\(\int \frac {1}{(b d+2 c d x)^{7/2} (a+b x+c x^2)} \, dx\) [118]

Optimal result

Integrand size = 26, antiderivative size = 161 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\frac {4}{5 \left (b^2-4 a c\right ) d (b d+2 c d x)^{5/2}}+\frac {4}{\left (b^2-4 a c\right )^2 d^3 \sqrt {b d+2 c d x}}+\frac {2 \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4} d^{7/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{9/4} d^{7/2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.36 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\frac {\left (\frac {1}{5}+\frac {i}{5}\right ) \left ((2-2 i) \sqrt [4]{b^2-4 a c} (b+2 c x) \left (b^2-4 a c+5 (b+2 c x)^2\right )-5 (b+2 c x)^{7/2} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+5 (b+2 c x)^{7/2} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )-5 (b+2 c x)^{7/2} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )\right )}{\left (b^2-4 a c\right )^{9/4} (d (b+2 c x))^{7/2}} \] Input:

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

Output:

((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))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1117, 1117, 1118, 27, 25, 266, 827, 216, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right ) (b d+2 c d x)^{7/2}} \, dx\)

\(\Big \downarrow \) 1117

\(\displaystyle \frac {\int \frac {1}{(b d+2 c x d)^{3/2} \left (c x^2+b x+a\right )}dx}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 1117

\(\displaystyle \frac {\frac {\int \frac {\sqrt {b d+2 c x d}}{c x^2+b x+a}dx}{d^2 \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 1118

\(\displaystyle \frac {\frac {\int \frac {4 c d^2 \sqrt {b d+2 c x d}}{\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2}d(b d+2 c x d)}{2 c d^3 \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \int -\frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}+\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {2 \int \frac {\sqrt {b d+2 c x d}}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d(b d+2 c x d)}{d \left (b^2-4 a c\right )}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \int \frac {b d+2 c x d}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}}{d \left (b^2-4 a c\right )}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {1}{2} \int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}\right )}{d \left (b^2-4 a c\right )}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {1}{2} \int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {4}{d \left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}-\frac {4 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}-\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 \sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{d \left (b^2-4 a c\right )}}{d^2 \left (b^2-4 a c\right )}+\frac {4}{5 d \left (b^2-4 a c\right ) (b d+2 c d x)^{5/2}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 1117

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> 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] + Simp[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] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & 
& (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) 
/2])
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[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] && EqQ[2*c*d - b*e, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(300\) vs. \(2(135)=270\).

Time = 1.51 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.87

method	result	size
derivativedivides	\(4 d \left (-\frac {1}{5 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}+\frac {1}{d^{4} \left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4} \left (4 a c -b^{2}\right )^{2} \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(301\)
default	\(4 d \left (-\frac {1}{5 d^{2} \left (4 a c -b^{2}\right ) \left (2 c d x +b d \right )^{\frac {5}{2}}}+\frac {1}{d^{4} \left (4 a c -b^{2}\right )^{2} \sqrt {2 c d x +b d}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4} \left (4 a c -b^{2}\right )^{2} \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}\right )\)	\(301\)
pseudoelliptic	\(\frac {\frac {\sqrt {2}\, \left (2 c x +b \right )^{2} \left (\ln \left (\frac {\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}+\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}\right )\right ) \sqrt {d \left (2 c x +b \right )}}{2}-\frac {16 \left (-5 c^{2} x^{2}+\left (-5 b x +a \right ) c -\frac {3 b^{2}}{2}\right ) \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}{5}}{16 \sqrt {d \left (2 c x +b \right )}\, \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \left (2 c x +b \right )^{2} \left (a c -\frac {b^{2}}{4}\right )^{2} d^{3}}\)	\(335\)

Input:

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

Output:

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 = 0.12 (sec) , antiderivative size = 1880, normalized size of antiderivative = 11.68 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-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 - 589824*a^7*b^4*c^7 + 589824*a^8*b^2*c^8 - 2 
62144*a^9*c^9)*d^14))^(1/4)*log((b^14 - 28*a*b^12*c + 336*a^2*b^10*c^2 - 2 
240*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 + 3225 
6*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^...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 605 vs. \(2 (135) = 270\).

Time = 0.16 (sec) , antiderivative size = 605, normalized size of antiderivative = 3.76 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=-\frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} - \frac {\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{b^{6} d^{5} - 12 \, a b^{4} c d^{5} + 48 \, a^{2} b^{2} c^{2} d^{5} - 64 \, a^{3} c^{3} d^{5}} + \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{5} - 12 \, \sqrt {2} a b^{4} c d^{5} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt {2} a^{3} c^{3} d^{5}} - \frac {{\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {3}{4}} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{\sqrt {2} b^{6} d^{5} - 12 \, \sqrt {2} a b^{4} c d^{5} + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d^{5} - 64 \, \sqrt {2} a^{3} c^{3} d^{5}} + \frac {4 \, {\left (b^{2} d^{2} - 4 \, a c d^{2} + 5 \, {\left (2 \, c d x + b d\right )}^{2}\right )}}{5 \, {\left (b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}\right )} {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}} \] Input:

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

Output:

-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) - 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) + (-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 = 5.35 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx=\frac {\frac {4}{5\,\left (b^2\,d-4\,a\,c\,d\right )}+\frac {4\,{\left (b\,d+2\,c\,d\,x\right )}^2}{d\,{\left (b^2\,d-4\,a\,c\,d\right )}^2}}{{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}+\frac {2\,\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}+16\,a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}-8\,a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}}+\frac {\mathrm {atan}\left (\frac {b^4\,\sqrt {b\,d+2\,c\,d\,x}\,1{}\mathrm {i}+a^2\,c^2\,\sqrt {b\,d+2\,c\,d\,x}\,16{}\mathrm {i}-a\,b^2\,c\,\sqrt {b\,d+2\,c\,d\,x}\,8{}\mathrm {i}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{9/4}}\right )\,2{}\mathrm {i}}{d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{9/4}} \] Input:

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

Output:

(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))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1021, normalized size of antiderivative = 6.34 \[ \int \frac {1}{(b d+2 c d x)^{7/2} \left (a+b x+c x^2\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(d)*( - 10*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c 
 - b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2 
)))*b**2 - 40*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - 
 b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)) 
)*b*c*x - 40*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - 
b**2)**(1/4)*sqrt(2) - 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2))) 
*c**2*x**2 + 10*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c 
 - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2 
)))*b**2 + 40*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - 
 b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)) 
)*b*c*x + 40*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*atan(((4*a*c - 
b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2))) 
*c**2*x**2 + 5*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log( - sqrt(b 
 + 2*c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)* 
b**2 + 20*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log( - sqrt(b + 2* 
c*x)*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*b*c*x 
 + 20*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log( - sqrt(b + 2*c*x) 
*(4*a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*c**2*x**2 
 - 5*sqrt(b + 2*c*x)*(4*a*c - b**2)**(3/4)*sqrt(2)*log(sqrt(b + 2*c*x)*(4* 
a*c - b**2)**(1/4)*sqrt(2) + sqrt(4*a*c - b**2) + b + 2*c*x)*b**2 - 20*...