\(\int \frac {1}{(b d+2 c d x)^{7/2} (a+b x+c x^2)} \, dx\) [1294]

Optimal result

Integrand size = 26, antiderivative size = 159 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {2 \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4} d^{7/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4} d^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 159, 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 = {707, 708, 335, 304, 209, 212} \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d^{7/2} \left (b^2-4 a c\right )^{9/4}}+\frac {4}{d^3 \left (b^2-4 a c\right )^2 \sqrt {b d+2 c d x}}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}} \]

[In]

Int[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)),x]

[Out]

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

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 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[-2*b*d*(d + e*x)^(m
+ 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Dist[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 -
4*a*c))), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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 {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {\int \frac {1}{(b d+2 c d x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) d^2} \\ & = \frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {\int \frac {\sqrt {b d+2 c d x}}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2 d^4} \\ & = \frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {\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 )}{2 c \left (b^2-4 a c\right )^2 d^5} \\ & = \frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {\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 )}{c \left (b^2-4 a c\right )^2 d^5} \\ & = \frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}-\frac {2 \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 (b^2-4 a c\right )^2 d^3}+\frac {2 \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 (b^2-4 a c\right )^2 d^3} \\ & = \frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {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 )^{9/4} d^{7/2}}-\frac {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 )^{9/4} d^{7/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.33 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\frac {\left (\frac {1}{5}+\frac {i}{5}\right ) \left ((2-2 i) \sqrt [4]{b^2-4 a c} (b+2 c x) \left (b^2-4 a c+5 (b+2 c x)^2\right )-5 (b+2 c x)^{7/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+5 (b+2 c x)^{7/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-5 (b+2 c x)^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{9/4} (d (b+2 c x))^{7/2}} \]

[In]

Integrate[1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(300\) vs. \(2(133)=266\).

Time = 2.53 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.89

method	result	size
derivativedivides	\(4 d \left (-\frac {1}{5 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}+\frac {1}{d^{4} \left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4} \left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(301\)
default	\(4 d \left (-\frac {1}{5 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}+\frac {1}{d^{4} \left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4} \left (4 a c -b^{2}\right )^{2} \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(301\)
pseudoelliptic	\(-\frac {-\frac {5 \sqrt {2}\, \left (2 c x +b \right )^{2} \left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )-2 \arctan \left (\frac {-\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right ) \sqrt {d \left (2 c x +b \right )}}{32}+\left (-5 c^{2} x^{2}+\left (-5 b x +a \right ) c -\frac {3 b^{2}}{2}\right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{5 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, d^{3} \left (2 c x +b \right )^{2} \left (-\frac {b^{2}}{4}+a c \right )^{2}}\)	\(333\)

[In]

int(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)

[Out]

4*d*(-1/5/d^2/(4*a*c-b^2)/(2*c*d*x+b*d)^(5/2)+1/d^4/(4*a*c-b^2)^2/(2*c*d*x+b*d)^(1/2)+1/8/d^4/(4*a*c-b^2)^2/(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 = 0.27 (sec) , antiderivative size = 1880, normalized size of antiderivative = 11.82 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \]

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

-1/5*(5*(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 +
6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^16
*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589
824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*log((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2
- 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^11*(1/((b^18
- 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b
^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(3/4) + sqrt(2*c*d*x + b*d)) + 5*(-8
*I*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 - 12*I*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 - 6*I*(b
^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x - I*(b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^16*c
+ 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 58982
4*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*log(I*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2
- 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^11*(1/((b^18
- 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b
^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(3/4) + sqrt(2*c*d*x + b*d)) + 5*(8*
I*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*I*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*I*(b^
6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + I*(b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^16*c
+ 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824
*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*log(-I*(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2
- 2240*a^3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^11*(1/((b^18
- 36*a*b^16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b
^6*c^6 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(3/4) + sqrt(2*c*d*x + b*d)) - 5*(8*
(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c -
8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16*a^2*b^3*c^2)*d^4)*(1/((b^18 - 36*a*b^16*c + 576*a^
2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 589824*a^7*b^4
*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(1/4)*log(-(b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2240*a^
3*b^8*c^3 + 8960*a^4*b^6*c^4 - 21504*a^5*b^4*c^5 + 28672*a^6*b^2*c^6 - 16384*a^7*c^7)*d^11*(1/((b^18 - 36*a*b^
16*c + 576*a^2*b^14*c^2 - 5376*a^3*b^12*c^3 + 32256*a^4*b^10*c^4 - 129024*a^5*b^8*c^5 + 344064*a^6*b^6*c^6 - 5
89824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 262144*a^9*c^9)*d^14))^(3/4) + sqrt(2*c*d*x + b*d)) - 8*(10*c^2*x^2 +
10*b*c*x + 3*b^2 - 2*a*c)*sqrt(2*c*d*x + b*d))/(8*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^4*x^3 + 12*(b^5*c^2
- 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^4*x^2 + 6*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*d^4*x + (b^7 - 8*a*b^5*c + 16
*a^2*b^3*c^2)*d^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]

[In]

integrate(1/(2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a),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. 605 vs. \(2 (133) = 266\).

Time = 0.30 (sec) , antiderivative size = 605, normalized size of antiderivative = 3.81 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=-\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{5} - 12 \, \sqrt {2} a b^{4} c d^{5} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt {2} a^{3} c^{3} d^{5}} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{5} - 12 \, \sqrt {2} a b^{4} c d^{5} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt {2} a^{3} c^{3} d^{5}} + \frac {4 \, {\left (b^{2} d^{2} - 4 \, a c d^{2} + 5 \, {\left (2 \, c d x + b d\right )}^{2}\right )}}{5 \, {\left (b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}\right )} {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \]

[In]

integrate(1/(2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

-sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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))/(b^6*d^5 - 12*a*b^4*c*d^5 + 48*a^2*b^2*c^2*d^5 - 64*a^3*c^3*d^5) - sqr
t(2)*(-b^2*d^2 + 4*a*c*d^2)^(3/4)*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))/(b^6*d^5 - 12*a*b^4*c*d^5 + 48*a^2*b^2*c^2*d^5 - 64*a^3*c^3*d^5) + (-b^2*
d^2 + 4*a*c*d^2)^(3/4)*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^6*d^5 - 12*sqrt(2)*a*b^4*c*d^5 + 48*sqrt(2)*a^2*b^2*c^2*d^5 - 64*sqrt(2)*a^3*c^
3*d^5) - (-b^2*d^2 + 4*a*c*d^2)^(3/4)*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^6*d^5 - 12*sqrt(2)*a*b^4*c*d^5 + 48*sqrt(2)*a^2*b^2*c^2*d^5 - 64
*sqrt(2)*a^3*c^3*d^5) + 4/5*(b^2*d^2 - 4*a*c*d^2 + 5*(2*c*d*x + b*d)^2)/((b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2
*d^3)*(2*c*d*x + b*d)^(5/2))

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\frac {\frac {4}{5\,\left (b^2\,d-4\,a\,c\,d\right )}+\frac {4\,{\left (b\,d+2\,c\,d\,x\right )}^2}{d\,{\left (b^2\,d-4\,a\,c\,d\right )}^2}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}+\frac {2\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}}+\frac {\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}+a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,16{}\mathrm {i}-a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}\,8{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,2{}\mathrm {i}}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}} \]

[In]

int(1/((b*d + 2*c*d*x)^(7/2)*(a + b*x + c*x^2)),x)

[Out]

(4/(5*(b^2*d - 4*a*c*d)) + (4*(b*d + 2*c*d*x)^2)/(d*(b^2*d - 4*a*c*d)^2))/(b*d + 2*c*d*x)^(5/2) + (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))))/(d^(7/2)*(b^2 - 4*a*c)^(9/4)) + (atan((b^4*(b*d + 2*c*d*x)^(1/2)*1i + a^2*c^2*(b*d + 2*c*d*x)^(
1/2)*16i - a*b^2*c*(b*d + 2*c*d*x)^(1/2)*8i)/(d^(1/2)*(b^2 - 4*a*c)^(9/4)))*2i)/(d^(7/2)*(b^2 - 4*a*c)^(9/4))