7\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt {b d+2 c d x}+\frac {52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}-\left (26 c \left (b^2-4 a c\right )^{5/2} d^8\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 (26 c \left (b^2-4 a c\right )^{5/2} d^8\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 )\\ &=52 c \left (b^2-4 a c\right )^2 d^7 \sqrt {b d+2 c d x}+\frac {52}{5} c \left (b^2-4 a c\right ) d^5 (b d+2 c d x)^{5/2}+\frac {52}{9} c d^3 (b d+2 c d x)^{9/2}-\frac {d (b d+2 c d x)^{13/2}}{a+b x+c x^2}-26 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-26 c \left (b^2-4 a c\right )^{9/4} d^{15/2} \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 = 0.60, size = 189, normalized size = 0.90 \[ -\frac {(d (b+2 c x))^{15/2} \left (-13 \left (b^2-4 a c\right ) \left (3 \left (b^2-4 a c\right ) \left (-30 \left (b^2-4 a c\right ) \sqrt {b+2 c x}-60 c \sqrt [4]{b^2-4 a c} (a+x (b+c x)) \left (\tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+\tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )+24 (b+2 c x)^{5/2}\right )+8 (b+2 c x)^{9/2}\right )-40 (b+2 c x)^{13/2}\right )}{90 (b+2 c x)^{15/2} (a+x (b+c x))} \]
Antiderivative was successfully verified.
[In]
Integrate[(b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x]
[Out]
-1/90*((d*(b + 2*c*x))^(15/2)*(-40*(b + 2*c*x)^(13/2) - 13*(b^2 - 4*a*c)*(8*(b + 2*c*x)^(9/2) + 3*(b^2 - 4*a*c
)*(-30*(b^2 - 4*a*c)*Sqrt[b + 2*c*x] + 24*(b + 2*c*x)^(5/2) - 60*c*(b^2 - 4*a*c)^(1/4)*(a + x*(b + c*x))*(ArcT
an[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)])))))/((b + 2*c*x)^(15/2
)*(a + x*(b + c*x)))
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fricas [B] time = 0.81, size = 1445, normalized size = 6.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
[Out]
-1/45*(2340*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^
5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)*(c*
x^2 + b*x + a)*arctan((((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8
- 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30
)^(3/4)*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d^7 - ((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14
*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11
+ 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(3/4)*sqrt(2*(b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^
3*b^2*c^6 + 256*a^4*c^7)*d^15*x + (b^9*c^2 - 16*a*b^7*c^3 + 96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*
d^15 + sqrt((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5
*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)))/((b^18*c
^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a
^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)) + 585*((b^18*c^4 - 36*a*b^16
*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 -
589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)*(c*x^2 + b*x + a)*log(13*(b^4*c - 8*a
*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d^7 + 13*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b
^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2
*c^12 - 262144*a^9*c^13)*d^30)^(1/4)) - 585*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7
+ 32256*a^4*b^10*c^8 - 129024*a^5*b^8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 -
262144*a^9*c^13)*d^30)^(1/4)*(c*x^2 + b*x + a)*log(13*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*sqrt(2*c*d*x + b*d)*d
^7 - 13*((b^18*c^4 - 36*a*b^16*c^5 + 576*a^2*b^14*c^6 - 5376*a^3*b^12*c^7 + 32256*a^4*b^10*c^8 - 129024*a^5*b^
8*c^9 + 344064*a^6*b^6*c^10 - 589824*a^7*b^4*c^11 + 589824*a^8*b^2*c^12 - 262144*a^9*c^13)*d^30)^(1/4)) - (128
0*c^6*d^7*x^6 + 3840*b*c^5*d^7*x^5 + 256*(22*b^2*c^4 - 13*a*c^5)*d^7*x^4 + 256*(19*b^3*c^3 - 26*a*b*c^4)*d^7*x
^3 + 96*(45*b^4*c^2 - 208*a*b^2*c^3 + 312*a^2*c^4)*d^7*x^2 + 32*(79*b^5*c - 520*a*b^3*c^2 + 936*a^2*b*c^3)*d^7
*x - (45*b^6 - 3068*a*b^4*c + 20592*a^2*b^2*c^2 - 37440*a^3*c^3)*d^7)*sqrt(2*c*d*x + b*d))/(c*x^2 + b*x + a)
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giac [B] time = 0.34, size = 876, normalized size = 4.17 \[ 48 \, \sqrt {2 \, c d x + b d} b^{4} c d^{7} - 384 \, \sqrt {2 \, c d x + b d} a b^{2} c^{2} d^{7} + 768 \, \sqrt {2 \, c d x + b d} a^{2} c^{3} d^{7} + \frac {32}{5} \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} b^{2} c d^{5} - \frac {128}{5} \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} a c^{2} d^{5} + \frac {16}{9} \, {\left (2 \, c d x + b d\right )}^{\frac {9}{2}} c d^{3} - 13 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{4} c d^{7} - 8 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a b^{2} c^{2} d^{7} + 16 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a^{2} c^{3} d^{7}\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 ) - 13 \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{4} c d^{7} - 8 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a b^{2} c^{2} d^{7} + 16 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a^{2} c^{3} d^{7}\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 {13}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{4} c d^{7} - 8 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a b^{2} c^{2} d^{7} + 16 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a^{2} c^{3} d^{7}\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 {13}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} b^{4} c d^{7} - 8 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a b^{2} c^{2} d^{7} + 16 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} a^{2} c^{3} d^{7}\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 (\sqrt {2 \, c d x + b d} b^{6} c d^{9} - 12 \, \sqrt {2 \, c d x + b d} a b^{4} c^{2} d^{9} + 48 \, \sqrt {2 \, c d x + b d} a^{2} b^{2} c^{3} d^{9} - 64 \, \sqrt {2 \, c d x + b d} a^{3} c^{4} d^{9}\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)^(15/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")
[Out]
48*sqrt(2*c*d*x + b*d)*b^4*c*d^7 - 384*sqrt(2*c*d*x + b*d)*a*b^2*c^2*d^7 + 768*sqrt(2*c*d*x + b*d)*a^2*c^3*d^7
+ 32/5*(2*c*d*x + b*d)^(5/2)*b^2*c*d^5 - 128/5*(2*c*d*x + b*d)^(5/2)*a*c^2*d^5 + 16/9*(2*c*d*x + b*d)^(9/2)*c
*d^3 - 13*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^4*c*d^7 - 8*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*b^2*c^2*d
^7 + 16*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a^2*c^3*d^7)*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)) - 13*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^4*c*d
^7 - 8*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*b^2*c^2*d^7 + 16*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a^2*c^3*d^
7)*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)) - 13/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^4*c*d^7 - 8*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*b^2*c^
2*d^7 + 16*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a^2*c^3*d^7)*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)) + 13/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*b^4*c*d
^7 - 8*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a*b^2*c^2*d^7 + 16*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*a^2*c^3*d^
7)*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*(sqrt(2*c*d*x + b*d)*b^6*c*d^9 - 12*sqrt(2*c*d*x + b*d)*a*b^4*c^2*d^9 + 48*sqrt(2*c*d*x + b*d)*a^2*b^2*c^3
*d^9 - 64*sqrt(2*c*d*x + b*d)*a^3*c^4*d^9)/(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 = 1512, normalized size = 7.20 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2*c*d*x+b*d)^(15/2)/(c*x^2+b*x+a)^2,x)
[Out]
16/9*c*d^3*(2*c*d*x+b*d)^(9/2)-128/5*c^2*d^5*(2*c*d*x+b*d)^(5/2)*a+32/5*c*d^5*(2*c*d*x+b*d)^(5/2)*b^2+768*c^3*
d^7*a^2*(2*c*d*x+b*d)^(1/2)-384*c^2*d^7*a*b^2*(2*c*d*x+b*d)^(1/2)+48*c*d^7*b^4*(2*c*d*x+b*d)^(1/2)+256*c^4*d^9
*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*a^3-192*c^3*d^9*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+
4*b*c*d^2*x+4*a*c*d^2)*a^2*b^2+48*c^2*d^9*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*a*b^4-4*c*
d^9*(2*c*d*x+b*d)^(1/2)/(4*c^2*d^2*x^2+4*b*c*d^2*x+4*a*c*d^2)*b^6-832*c^4*d^9/(4*a*c*d^2-b^2*d^2)^(3/4)*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^3+624*c^3*d^9/(4*a*c*d^2-b^2*d^2)^(3/4)*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*b^2-156*c^2*d^9/(4*a*c*d^2-b^2*d^2)^
(3/4)*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^4+13*c*d^9/(4*a*c*d^2-b^2*d^
2)^(3/4)*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^6+832*c^4*d^9/(4*a*c*d^2-b^
2*d^2)^(3/4)*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^3-624*c^3*d^9/(4*a*c*d
^2-b^2*d^2)^(3/4)*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*b^2+156*c^2*d^9
/(4*a*c*d^2-b^2*d^2)^(3/4)*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^4-13*c
*d^9/(4*a*c*d^2-b^2*d^2)^(3/4)*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^6-41
6*c^4*d^9/(4*a*c*d^2-b^2*d^2)^(3/4)*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^3+312*c^3*d^9/(4*a*c*d^2-b^2*d^2)^(3/4)*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*b^2-78*c^2*d^9/(4*a*c*d^2-b^2*d^2)^(3/4)*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^4+13/2*c*d^9/(4*a*c*d^2-b^2*d^2)^(3/4)*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^6
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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)^(15/2)/(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 = 0.69, size = 1060, normalized size = 5.05 \[ \frac {16\,c\,d^3\,{\left (b\,d+2\,c\,d\,x\right )}^{9/2}}{9}-\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (-256\,a^3\,c^4\,d^9+192\,a^2\,b^2\,c^3\,d^9-48\,a\,b^4\,c^2\,d^9+4\,b^6\,c\,d^9\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}+48\,c\,d^7\,\sqrt {b\,d+2\,c\,d\,x}\,{\left (4\,a\,c-b^2\right )}^2-26\,c\,d^{15/2}\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{9/4}-\frac {32\,c\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}\,\left (4\,a\,c-b^2\right )}{5}-c\,d^{15/2}\,\mathrm {atan}\left (\frac {c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (44302336\,a^6\,c^8\,d^{18}-66453504\,a^5\,b^2\,c^7\,d^{18}+41533440\,a^4\,b^4\,c^6\,d^{18}-13844480\,a^3\,b^6\,c^5\,d^{18}+2595840\,a^2\,b^8\,c^4\,d^{18}-259584\,a\,b^{10}\,c^3\,d^{18}+10816\,b^{12}\,c^2\,d^{18}\right )-13\,c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (212992\,a^4\,c^5\,d^{11}-212992\,a^3\,b^2\,c^4\,d^{11}+79872\,a^2\,b^4\,c^3\,d^{11}-13312\,a\,b^6\,c^2\,d^{11}+832\,b^8\,c\,d^{11}\right )\right )\,13{}\mathrm {i}+c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (44302336\,a^6\,c^8\,d^{18}-66453504\,a^5\,b^2\,c^7\,d^{18}+41533440\,a^4\,b^4\,c^6\,d^{18}-13844480\,a^3\,b^6\,c^5\,d^{18}+2595840\,a^2\,b^8\,c^4\,d^{18}-259584\,a\,b^{10}\,c^3\,d^{18}+10816\,b^{12}\,c^2\,d^{18}\right )+13\,c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (212992\,a^4\,c^5\,d^{11}-212992\,a^3\,b^2\,c^4\,d^{11}+79872\,a^2\,b^4\,c^3\,d^{11}-13312\,a\,b^6\,c^2\,d^{11}+832\,b^8\,c\,d^{11}\right )\right )\,13{}\mathrm {i}}{13\,c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (44302336\,a^6\,c^8\,d^{18}-66453504\,a^5\,b^2\,c^7\,d^{18}+41533440\,a^4\,b^4\,c^6\,d^{18}-13844480\,a^3\,b^6\,c^5\,d^{18}+2595840\,a^2\,b^8\,c^4\,d^{18}-259584\,a\,b^{10}\,c^3\,d^{18}+10816\,b^{12}\,c^2\,d^{18}\right )-13\,c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (212992\,a^4\,c^5\,d^{11}-212992\,a^3\,b^2\,c^4\,d^{11}+79872\,a^2\,b^4\,c^3\,d^{11}-13312\,a\,b^6\,c^2\,d^{11}+832\,b^8\,c\,d^{11}\right )\right )-13\,c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (44302336\,a^6\,c^8\,d^{18}-66453504\,a^5\,b^2\,c^7\,d^{18}+41533440\,a^4\,b^4\,c^6\,d^{18}-13844480\,a^3\,b^6\,c^5\,d^{18}+2595840\,a^2\,b^8\,c^4\,d^{18}-259584\,a\,b^{10}\,c^3\,d^{18}+10816\,b^{12}\,c^2\,d^{18}\right )+13\,c\,d^{15/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}\,\left (212992\,a^4\,c^5\,d^{11}-212992\,a^3\,b^2\,c^4\,d^{11}+79872\,a^2\,b^4\,c^3\,d^{11}-13312\,a\,b^6\,c^2\,d^{11}+832\,b^8\,c\,d^{11}\right )\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{9/4}\,26{}\mathrm {i} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*d + 2*c*d*x)^(15/2)/(a + b*x + c*x^2)^2,x)
[Out]
(16*c*d^3*(b*d + 2*c*d*x)^(9/2))/9 - ((b*d + 2*c*d*x)^(1/2)*(4*b^6*c*d^9 - 256*a^3*c^4*d^9 - 48*a*b^4*c^2*d^9
+ 192*a^2*b^2*c^3*d^9))/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2) + 48*c*d^7*(b*d + 2*c*d*x)^(1/2)*(4*a*c - b^
2)^2 - 26*c*d^(15/2)*atan((b^4*(b*d + 2*c*d*x)^(1/2) + 16*a^2*c^2*(b*d + 2*c*d*x)^(1/2) - 8*a*b^2*c*(b*d + 2*c
*d*x)^(1/2))/(d^(1/2)*(b^2 - 4*a*c)^(9/4)))*(b^2 - 4*a*c)^(9/4) - c*d^(15/2)*atan((c*d^(15/2)*(b^2 - 4*a*c)^(9
/4)*((b*d + 2*c*d*x)^(1/2)*(44302336*a^6*c^8*d^18 + 10816*b^12*c^2*d^18 - 259584*a*b^10*c^3*d^18 + 2595840*a^2
*b^8*c^4*d^18 - 13844480*a^3*b^6*c^5*d^18 + 41533440*a^4*b^4*c^6*d^18 - 66453504*a^5*b^2*c^7*d^18) - 13*c*d^(1
5/2)*(b^2 - 4*a*c)^(9/4)*(832*b^8*c*d^11 + 212992*a^4*c^5*d^11 - 13312*a*b^6*c^2*d^11 + 79872*a^2*b^4*c^3*d^11
- 212992*a^3*b^2*c^4*d^11))*13i + c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*((b*d + 2*c*d*x)^(1/2)*(44302336*a^6*c^8*d^1
8 + 10816*b^12*c^2*d^18 - 259584*a*b^10*c^3*d^18 + 2595840*a^2*b^8*c^4*d^18 - 13844480*a^3*b^6*c^5*d^18 + 4153
3440*a^4*b^4*c^6*d^18 - 66453504*a^5*b^2*c^7*d^18) + 13*c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*(832*b^8*c*d^11 + 21299
2*a^4*c^5*d^11 - 13312*a*b^6*c^2*d^11 + 79872*a^2*b^4*c^3*d^11 - 212992*a^3*b^2*c^4*d^11))*13i)/(13*c*d^(15/2)
*(b^2 - 4*a*c)^(9/4)*((b*d + 2*c*d*x)^(1/2)*(44302336*a^6*c^8*d^18 + 10816*b^12*c^2*d^18 - 259584*a*b^10*c^3*d
^18 + 2595840*a^2*b^8*c^4*d^18 - 13844480*a^3*b^6*c^5*d^18 + 41533440*a^4*b^4*c^6*d^18 - 66453504*a^5*b^2*c^7*
d^18) - 13*c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*(832*b^8*c*d^11 + 212992*a^4*c^5*d^11 - 13312*a*b^6*c^2*d^11 + 79872
*a^2*b^4*c^3*d^11 - 212992*a^3*b^2*c^4*d^11)) - 13*c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*((b*d + 2*c*d*x)^(1/2)*(4430
2336*a^6*c^8*d^18 + 10816*b^12*c^2*d^18 - 259584*a*b^10*c^3*d^18 + 2595840*a^2*b^8*c^4*d^18 - 13844480*a^3*b^6
*c^5*d^18 + 41533440*a^4*b^4*c^6*d^18 - 66453504*a^5*b^2*c^7*d^18) + 13*c*d^(15/2)*(b^2 - 4*a*c)^(9/4)*(832*b^
8*c*d^11 + 212992*a^4*c^5*d^11 - 13312*a*b^6*c^2*d^11 + 79872*a^2*b^4*c^3*d^11 - 212992*a^3*b^2*c^4*d^11))))*(
b^2 - 4*a*c)^(9/4)*26i - (32*c*d^5*(b*d + 2*c*d*x)^(5/2)*(4*a*c - b^2))/5
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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)**(15/2)/(c*x**2+b*x+a)**2,x)
[Out]
Timed out
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