Integrand size = 26, antiderivative size = 152 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx=20 c d^3 \sqrt {b d+2 c d x}-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )-10 c \sqrt [4]{b^2-4 a c} d^{7/2} \text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ) \] Output:
20*c*d^3*(2*c*d*x+b*d)^(1/2)-d*(2*c*d*x+b*d)^(5/2)/(c*x^2+b*x+a)-10*c*(-4* a*c+b^2)^(1/4)*d^(7/2)*arctan((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/ 2))-10*c*(-4*a*c+b^2)^(1/4)*d^(7/2)*arctanh((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^ 2)^(1/4)/d^(1/2))
Result contains complex when optimal does not.
Time = 0.88 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.68 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx=(1+i) c (d (b+2 c x))^{7/2} \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-5 b^2+20 a c+4 (b+2 c x)^2\right )}{c (b+2 c x)^3 (a+x (b+c x))}-\frac {5 i \sqrt [4]{b^2-4 a c} \arctan \left (1-\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{7/2}}+\frac {5 i \sqrt [4]{b^2-4 a c} \arctan \left (1+\frac {(1+i) \sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )}{(b+2 c x)^{7/2}}+\frac {5 i \sqrt [4]{b^2-4 a c} \text {arctanh}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x}}{\sqrt {b^2-4 a c}+i (b+2 c x)}\right )}{(b+2 c x)^{7/2}}\right ) \] Input:
Integrate[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^2,x]
Output:
(1 + I)*c*(d*(b + 2*c*x))^(7/2)*(((1/2 - I/2)*(-5*b^2 + 20*a*c + 4*(b + 2* c*x)^2))/(c*(b + 2*c*x)^3*(a + x*(b + c*x))) - ((5*I)*(b^2 - 4*a*c)^(1/4)* ArcTan[1 - ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(7/ 2) + ((5*I)*(b^2 - 4*a*c)^(1/4)*ArcTan[1 + ((1 + I)*Sqrt[b + 2*c*x])/(b^2 - 4*a*c)^(1/4)])/(b + 2*c*x)^(7/2) + ((5*I)*(b^2 - 4*a*c)^(1/4)*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*c*x)^(7/2))
Time = 0.34 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1110, 1116, 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.
| \(\displaystyle \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 1110 |
| \(\displaystyle 5 c d^2 \int \frac {(b d+2 c x d)^{3/2}}{c x^2+b x+a}dx-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 1116 |
| \(\displaystyle 5 c d^2 \left (d^2 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \left (c x^2+b x+a\right )}dx+4 d \sqrt {b d+2 c d x}\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle 5 c d^2 \left (\frac {d \left (b^2-4 a c\right ) \int \frac {4 c d^2}{\sqrt {b d+2 c x d} \left (\left (4 a-\frac {b^2}{c}\right ) c d^2+(b d+2 c x d)^2\right )}d(b d+2 c x d)}{2 c}+4 d \sqrt {b d+2 c d x}\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 5 c d^2 \left (2 d^3 \left (b^2-4 a c\right ) \int -\frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)+4 d \sqrt {b d+2 c d x}\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 5 c d^2 \left (4 d \sqrt {b d+2 c d x}-2 d^3 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {b d+2 c x d} \left (\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2\right )}d(b d+2 c x d)\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle 5 c d^2 \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \int \frac {1}{\left (b^2-4 a c\right ) d^2-(b d+2 c x d)^2}d\sqrt {b d+2 c x d}\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle 5 c d^2 \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\int \frac {1}{b d+2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}\right )\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle 5 c d^2 \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\int \frac {1}{-b d-2 c x d+\sqrt {b^2-4 a c} d}d\sqrt {b d+2 c x d}}{2 d \sqrt {b^2-4 a c}}+\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle 5 c d^2 \left (4 d \sqrt {b d+2 c d x}-4 d^3 \left (b^2-4 a c\right ) \left (\frac {\arctan \left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}+\frac {\text {arctanh}\left (\frac {\sqrt {b d+2 c d x}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{2 d^{3/2} \left (b^2-4 a c\right )^{3/4}}\right )\right )-\frac {d (b d+2 c d x)^{5/2}}{a+b x+c x^2}\) |
Input:
Int[(b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^2,x]
Output:
-((d*(b*d + 2*c*d*x)^(5/2))/(a + b*x + c*x^2)) + 5*c*d^2*(4*d*Sqrt[b*d + 2 *c*d*x] - 4*(b^2 - 4*a*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[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))))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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]
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]
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]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))) 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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
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]
Leaf count of result is larger than twice the leaf count of optimal. \(301\) vs. \(2(128)=256\).
Time = 2.24 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.99
| method | result | size |
| pseudoelliptic | \(\frac {40 \left (\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (\frac {4 c^{2} x^{2}}{5}+\left (\frac {4 b x}{5}+a \right ) c -\frac {b^{2}}{20}\right ) \sqrt {d \left (2 c x +b \right )}-\frac {c \,d^{2} \sqrt {2}\, \left (a c -\frac {b^{2}}{4}\right ) \left (c \,x^{2}+b x +a \right ) \left (\ln \left (\frac {\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+\sqrt {d^{2} \left (4 a c -b^{2}\right )}+d \left (2 c x +b \right )}{\sqrt {d^{2} \left (4 a c -b^{2}\right )}-\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}} \sqrt {d \left (2 c x +b \right )}\, \sqrt {2}+d \left (2 c x +b \right )}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}-1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \left (2 c x +b \right )}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {1}{4}}}+1\right )\right )}{2}\right ) d^{3}}{\left (d^{2} \left (4 a c -b^{2}\right )\right )^{\frac {3}{4}} \left (2 c \,x^{2}+2 b x +2 a \right )}\) | \(302\) |
| derivativedivides | \(16 c \,d^{3} \left (\sqrt {2 c d x +b d}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \sqrt {2 c d x +b d}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {5 \left (a c -\frac {b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\) | \(312\) |
| default | \(16 c \,d^{3} \left (\sqrt {2 c d x +b d}-d^{2} \left (\frac {\left (-\frac {a c}{4}+\frac {b^{2}}{16}\right ) \sqrt {2 c d x +b d}}{a \,d^{2} c -\frac {b^{2} d^{2}}{4}+\frac {\left (2 c d x +b d \right )^{2}}{4}}+\frac {5 \left (a c -\frac {b^{2}}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {2 c d x +b d +\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}{2 c d x +b d -\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a \,d^{2} c -b^{2} d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (4 a \,d^{2} c -b^{2} d^{2}\right )^{\frac {3}{4}}}\right )\right )\) | \(312\) |
Input:
int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
40*((d^2*(4*a*c-b^2))^(3/4)*(4/5*c^2*x^2+(4/5*b*x+a)*c-1/20*b^2)*(d*(2*c*x +b))^(1/2)-1/2*c*d^2*2^(1/2)*(a*c-1/4*b^2)*(c*x^2+b*x+a)*(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)))+2*arctan(2^(1/2)/(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b)) ^(1/2)-1)+2*arctan(2^(1/2)/(d^2*(4*a*c-b^2))^(1/4)*(d*(2*c*x+b))^(1/2)+1)) )*d^3/(d^2*(4*a*c-b^2))^(3/4)/(2*c*x^2+2*b*x+2*a)
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.38 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx=-\frac {5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} + 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (-i \, c x^{2} - i \, b x - i \, a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} + 5 i \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (i \, c x^{2} + i \, b x + i \, a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} - 5 i \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}} {\left (c x^{2} + b x + a\right )} \log \left (5 \, \sqrt {2 \, c d x + b d} c d^{3} - 5 \, \left ({\left (b^{2} c^{4} - 4 \, a c^{5}\right )} d^{14}\right )^{\frac {1}{4}}\right ) - {\left (16 \, c^{2} d^{3} x^{2} + 16 \, b c d^{3} x - {\left (b^{2} - 20 \, a c\right )} d^{3}\right )} \sqrt {2 \, c d x + b d}}{c x^{2} + b x + a} \] Input:
integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
-(5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt(2*c*d*x + b*d)*c*d^3 + 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - 5*((b^2*c^4 - 4*a*c^5 )*d^14)^(1/4)*(-I*c*x^2 - I*b*x - I*a)*log(5*sqrt(2*c*d*x + b*d)*c*d^3 + 5 *I*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)* (I*c*x^2 + I*b*x + I*a)*log(5*sqrt(2*c*d*x + b*d)*c*d^3 - 5*I*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)*(c*x^2 + b*x + a)*log(5*sqrt(2*c*d*x + b*d)*c*d^3 - 5*((b^2*c^4 - 4*a*c^5)*d^14)^(1/4)) - (16*c^2*d^3*x^2 + 16*b*c*d^3*x - (b^2 - 20*a*c)*d^3)*sqrt(2*c*d*x + b*d)) /(c*x^2 + b*x + a)
Timed out. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((2*c*d*x+b*d)**(7/2)/(c*x**2+b*x+a)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,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
Leaf count of result is larger than twice the leaf count of optimal. 441 vs. \(2 (128) = 256\).
Time = 0.13 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.90 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx=-5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - 5 \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right ) - \frac {5}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + \frac {5}{2} \, \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c d^{3} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right ) + 16 \, \sqrt {2 \, c d x + b d} c d^{3} + \frac {4 \, {\left (\sqrt {2 \, c d x + b d} b^{2} c d^{5} - 4 \, \sqrt {2 \, c d x + b d} a c^{2} d^{5}\right )}}{b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}} \] Input:
integrate((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
-5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*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)) - 5*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*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)) - 5/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c*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)) + 5/2*sqrt(2)*(-b^2*d^2 + 4*a*c*d^2)^(1/4)*c *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)) + 16*sqrt(2*c*d*x + b*d)*c*d^3 + 4*( sqrt(2*c*d*x + b*d)*b^2*c*d^5 - 4*sqrt(2*c*d*x + b*d)*a*c^2*d^5)/(b^2*d^2 - 4*a*c*d^2 - (2*c*d*x + b*d)^2)
Time = 5.36 (sec) , antiderivative size = 597, normalized size of antiderivative = 3.93 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx=16\,c\,d^3\,\sqrt {b\,d+2\,c\,d\,x}+\frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (16\,a\,c^2\,d^5-4\,b^2\,c\,d^5\right )}{{\left (b\,d+2\,c\,d\,x\right )}^2-b^2\,d^2+4\,a\,c\,d^2}-10\,c\,d^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-c\,d^{7/2}\,\mathrm {atan}\left (\frac {c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )-5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,5{}\mathrm {i}+c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )+5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,5{}\mathrm {i}}{5\,c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )-5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}-5\,c\,d^{7/2}\,\left (\sqrt {b\,d+2\,c\,d\,x}\,\left (25600\,a^2\,c^4\,d^{10}-12800\,a\,b^2\,c^3\,d^{10}+1600\,b^4\,c^2\,d^{10}\right )+5\,c\,d^{7/2}\,{\left (b^2-4\,a\,c\right )}^{1/4}\,\left (5120\,a^2\,c^3\,d^7-2560\,a\,b^2\,c^2\,d^7+320\,b^4\,c\,d^7\right )\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}}\right )\,{\left (b^2-4\,a\,c\right )}^{1/4}\,10{}\mathrm {i} \] Input:
int((b*d + 2*c*d*x)^(7/2)/(a + b*x + c*x^2)^2,x)
Output:
16*c*d^3*(b*d + 2*c*d*x)^(1/2) + ((b*d + 2*c*d*x)^(1/2)*(16*a*c^2*d^5 - 4* b^2*c*d^5))/((b*d + 2*c*d*x)^2 - b^2*d^2 + 4*a*c*d^2) - c*d^(7/2)*atan((c* d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 1 2800*a*b^2*c^3*d^10) - 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 51 20*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^2 - 4*a*c)^(1/4)*5i + c*d^(7/2)*( (b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*a*b^ 2*c^3*d^10) + 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^ 3*d^7 - 2560*a*b^2*c^2*d^7))*(b^2 - 4*a*c)^(1/4)*5i)/(5*c*d^(7/2)*((b*d + 2*c*d*x)^(1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*a*b^2*c^3*d ^10) - 5*c*d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^2*c^2*d^7))*(b^2 - 4*a*c)^(1/4) - 5*c*d^(7/2)*((b*d + 2*c*d*x)^( 1/2)*(25600*a^2*c^4*d^10 + 1600*b^4*c^2*d^10 - 12800*a*b^2*c^3*d^10) + 5*c *d^(7/2)*(b^2 - 4*a*c)^(1/4)*(320*b^4*c*d^7 + 5120*a^2*c^3*d^7 - 2560*a*b^ 2*c^2*d^7))*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(1/4)*10i - 10*c*d^(7/2)*a tan((b*d + 2*c*d*x)^(1/2)/(d^(1/2)*(b^2 - 4*a*c)^(1/4)))*(b^2 - 4*a*c)^(1/ 4)
Time = 0.21 (sec) , antiderivative size = 810, normalized size of antiderivative = 5.33 \[ \int \frac {(b d+2 c d x)^{7/2}}{\left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
int((2*c*d*x+b*d)^(7/2)/(c*x^2+b*x+a)^2,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*c + 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*c*x + 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**2*x**2 - 10*(4*a*c - b**2)**(1/4)*sqrt(2)*a tan(((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 - 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*c* x - 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**2*x**2 + 5*(4*a*c - b**2)**(1/4)*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)*a*c + 5*(4*a*c - b**2)**(1/4)*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 + 5*(4*a*c - b**2)**(1/4)*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*(4*a*c - b**2)**(1/4)*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)*a*c - 5*(4*a*c - b**2)**(1/4 )*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 - 5*(4*a*c - b**2)**(1/4)*sqrt(2)*log(sqrt(b ...