3.12.19 \(\int \frac {(b d+2 c d x)^{17/2}}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=222 \[ 165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+110 c^2 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2} \]

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Rubi [A]  time = 0.19, antiderivative size = 222, 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, 298, 203, 206} \begin {gather*} 110 c^2 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}+165 c^2 d^{17/2} \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 )-165 c^2 d^{17/2} \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 {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*d + 2*c*d*x)^(17/2)/(a + b*x + c*x^2)^3,x]

[Out]

110*c^2*(b^2 - 4*a*c)*d^7*(b*d + 2*c*d*x)^(3/2) + (330*c^2*d^5*(b*d + 2*c*d*x)^(7/2))/7 - (d*(b*d + 2*c*d*x)^(
15/2))/(2*(a + b*x + c*x^2)^2) - (15*c*d^3*(b*d + 2*c*d*x)^(11/2))/(2*(a + b*x + c*x^2)) + 165*c^2*(b^2 - 4*a*
c)^(7/4)*d^(17/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])] - 165*c^2*(b^2 - 4*a*c)^(7/4)*d^(1
7/2)*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)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (15 c d^2\right ) \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)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (165 c^2 d^4\right ) \int \frac {(b d+2 c d x)^{9/2}}{a+b x+c x^2} \, dx\\ &=\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (165 c^2 \left (b^2-4 a c\right ) d^6\right ) \int \frac {(b d+2 c d x)^{5/2}}{a+b x+c x^2} \, dx\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (165 c^2 \left (b^2-4 a c\right )^2 d^8\right ) \int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{4} \left (165 c \left (b^2-4 a c\right )^2 d^7\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 )\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (165 c \left (b^2-4 a c\right )^2 d^7\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 )\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}-\left (165 c^2 \left (b^2-4 a c\right )^2 d^9\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 (165 c^2 \left (b^2-4 a c\right )^2 d^9\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 )\\ &=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+165 c^2 \left (b^2-4 a c\right )^{7/4} d^{17/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-165 c^2 \left (b^2-4 a c\right )^{7/4} d^{17/2} \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 = 0.31, size = 138, normalized size = 0.62 \begin {gather*} \frac {4 (d (b+2 c x))^{17/2} \left (\left (b^2-4 a c\right ) \left (77 \left (\left (b^2-4 a c\right )^2-16 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 )\right )-55 \left (b^2-4 a c\right ) (b+2 c x)^2+5 (b+2 c x)^4\right )+(b+2 c x)^6\right )}{7 (b+2 c x)^7 (a+x (b+c x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*d + 2*c*d*x)^(17/2)/(a + b*x + c*x^2)^3,x]

[Out]

(4*(d*(b + 2*c*x))^(17/2)*((b + 2*c*x)^6 + (b^2 - 4*a*c)*(-55*(b^2 - 4*a*c)*(b + 2*c*x)^2 + 5*(b + 2*c*x)^4 +
77*((b^2 - 4*a*c)^2 - 16*c^2*(a + x*(b + c*x))^2*Hypergeometric2F1[3/4, 3, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)]))
))/(7*(b + 2*c*x)^7*(a + x*(b + c*x))^2)

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IntegrateAlgebraic [C]  time = 1.68, size = 405, normalized size = 1.82 \begin {gather*} -\frac {2 c^2 d^5 (b d+2 c d x)^{3/2} \left (24640 a^3 c^3 d^6-18480 a^2 b^2 c^2 d^6+9680 a^2 c^2 d^4 (b d+2 c d x)^2+4620 a b^4 c d^6-4840 a b^2 c d^4 (b d+2 c d x)^2+640 a c d^2 (b d+2 c d x)^4-385 b^6 d^6+605 b^4 d^4 (b d+2 c d x)^2-160 b^2 d^2 (b d+2 c d x)^4-32 (b d+2 c d x)^6\right )}{7 \left (4 a c d^2-b^2 d^2+(b d+2 c d x)^2\right )^2}+\left (-\frac {165}{2}-\frac {165 i}{2}\right ) c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tan ^{-1}\left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {d} \sqrt [4]{b^2-4 a c}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b d+2 c d x)}{\sqrt {d} \sqrt [4]{b^2-4 a c}}}{\sqrt {b d+2 c d x}}\right )-\left (\frac {165}{2}+\frac {165 i}{2}\right ) c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \tanh ^{-1}\left (\frac {(1+i) \sqrt {d} \sqrt [4]{b^2-4 a c} \sqrt {b d+2 c d x}}{d \sqrt {b^2-4 a c}+i (b d+2 c d x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*d + 2*c*d*x)^(17/2)/(a + b*x + c*x^2)^3,x]

[Out]

(-2*c^2*d^5*(b*d + 2*c*d*x)^(3/2)*(-385*b^6*d^6 + 4620*a*b^4*c*d^6 - 18480*a^2*b^2*c^2*d^6 + 24640*a^3*c^3*d^6
+ 605*b^4*d^4*(b*d + 2*c*d*x)^2 - 4840*a*b^2*c*d^4*(b*d + 2*c*d*x)^2 + 9680*a^2*c^2*d^4*(b*d + 2*c*d*x)^2 - 1
60*b^2*d^2*(b*d + 2*c*d*x)^4 + 640*a*c*d^2*(b*d + 2*c*d*x)^4 - 32*(b*d + 2*c*d*x)^6))/(7*(-(b^2*d^2) + 4*a*c*d
^2 + (b*d + 2*c*d*x)^2)^2) - (165/2 + (165*I)/2)*c^2*(b^2 - 4*a*c)^(7/4)*d^(17/2)*ArcTan[((1/2 - I/2)*(b^2 - 4
*a*c)^(1/4)*Sqrt[d] - ((1/2 + I/2)*(b*d + 2*c*d*x))/((b^2 - 4*a*c)^(1/4)*Sqrt[d]))/Sqrt[b*d + 2*c*d*x]] - (165
/2 + (165*I)/2)*c^2*(b^2 - 4*a*c)^(7/4)*d^(17/2)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[d]*Sqrt[b*d + 2*c*d
*x])/(Sqrt[b^2 - 4*a*c]*d + I*(b*d + 2*c*d*x))]

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fricas [B]  time = 0.49, size = 1713, normalized size = 7.72

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(17/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/14*(4620*((b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^5
*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34)^(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^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 215
04*a^5*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34)^(1/4)*(b^10*c^6 - 20*a*b^8*c^7 + 160*a^2*b^6*c^8
- 640*a^3*b^4*c^9 + 1280*a^4*b^2*c^10 - 1024*a^5*c^11)*sqrt(2*c*d*x + b*d)*d^25 + sqrt(2*(b^20*c^13 - 40*a*b^1
8*c^14 + 720*a^2*b^16*c^15 - 7680*a^3*b^14*c^16 + 53760*a^4*b^12*c^17 - 258048*a^5*b^10*c^18 + 860160*a^6*b^8*
c^19 - 1966080*a^7*b^6*c^20 + 2949120*a^8*b^4*c^21 - 2621440*a^9*b^2*c^22 + 1048576*a^10*c^23)*d^51*x + (b^21*
c^12 - 40*a*b^19*c^13 + 720*a^2*b^17*c^14 - 7680*a^3*b^15*c^15 + 53760*a^4*b^13*c^16 - 258048*a^5*b^11*c^17 +
860160*a^6*b^9*c^18 - 1966080*a^7*b^7*c^19 + 2949120*a^8*b^5*c^20 - 2621440*a^9*b^3*c^21 + 1048576*a^10*b*c^22
)*d^51 + sqrt((b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^
5*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34)*(b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a
^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^5*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34)*((b^14*c^8 -
28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^5*b^4*c^13 + 28672*a^6*b^
2*c^14 - 16384*a^7*c^15)*d^34)^(1/4))/((b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 896
0*a^4*b^6*c^12 - 21504*a^5*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34)) - 1155*((b^14*c^8 - 28*a*b^1
2*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^5*b^4*c^13 + 28672*a^6*b^2*c^14 -
16384*a^7*c^15)*d^34)^(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(-4492125*(b^10*c^6 -
20*a*b^8*c^7 + 160*a^2*b^6*c^8 - 640*a^3*b^4*c^9 + 1280*a^4*b^2*c^10 - 1024*a^5*c^11)*sqrt(2*c*d*x + b*d)*d^2
5 + 4492125*((b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^5
*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34)^(3/4)) + 1155*((b^14*c^8 - 28*a*b^12*c^9 + 336*a^2*b^10
*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^5*b^4*c^13 + 28672*a^6*b^2*c^14 - 16384*a^7*c^15)*d^34
)^(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(-4492125*(b^10*c^6 - 20*a*b^8*c^7 + 160*
a^2*b^6*c^8 - 640*a^3*b^4*c^9 + 1280*a^4*b^2*c^10 - 1024*a^5*c^11)*sqrt(2*c*d*x + b*d)*d^25 - 4492125*((b^14*c
^8 - 28*a*b^12*c^9 + 336*a^2*b^10*c^10 - 2240*a^3*b^8*c^11 + 8960*a^4*b^6*c^12 - 21504*a^5*b^4*c^13 + 28672*a^
6*b^2*c^14 - 16384*a^7*c^15)*d^34)^(3/4)) - (1024*c^7*d^8*x^7 + 3584*b*c^6*d^8*x^6 + 512*(13*b^2*c^5 - 10*a*c^
6)*d^8*x^5 + 2560*(3*b^3*c^4 - 5*a*b*c^5)*d^8*x^4 + 10*(423*b^4*c^3 - 312*a*b^2*c^4 - 1936*a^2*c^5)*d^8*x^3 +
(457*b^5*c^2 + 8120*a*b^3*c^3 - 29040*a^2*b*c^4)*d^8*x^2 - (203*b^6*c - 3350*a*b^4*c^2 + 5280*a^2*b^2*c^3 + 12
320*a^3*c^4)*d^8*x - (7*b^7 + 105*a*b^5*c - 2200*a^2*b^3*c^2 + 6160*a^3*b*c^3)*d^8)*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 = 0.41, size = 753, normalized size = 3.39 \begin {gather*} 64 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{2} c^{2} d^{7} - 256 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a c^{3} d^{7} + \frac {64}{7} \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} c^{2} d^{5} - \frac {165}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c^{2} d^{7} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a 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 {165}{2} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c^{2} d^{7} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a 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 {165}{4} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c^{2} d^{7} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a 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 {165}{4} \, {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} b^{2} c^{2} d^{7} - 4 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} a 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 {2 \, {\left (23 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b^{6} c^{2} d^{11} - 276 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a b^{4} c^{3} d^{11} + 1104 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a^{2} b^{2} c^{4} d^{11} - 1472 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} a^{3} c^{5} d^{11} - 27 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} b^{4} c^{2} d^{9} + 216 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} a b^{2} c^{3} d^{9} - 432 \, {\left (2 \, c d x + b d\right )}^{\frac {7}{2}} a^{2} c^{4} d^{9}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*d*x+b*d)^(17/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

64*(2*c*d*x + b*d)^(3/2)*b^2*c^2*d^7 - 256*(2*c*d*x + b*d)^(3/2)*a*c^3*d^7 + 64/7*(2*c*d*x + b*d)^(7/2)*c^2*d^
5 - 165/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c^2*d^7 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*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)) - 165/2*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c^2*d^7 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*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)) + 165/4*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c^2*d^7 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*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)) - 165/4*(sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*b^2*c^2*d^7 - 4*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*a*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))
+ 2*(23*(2*c*d*x + b*d)^(3/2)*b^6*c^2*d^11 - 276*(2*c*d*x + b*d)^(3/2)*a*b^4*c^3*d^11 + 1104*(2*c*d*x + b*d)^
(3/2)*a^2*b^2*c^4*d^11 - 1472*(2*c*d*x + b*d)^(3/2)*a^3*c^5*d^11 - 27*(2*c*d*x + b*d)^(7/2)*b^4*c^2*d^9 + 216*
(2*c*d*x + b*d)^(7/2)*a*b^2*c^3*d^9 - 432*(2*c*d*x + b*d)^(7/2)*a^2*c^4*d^9)/(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 = 1310, normalized size = 5.90

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*d*x+b*d)^(17/2)/(c*x^2+b*x+a)^3,x)

[Out]

64/7*c^2*d^5*(2*c*d*x+b*d)^(7/2)-256*c^3*d^7*(2*c*d*x+b*d)^(3/2)*a+64*c^2*d^7*(2*c*d*x+b*d)^(3/2)*b^2-864*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)^(7/2)*a^2+432*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)^(7/2)*a*b^2-54*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)^(7/2)*b^4
-2944*c^5*d^11/(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^3+2208*c^4*d^11/(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*b^2-552*c^3*d^11/(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^4+46*c^2*d^11/(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^6+660*c^4*d^9*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)))+1320*c^4*d^9*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)-1320*c^4*d^9*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)-330*c^3*d^9*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)))-660*c^3*d^9*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)+660*c^3*d^9*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)+165/4*c^2*d^9*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)))+165/2*c^2*d^9*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)-165/2*c^2*d^9*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 \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)^(17/2)/(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 = 0.65, size = 364, normalized size = 1.64 \begin {gather*} \frac {64\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{7}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^{7/2}\,\left (864\,a^2\,c^4\,d^9-432\,a\,b^2\,c^3\,d^9+54\,b^4\,c^2\,d^9\right )+{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (2944\,a^3\,c^5\,d^{11}-2208\,a^2\,b^2\,c^4\,d^{11}+552\,a\,b^4\,c^3\,d^{11}-46\,b^6\,c^2\,d^{11}\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}-64\,c^2\,d^7\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (4\,a\,c-b^2\right )+165\,c^2\,d^{17/2}\,\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}+c^2\,d^{17/2}\,\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}\,165{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*d + 2*c*d*x)^(17/2)/(a + b*x + c*x^2)^3,x)

[Out]

(64*c^2*d^5*(b*d + 2*c*d*x)^(7/2))/7 - ((b*d + 2*c*d*x)^(7/2)*(864*a^2*c^4*d^9 + 54*b^4*c^2*d^9 - 432*a*b^2*c^
3*d^9) + (b*d + 2*c*d*x)^(3/2)*(2944*a^3*c^5*d^11 - 46*b^6*c^2*d^11 + 552*a*b^4*c^3*d^11 - 2208*a^2*b^2*c^4*d^
11))/((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^7*(b*d + 2*c*d*x)^(3/2)*(4*a*c - b^2) + 165*c^2*d^(17/2)*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^2*d^(17/2)*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)*165i

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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)**(17/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

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