\(\int \frac {(b d+2 c d x)^{17/2}}{(a+b x+c x^2)^3} \, dx\) [1307]

Optimal result

Integrand size = 26, antiderivative size = 222 \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=110 c^2 \left (b^2-4 a c\right ) d^7 (b d+2 c d x)^{3/2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}+165 c^2 \left (b^2-4 a c\right )^{7/4} d^{17/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-165 c^2 \left (b^2-4 a c\right )^{7/4} d^{17/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \]

[Out]

110*c^2*(-4*a*c+b^2)*d^7*(2*c*d*x+b*d)^(3/2)+330/7*c^2*d^5*(2*c*d*x+b*d)^(7/2)-1/2*d*(2*c*d*x+b*d)^(15/2)/(c*x
^2+b*x+a)^2-15/2*c*d^3*(2*c*d*x+b*d)^(11/2)/(c*x^2+b*x+a)+165*c^2*(-4*a*c+b^2)^(7/4)*d^(17/2)*arctan((d*(2*c*x
+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))-165*c^2*(-4*a*c+b^2)^(7/4)*d^(17/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c
+b^2)^(1/4)/d^(1/2))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {700, 706, 708, 335, 304, 209, 212} \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )-165 c^2 d^{17/2} \left (b^2-4 a c\right )^{7/4} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )+110 c^2 d^7 \left (b^2-4 a c\right ) (b d+2 c d x)^{3/2}-\frac {15 c d^3 (b d+2 c d x)^{11/2}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{15/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {330}{7} c^2 d^5 (b d+2 c d x)^{7/2} \]

[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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 304

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 335

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 700

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 706

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 708

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*} \text {integral}& = -\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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.64 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.55 \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {1}{14} (d (b+2 c x))^{17/2} \left (\frac {-7 b^6-189 b^5 c x+40 b^3 c^2 x \left (89 a+64 c x^2\right )+5 b^4 c \left (-21 a+167 c x^2\right )+40 b^2 c^2 \left (55 a^2+25 a c x^2+64 c^2 x^4\right )+16 b c^3 x \left (-605 a^2-320 a c x^2+96 c^2 x^4\right )-16 c^3 \left (385 a^3+605 a^2 c x^2+160 a c^2 x^4-32 c^3 x^6\right )}{(b+2 c x)^7 (a+x (b+c x))^2}+\frac {(1155+1155 i) c^2 \left (b^2-4 a c\right )^{7/4} \arctan \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \left (b+i \sqrt {b^2-4 a c}+2 c x\right )}{\sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}\right )}{(b+2 c x)^{17/2}}-\frac {(1155+1155 i) c^2 \left (b^2-4 a c\right )^{7/4} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{i b+\sqrt {b^2-4 a c}+2 i c x}\right )}{(b+2 c x)^{17/2}}\right ) \]

[In]

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

[Out]

((d*(b + 2*c*x))^(17/2)*((-7*b^6 - 189*b^5*c*x + 40*b^3*c^2*x*(89*a + 64*c*x^2) + 5*b^4*c*(-21*a + 167*c*x^2)
+ 40*b^2*c^2*(55*a^2 + 25*a*c*x^2 + 64*c^2*x^4) + 16*b*c^3*x*(-605*a^2 - 320*a*c*x^2 + 96*c^2*x^4) - 16*c^3*(3
85*a^3 + 605*a^2*c*x^2 + 160*a*c^2*x^4 - 32*c^3*x^6))/((b + 2*c*x)^7*(a + x*(b + c*x))^2) + ((1155 + 1155*I)*c
^2*(b^2 - 4*a*c)^(7/4)*ArcTan[((1/2 + I/2)*(b + I*Sqrt[b^2 - 4*a*c] + 2*c*x))/((b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*
c*x])])/(b + 2*c*x)^(17/2) - ((1155 + 1155*I)*c^2*(b^2 - 4*a*c)^(7/4)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqr
t[b + 2*c*x])/(I*b + Sqrt[b^2 - 4*a*c] + (2*I)*c*x)])/(b + 2*c*x)^(17/2)))/14

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(431\) vs. \(2(188)=376\).

Time = 2.79 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.95

method	result	size
derivativedivides	\(64 c^{2} d^{5} \left (-4 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}+b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+d^{4} \left (\frac {16 \left (-\frac {27}{32} a^{2} c^{2}+\frac {27}{64} a \,b^{2} c -\frac {27}{512} b^{4}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}+16 \left (-\frac {23}{8} a^{3} c^{3} d^{2}+\frac {69}{32} a^{2} b^{2} c^{2} d^{2}-\frac {69}{128} a \,b^{4} c \,d^{2}+\frac {23}{512} b^{6} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {\left (\frac {165}{2} a^{2} c^{2}-\frac {165}{4} a \,b^{2} c +\frac {165}{32} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(432\)
default	\(64 c^{2} d^{5} \left (-4 a c \,d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}+b^{2} d^{2} \left (2 c d x +b d \right )^{\frac {3}{2}}+\frac {\left (2 c d x +b d \right )^{\frac {7}{2}}}{7}+d^{4} \left (\frac {16 \left (-\frac {27}{32} a^{2} c^{2}+\frac {27}{64} a \,b^{2} c -\frac {27}{512} b^{4}\right ) \left (2 c d x +b d \right )^{\frac {7}{2}}+16 \left (-\frac {23}{8} a^{3} c^{3} d^{2}+\frac {69}{32} a^{2} b^{2} c^{2} d^{2}-\frac {69}{128} a \,b^{4} c \,d^{2}+\frac {23}{512} b^{6} d^{2}\right ) \left (2 c d x +b d \right )^{\frac {3}{2}}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {\left (\frac {165}{2} a^{2} c^{2}-\frac {165}{4} a \,b^{2} c +\frac {165}{32} b^{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\right )\)	\(432\)
pseudoelliptic	\(\frac {660 \left (-\frac {2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} d^{2} \left (\frac {32 c^{4} x^{4}}{55}+\frac {64 x^{2} \left (b x +a \right ) c^{3}}{55}+\left (\frac {32}{55} b^{2} x^{2}+\frac {64}{55} a b x +a^{2}\right ) c^{2}-\frac {23 a \,b^{2} c}{110}+\frac {23 b^{4}}{880}\right ) \left (-\frac {b^{2}}{4}+a c \right ) \left (d \left (2 c x +b \right )\right )^{\frac {3}{2}}}{3}-\frac {157 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (-\frac {32 c^{4} x^{4}}{157}-\frac {64 x^{2} \left (b x +a \right ) c^{3}}{157}+\left (-\frac {32}{157} b^{2} x^{2}-\frac {64}{157} a b x +a^{2}\right ) c^{2}-\frac {189 a \,b^{2} c}{314}+\frac {189 b^{4}}{2512}\right ) \left (d \left (2 c x +b \right )\right )^{\frac {7}{2}}}{2310}+\sqrt {2}\, c^{2} d^{4} \left (c \,x^{2}+b x +a \right )^{2} \left (-\frac {b^{2}}{4}+a c \right )^{2} \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right )\right ) d^{5}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (c \,x^{2}+b x +a \right )^{2}}\)	\(465\)

[In]

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

[Out]

64*c^2*d^5*(-4*a*c*d^2*(2*c*d*x+b*d)^(3/2)+b^2*d^2*(2*c*d*x+b*d)^(3/2)+1/7*(2*c*d*x+b*d)^(7/2)+d^4*(16*((-27/3
2*a^2*c^2+27/64*a*b^2*c-27/512*b^4)*(2*c*d*x+b*d)^(7/2)+(-23/8*a^3*c^3*d^2+69/32*a^2*b^2*c^2*d^2-69/128*a*b^4*
c*d^2+23/512*b^6*d^2)*(2*c*d*x+b*d)^(3/2))/((2*c*d*x+b*d)^2+4*a*c*d^2-b^2*d^2)^2+1/8*(165/2*a^2*c^2-165/4*a*b^
2*c+165/32*b^4)/(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)))+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)-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))))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 1426, normalized size of antiderivative = 6.42 \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/14*(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)) - 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)*(I*c^2*x^4 + 2*I*b*c*x^3 + 2*I*a*b*x + I*(b^2 + 2*a*c)*x^2 + I*a^2)*lo
g(-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)*s
qrt(2*c*d*x + b*d)*d^25 + 4492125*I*((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^1
4 - 16384*a^7*c^15)*d^34)^(1/4)*(-I*c^2*x^4 - 2*I*b*c*x^3 - 2*I*a*b*x - I*(b^2 + 2*a*c)*x^2 - I*a^2)*log(-4492
125*(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*I*((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 - 163
84*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 + 12320*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)

Sympy [F(-1)]

Timed out. \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[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 deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 753 vs. \(2 (188) = 376\).

Time = 0.34 (sec) , antiderivative size = 753, normalized size of antiderivative = 3.39 \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=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}} \]

[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

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.64 \[ \int \frac {(b d+2 c d x)^{17/2}}{\left (a+b x+c x^2\right )^3} \, dx=\frac {64\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{7/2}}{7}-\frac {{\left (b\,d+2\,c\,d\,x\right )}^{7/2}\,\left (864\,a^2\,c^4\,d^9-432\,a\,b^2\,c^3\,d^9+54\,b^4\,c^2\,d^9\right )+{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (2944\,a^3\,c^5\,d^{11}-2208\,a^2\,b^2\,c^4\,d^{11}+552\,a\,b^4\,c^3\,d^{11}-46\,b^6\,c^2\,d^{11}\right )}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}-64\,c^2\,d^7\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}\,\left (4\,a\,c-b^2\right )+165\,c^2\,d^{17/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}+c^2\,d^{17/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{7/4}\,1{}\mathrm {i}}{\sqrt {d}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,{\left (b^2-4\,a\,c\right )}^{7/4}\,165{}\mathrm {i} \]

[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