\(\int \frac {(b d+2 c d x)^{7/2}}{(a+b x+c x^2)^3} \, dx\) [136]

Optimal result

Integrand size = 26, antiderivative size = 172 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}-\frac {5 c d^3 \sqrt {b d+2 c d x}}{2 \left (a+b x+c x^2\right )}-\frac {5 c^2 d^{7/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}}-\frac {5 c^2 d^{7/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{3/4}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.41 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.53 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=\left (\frac {1}{2}+\frac {i}{2}\right ) c^2 (d (b+2 c x))^{7/2} \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (b^2+9 b c x+c \left (5 a+9 c x^2\right )\right )}{c^2 (b+2 c x)^3 (a+x (b+c x))^2}-\frac {5 i \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4} (b+2 c x)^{7/2}}+\frac {5 i \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/4} (b+2 c x)^{7/2}}+\frac {5 i \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{3/4} (b+2 c x)^{7/2}}\right ) \] Input:

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

Output:

(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)))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1110, 1110, 1118, 27, 25, 266, 756, 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.

\(\displaystyle \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx\)

\(\Big \downarrow \) 1110

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

\(\Big \downarrow \) 1110

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 756

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {5}{2} c d^2 \left (-4 c d^3 \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )-\frac {d \sqrt {b d+2 c d x}}{a+b x+c x^2}\right )-\frac {d (b d+2 c d x)^{5/2}}{2 \left (a+b x+c x^2\right )^2}\)

Input:

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

Output:

-1/2*(d*(b*d + 2*c*d*x)^(5/2))/(a + b*x + c*x^2)^2 + (5*c*d^2*(-((d*Sqrt[b 
*d + 2*c*d*x])/(a + b*x + c*x^2)) - 4*c*d^3*(ArcTan[Sqrt[b*d + 2*c*d*x]/(( 
b^2 - 4*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 4*a*c)^(3/4)*d^(3/2)) + ArcTanh[Sqr 
t[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])]/(2*(b^2 - 4*a*c)^(3/4)*d^( 
3/2)))))/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 756

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

rule 1110

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), 
x] - Simp[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] && EqQ[2*c*d - b*e, 0] && N 
eQ[m + 2*p + 3, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

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. \(304\) vs. \(2(148)=296\).

Time = 1.48 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.77

method	result	size
pseudoelliptic	\(\frac {c^{2} d^{5} \left (-\frac {\sqrt {d \left (2 c x +b \right )}\, \left (9 c^{2} x^{2}+9 c b x +5 a c +b^{2}\right )}{d^{2} c^{2} \left (c \,x^{2}+b x +a \right )^{2}}+\frac {5 \ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right ) \sqrt {2}}{2 \left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}+\frac {5 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right ) \sqrt {2}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}+\frac {5 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right ) \sqrt {2}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}}}\right )}{2}\)	\(305\)
derivativedivides	\(64 c^{2} d^{5} \left (\frac {-\frac {9 \left (2 c d x +b d \right )^{\frac {5}{2}}}{32}+16 \left (-\frac {5}{128} a \,d^{2} c +\frac {5}{512} b^{2} d^{2}\right ) \sqrt {2 c d x +b d}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(309\)
default	\(64 c^{2} d^{5} \left (\frac {-\frac {9 \left (2 c d x +b d \right )^{\frac {5}{2}}}{32}+16 \left (-\frac {5}{128} a \,d^{2} c +\frac {5}{512} b^{2} d^{2}\right ) \sqrt {2 c d x +b d}}{\left (\left (2 c d x +b d \right )^{2}+4 a \,d^{2} c -b^{2} d^{2}\right )^{2}}+\frac {5 \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{256 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\)	\(309\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 696, normalized size of antiderivative = 4.05 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-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 - 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)) + 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. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^3,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. 510 vs. \(2 (144) = 288\).

Time = 0.15 (sec) , antiderivative size = 510, normalized size of antiderivative = 2.97 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{2} - 4 \, \sqrt {2} a c} - \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{2} - 4 \, \sqrt {2} a c\right )}} + \frac {5 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{2} - 4 \, \sqrt {2} a c\right )}} + \frac {2 \, {\left (5 \, \sqrt {2 \, c d x + b d} b^{2} c^{2} d^{7} - 20 \, \sqrt {2 \, c d x + b d} a c^{3} d^{7} - 9 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d^{5}\right )}}{{\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \] Input:

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

Output:

-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))/(s 
qrt(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 = 0.08 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.80 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx=-\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (40\,a\,c^3\,d^7-10\,b^2\,c^2\,d^7\right )+18\,c^2\,d^5\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}-\frac {5\,c^2\,d^{7/2}\,\mathrm {atan}\left (\frac {2000\,c^6\,d^{27/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {2000\,b^2\,c^6\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {8000\,a\,c^7\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}}-\frac {5\,c^2\,d^{7/2}\,\mathrm {atanh}\left (\frac {2000\,c^6\,d^{27/2}\,\sqrt {b\,d+2\,c\,d\,x}}{\left (\frac {2000\,b^2\,c^6\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}-\frac {8000\,a\,c^7\,d^{14}}{{\left (b^2-4\,a\,c\right )}^{3/2}}\right )\,{\left (b^2-4\,a\,c\right )}^{3/4}}\right )}{{\left (b^2-4\,a\,c\right )}^{3/4}} \] Input:

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

Output:

- ((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)*atan( 
(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)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1761, normalized size of antiderivative = 10.24 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(d)*d**3*( - 10*(4*a*c - b**2)**(1/4)*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)))*a**2*c* 
*2 - 20*(4*a*c - b**2)**(1/4)*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)))*a*b*c**2*x - 20*(4*a 
*c - b**2)**(1/4)*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)))*a*c**3*x**2 - 10*(4*a*c - b**2)* 
*(1/4)*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*c**2*x**2 - 20*(4*a*c - b**2)**(1/4)*s 
qrt(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**3*x**3 - 10*(4*a*c - b**2)**(1/4)*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**4*x**4 + 10*(4*a*c - b**2)**(1/4)*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)))*a 
**2*c**2 + 20*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sq 
rt(2) + 2*sqrt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*b*c**2*x + 2 
0*(4*a*c - b**2)**(1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sq 
rt(b + 2*c*x))/((4*a*c - b**2)**(1/4)*sqrt(2)))*a*c**3*x**2 + 10*(4*a*c - 
b**2)**(1/4)*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*c**2*x**2 + 20*(4*a*c - b**2)**( 
1/4)*sqrt(2)*atan(((4*a*c - b**2)**(1/4)*sqrt(2) + 2*sqrt(b + 2*c*x))/(...