Integrand size = 25, antiderivative size = 126 \[ \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {B e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}} \]
B*e*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)+2*(2*a*c*(A *e+B*d)-b*(A*c*d+B*a*e)-(b^2*B*e-b*c*(A*e+B*d)+2*c*(A*c*d-B*a*e))*x)/c/(-4 *a*c+b^2)/(c*x^2+b*x+a)^(1/2)
Time = 0.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {\frac {2 \sqrt {c} (A c (-2 a e+2 c d x+b (d-e x))+B (a b e+b (-c d+b e) x-2 a c (d+e x)))}{\sqrt {a+x (b+c x)}}-B \left (b^2-4 a c\right ) e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{c^{3/2} \left (-b^2+4 a c\right )} \]
((2*Sqrt[c]*(A*c*(-2*a*e + 2*c*d*x + b*(d - e*x)) + B*(a*b*e + b*(-(c*d) + b*e)*x - 2*a*c*(d + e*x))))/Sqrt[a + x*(b + c*x)] - B*(b^2 - 4*a*c)*e*Arc Tanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(c^(3/2)*(-b^2 + 4*a* c))
Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1224, 1092, 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.
\(\displaystyle \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1224 |
\(\displaystyle \frac {B e \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{c}+\frac {2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {2 B e \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c}+\frac {2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {B e \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}\) |
(2*(2*a*c*(B*d + A*e) - b*(A*c*d + a*B*e) - (b^2*B*e - b*c*(B*d + A*e) + 2 *c*(A*c*d - a*B*e))*x))/(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (B*e*Arc Tanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/c^(3/2)
3.25.75.3.1 Defintions of rubi rules used
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[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - ( b^2*e*g - b*c*(e*f + d*g) + 2*c*(c*d*f - a*e*g))*x))*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c *(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(c*(p + 1)*(b^2 - 4*a*c)) Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, - 1] && !(IntegerQ[p] && NeQ[a, 0] && NiceSqrtQ[b^2 - 4*a*c])
Time = 0.51 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.66
method | result | size |
default | \(\frac {2 d A \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+B e \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )+\left (A e +B d \right ) \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )\) | \(209\) |
2*d*A*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+B*e*(-x/c/(c*x^2+b*x+a)^(1 /2)-1/2*b/c*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x +a)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))+(A*e+B*d )*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (116) = 232\).
Time = 1.00 (sec) , antiderivative size = 489, normalized size of antiderivative = 3.88 \[ \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (B b^{2} c - 4 \, B a c^{2}\right )} e x^{2} + {\left (B b^{3} - 4 \, B a b c\right )} e x + {\left (B a b^{2} - 4 \, B a^{2} c\right )} e\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left ({\left (2 \, B a - A b\right )} c^{2} d - {\left (B a b c - 2 \, A a c^{2}\right )} e + {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )}}, -\frac {{\left ({\left (B b^{2} c - 4 \, B a c^{2}\right )} e x^{2} + {\left (B b^{3} - 4 \, B a b c\right )} e x + {\left (B a b^{2} - 4 \, B a^{2} c\right )} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left ({\left (2 \, B a - A b\right )} c^{2} d - {\left (B a b c - 2 \, A a c^{2}\right )} e + {\left ({\left (B b c^{2} - 2 \, A c^{3}\right )} d - {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} e\right )} x\right )} \sqrt {c x^{2} + b x + a}}{a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x}\right ] \]
[1/2*(((B*b^2*c - 4*B*a*c^2)*e*x^2 + (B*b^3 - 4*B*a*b*c)*e*x + (B*a*b^2 - 4*B*a^2*c)*e)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*((2*B*a - A*b)*c^2*d - (B*a*b*c - 2* A*a*c^2)*e + ((B*b*c^2 - 2*A*c^3)*d - (B*b^2*c - (2*B*a + A*b)*c^2)*e)*x)* sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x), -(((B*b^2*c - 4*B*a*c^2)*e*x^2 + (B*b^3 - 4*B*a* b*c)*e*x + (B*a*b^2 - 4*B*a^2*c)*e)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*((2*B*a - A*b)*c^2*d - (B*a*b*c - 2*A*a*c^2)*e + ((B*b*c^2 - 2*A*c^3)*d - (B*b^2*c - (2*B*a + A*b)*c^2)*e)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^2 + (b^3*c^2 - 4*a*b*c^3)*x)]
\[ \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {B e \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}} + \frac {2 \, {\left (\frac {{\left (B b c d - 2 \, A c^{2} d - B b^{2} e + 2 \, B a c e + A b c e\right )} x}{b^{2} c - 4 \, a c^{2}} + \frac {2 \, B a c d - A b c d - B a b e + 2 \, A a c e}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt {c x^{2} + b x + a}} \]
-B*e*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(3/2) + 2*((B*b*c*d - 2*A*c^2*d - B*b^2*e + 2*B*a*c*e + A*b*c*e)*x/(b^2*c - 4*a*c ^2) + (2*B*a*c*d - A*b*c*d - B*a*b*e + 2*A*a*c*e)/(b^2*c - 4*a*c^2))/sqrt( c*x^2 + b*x + a)
Time = 12.23 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.29 \[ \int \frac {(A+B x) (d+e x)}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {B\,e\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{c^{3/2}}-\frac {4\,A\,a\,e-2\,A\,b\,d+2\,A\,b\,e\,x-4\,A\,c\,d\,x}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}}-\frac {B\,d\,\left (4\,a+2\,b\,x\right )}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}}+\frac {B\,e\,\left (\frac {a\,b}{2}-x\,\left (a\,c-\frac {b^2}{2}\right )\right )}{c\,\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}} \]
(B*e*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2)))/c^(3/2) - (4*A*a* e - 2*A*b*d + 2*A*b*e*x - 4*A*c*d*x)/((4*a*c - b^2)*(a + b*x + c*x^2)^(1/2 )) - (B*d*(4*a + 2*b*x))/((4*a*c - b^2)*(a + b*x + c*x^2)^(1/2)) + (B*e*(( a*b)/2 - x*(a*c - b^2/2)))/(c*(a*c - b^2/4)*(a + b*x + c*x^2)^(1/2))