\(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^3} \, dx\) [1312]

Optimal result

Integrand size = 26, antiderivative size = 170 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {5 c^2 d^{7/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c^2 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}} \]

[Out]

-1/2*d*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)^2-5*c^2*d^(7/2)*arctan((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2)
)/(-4*a*c+b^2)^(3/4)-5*c^2*d^(7/2)*arctanh((d*(2*c*x+b))^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2))/(-4*a*c+b^2)^(3/4)-
5/2*c*d^3*(2*c*d*x+b*d)^(1/2)/(c*x^2+b*x+a)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {700, 708, 335, 218, 212, 209} \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {5 c^2 d^{7/2} \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c^2 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2} \]

[In]

Int[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^3,x]

[Out]

-1/2*(d*(b*d + 2*c*d*x)^(5/2))/(a + b*x + c*x^2)^2 - (5*c*d^3*Sqrt[b*d + 2*c*d*x])/(2*(a + b*x + c*x^2)) - (5*
c^2*d^(7/2)*ArcTan[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(3/4) - (5*c^2*d^(7/2)*Ar
cTanh[Sqrt[d*(b + 2*c*x)]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])])/(b^2 - 4*a*c)^(3/4)

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 218

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] &&  !Gt
Q[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 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)^{5/2}}{2 \left (a+b x+c x^2\right )^2}+\frac {1}{2} \left (5 c d^2\right ) \int \frac {(b d+2 c d x)^{3/2}}{\left (a+b x+c x^2\right )^2} \, dx \\ & = -\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (5 c^2 d^4\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx \\ & = -\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}+\frac {1}{4} \left (5 c d^3\right ) \text {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 ) \\ & = -\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}+\frac {1}{2} \left (5 c d^3\right ) \text {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 ) \\ & = -\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {\left (5 c^2 d^4\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 )}{\sqrt {b^2-4 a c}}-\frac {\left (5 c^2 d^4\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 )}{\sqrt {b^2-4 a c}} \\ & = -\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {5 c^2 d^{7/2} \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c^2 d^{7/2} \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.11 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.55 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=\left (\frac {1}{2}+\frac {i}{2}\right ) c^2 (d (b+2 c x))^{7/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (b^2+9 b c x+c \left (5 a+9 c x^2\right )\right )}{c^2 (b+2 c x)^3 (a+x (b+c x))^2}-\frac {5 i \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4} (b+2 c x)^{7/2}}+\frac {5 i \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4} (b+2 c x)^{7/2}}+\frac {5 i \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{3/4} (b+2 c x)^{7/2}}\right ) \]

[In]

Integrate[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^3,x]

[Out]

(1/2 + I/2)*c^2*(d*(b + 2*c*x))^(7/2)*(((-1/2 + I/2)*(b^2 + 9*b*c*x + c*(5*a + 9*c*x^2)))/(c^2*(b + 2*c*x)^3*(
a + x*(b + c*x))^2) - ((5*I)*ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)^(3/4)*(
b + 2*c*x)^(7/2)) + ((5*I)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/((b^2 - 4*a*c)^(3/4)*(b
+ 2*c*x)^(7/2)) + ((5*I)*ArcTanh[((1 + I)*(b^2 - 4*a*c)^(1/4)*Sqrt[b + 2*c*x])/(Sqrt[b^2 - 4*a*c] + I*(b + 2*c
*x))])/((b^2 - 4*a*c)^(3/4)*(b + 2*c*x)^(7/2)))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(308\) vs. \(2(142)=284\).

Time = 2.77 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.82

method	result	size
derivativedivides	\(64 c^{2} d^{5} \left (\frac {-\frac {9 \left (2 c d x +b d \right )^{\frac {5}{2}}}{32}+16 \left (-\frac {5}{128} a c \,d^{2}+\frac {5}{512} b^{2} d^{2}\right ) \sqrt {2 c d x +b d}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(309\)
default	\(64 c^{2} d^{5} \left (\frac {-\frac {9 \left (2 c d x +b d \right )^{\frac {5}{2}}}{32}+16 \left (-\frac {5}{128} a c \,d^{2}+\frac {5}{512} b^{2} d^{2}\right ) \sqrt {2 c d x +b d}}{\left (\left (2 c d x +b d \right )^{2}+4 a c \,d^{2}-b^{2} d^{2}\right )^{2}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(309\)
pseudoelliptic	\(-\frac {5 d^{3} \left (-\frac {\sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2} \ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )}{2}+\arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2}-\arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right ) \sqrt {2}\, c^{2} d^{2} \left (c \,x^{2}+b x +a \right )^{2}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (\frac {9 c^{2} x^{2}}{5}+\left (\frac {9 b x}{5}+a \right ) c +\frac {b^{2}}{5}\right ) \sqrt {d \left (2 c x +b \right )}\right )}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (c \,x^{2}+b x +a \right )^{2}}\)	\(366\)

[In]

int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

64*c^2*d^5*(16*(-9/512*(2*c*d*x+b*d)^(5/2)+(-5/128*a*c*d^2+5/512*b^2*d^2)*(2*c*d*x+b*d)^(1/2))/((2*c*d*x+b*d)^
2+4*a*c*d^2-b^2*d^2)^2+5/256/(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)))+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 = 0.38 (sec) , antiderivative size = 696, normalized size of antiderivative = 4.09 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c^{2} d^{3} + 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) - 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c^{2} d^{3} - 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (b^{2} - 4 \, a c\right )}\right ) + 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (-i \, c^{2} x^{4} - 2 i \, b c x^{3} - 2 i \, a b x - i \, {\left (b^{2} + 2 \, a c\right )} x^{2} - i \, a^{2}\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c^{2} d^{3} - 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (i \, b^{2} - 4 i \, a c\right )}\right ) + 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (i \, c^{2} x^{4} + 2 i \, b c x^{3} + 2 i \, a b x + i \, {\left (b^{2} + 2 \, a c\right )} x^{2} + i \, a^{2}\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c^{2} d^{3} - 5 \, \left (\frac {c^{8} d^{14}}{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}\right )^{\frac {1}{4}} {\left (-i \, b^{2} + 4 i \, a c\right )}\right ) + {\left (9 \, c^{2} d^{3} x^{2} + 9 \, b c d^{3} x + {\left (b^{2} + 5 \, a c\right )} d^{3}\right )} \sqrt {2 \, c d x + b d}}{2 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

[In]

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(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(5*sqrt(2*c*d*x + b*d)*c^2*d^3 + 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*
a^3*c^3))^(1/4)*(b^2 - 4*a*c)) - 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(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(5*sqrt(2*c*d*x + b*d)*c^2*d^3 - 5*(c^8*d^14/(b^6 - 12*a*b
^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(b^2 - 4*a*c)) + 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 6
4*a^3*c^3))^(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(5*sqrt(2*c*d*x + b*
d)*c^2*d^3 - 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(1/4)*(I*b^2 - 4*I*a*c)) + 5*(c^8*d
^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3))^(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(5*sqrt(2*c*d*x + b*d)*c^2*d^3 - 5*(c^8*d^14/(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*
c^3))^(1/4)*(-I*b^2 + 4*I*a*c)) + (9*c^2*d^3*x^2 + 9*b*c*d^3*x + (b^2 + 5*a*c)*d^3)*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. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \]

[In]

integrate((2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**3,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((2*c*d*x+b*d)^(7/2)/(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. 510 vs. \(2 (142) = 284\).

Time = 0.31 (sec) , antiderivative size = 510, normalized size of antiderivative = 3.00 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{2} - 4 \, \sqrt {2} a c\right )}} + \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{2} - 4 \, \sqrt {2} a c\right )}} + \frac {2 \, {\left (5 \, \sqrt {2 \, c d x + b d} b^{2} c^{2} d^{7} - 20 \, \sqrt {2 \, c d x + b d} a c^{3} d^{7} - 9 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d^{5}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \]

[In]

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^3,x, algorithm="giac")

[Out]

-5*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^3*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 5*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d
^3*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) - 5/2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^3*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) + 5/
2*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c^2*d^3*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))/(sqrt(2)*b^2 - 4*sqrt(2)*a*c) + 2*(5*sqrt(2*c*d*x + b*d)*b^2*c^2*d^7 - 20*
sqrt(2*c*d*x + b*d)*a*c^3*d^7 - 9*(2*c*d*x + b*d)^(5/2)*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)^2

Mupad [B] (verification not implemented)

Time = 9.41 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.82 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (40\,a\,c^3\,d^7-10\,b^2\,c^2\,d^7\right )+18\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}-\frac {5\,c^2\,d^{7/2}\,\mathrm {atan}\left (\frac {2000\,c^6\,d^{27/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {2000\,b^2\,c^6\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {8000\,a\,c^7\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}}-\frac {5\,c^2\,d^{7/2}\,\mathrm {atanh}\left (\frac {2000\,c^6\,d^{27/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {2000\,b^2\,c^6\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {8000\,a\,c^7\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}} \]

[In]

int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^3,x)

[Out]

- ((b*d + 2*c*d*x)^(1/2)*(40*a*c^3*d^7 - 10*b^2*c^2*d^7) + 18*c^2*d^5*(b*d + 2*c*d*x)^(5/2))/((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) - (5*c^2*d^(7/2)*ata
n((2000*c^6*d^(27/2)*(b*d + 2*c*d*x)^(1/2))/(((2000*b^2*c^6*d^14)/(b^2 - 4*a*c)^(3/2) - (8000*a*c^7*d^14)/(b^2
- 4*a*c)^(3/2))*(b^2 - 4*a*c)^(3/4))))/(b^2 - 4*a*c)^(3/4) - (5*c^2*d^(7/2)*atanh((2000*c^6*d^(27/2)*(b*d + 2
*c*d*x)^(1/2))/(((2000*b^2*c^6*d^14)/(b^2 - 4*a*c)^(3/2) - (8000*a*c^7*d^14)/(b^2 - 4*a*c)^(3/2))*(b^2 - 4*a*c
)^(3/4))))/(b^2 - 4*a*c)^(3/4)