Integrand size = 46, antiderivative size = 233 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (2 c d-b e) (d+e x)^{5/2}}-\frac {(3 c e f+5 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{4 e^2 (2 c d-b e)^2 (d+e x)^{3/2}}-\frac {c (3 c e f+5 c d g-4 b e g) \text {arctanh}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {2 c d-b e} \sqrt {d+e x}}\right )}{4 e^2 (2 c d-b e)^{5/2}} \]
-1/4*c*(-4*b*e*g+5*c*d*g+3*c*e*f)*arctanh((d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2) ^(1/2)/(-b*e+2*c*d)^(1/2)/(e*x+d)^(1/2))/e^2/(-b*e+2*c*d)^(5/2)-1/2*(-d*g+ e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/e^2/(-b*e+2*c*d)/(e*x+d)^(5/2) -1/4*(-4*b*e*g+5*c*d*g+3*c*e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/e^2 /(-b*e+2*c*d)^2/(e*x+d)^(3/2)
Time = 0.49 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.84 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {c \sqrt {d+e x} \left (-\frac {(-b e+c (d-e x)) \left (-2 b e (d g+e (f+2 g x))+c \left (d^2 g+3 e^2 f x+d e (7 f+5 g x)\right )\right )}{c (-2 c d+b e)^2 (d+e x)^2}+\frac {(3 c e f+5 c d g-4 b e g) \sqrt {-b e+c (d-e x)} \arctan \left (\frac {\sqrt {-b e+c (d-e x)}}{\sqrt {-2 c d+b e}}\right )}{(-2 c d+b e)^{5/2}}\right )}{4 e^2 \sqrt {(d+e x) (-b e+c (d-e x))}} \]
(c*Sqrt[d + e*x]*(-(((-(b*e) + c*(d - e*x))*(-2*b*e*(d*g + e*(f + 2*g*x)) + c*(d^2*g + 3*e^2*f*x + d*e*(7*f + 5*g*x))))/(c*(-2*c*d + b*e)^2*(d + e*x )^2)) + ((3*c*e*f + 5*c*d*g - 4*b*e*g)*Sqrt[-(b*e) + c*(d - e*x)]*ArcTan[S qrt[-(b*e) + c*(d - e*x)]/Sqrt[-2*c*d + b*e]])/(-2*c*d + b*e)^(5/2)))/(4*e ^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])
Time = 0.45 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1220, 1135, 1136, 221}
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.
| \(\displaystyle \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2}} \, dx\) |
| \(\Big \downarrow \) 1220 |
| \(\displaystyle \frac {(-4 b e g+5 c d g+3 c e f) \int \frac {1}{(d+e x)^{3/2} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\) |
| \(\Big \downarrow \) 1135 |
| \(\displaystyle \frac {(-4 b e g+5 c d g+3 c e f) \left (\frac {c \int \frac {1}{\sqrt {d+e x} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{2 (2 c d-b e)}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2} (2 c d-b e)}\right )}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\) |
| \(\Big \downarrow \) 1136 |
| \(\displaystyle \frac {(-4 b e g+5 c d g+3 c e f) \left (\frac {c e \int \frac {1}{\frac {e^2 \left (-c x^2 e^2-b x e^2+d (c d-b e)\right )}{d+e x}-e^2 (2 c d-b e)}d\frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{\sqrt {d+e x}}}{2 c d-b e}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2} (2 c d-b e)}\right )}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (-\frac {c \text {arctanh}\left (\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{\sqrt {d+e x} \sqrt {2 c d-b e}}\right )}{e (2 c d-b e)^{3/2}}-\frac {\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)^{3/2} (2 c d-b e)}\right ) (-4 b e g+5 c d g+3 c e f)}{4 e (2 c d-b e)}-\frac {(e f-d g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{2 e^2 (d+e x)^{5/2} (2 c d-b e)}\) |
-1/2*((e*f - d*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(e^2*(2*c*d - b*e)*(d + e*x)^(5/2)) + ((3*c*e*f + 5*c*d*g - 4*b*e*g)*(-(Sqrt[d*(c*d - b *e) - b*e^2*x - c*e^2*x^2]/(e*(2*c*d - b*e)*(d + e*x)^(3/2))) - (c*ArcTanh [Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2]/(Sqrt[2*c*d - b*e]*Sqrt[d + e*x ])])/(e*(2*c*d - b*e)^(3/2))))/(4*e*(2*c*d - b*e))
3.23.67.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e))) Int [(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I ntegerQ[2*p]
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x _Symbol] :> Simp[2*e Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Leaf count of result is larger than twice the leaf count of optimal. \(621\) vs. \(2(211)=422\).
Time = 0.36 (sec) , antiderivative size = 622, normalized size of antiderivative = 2.67
| method | result | size |
| default | \(-\frac {\sqrt {-\left (e x +d \right ) \left (x c e +b e -c d \right )}\, \left (4 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,e^{3} g \,x^{2}-5 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d \,e^{2} g \,x^{2}-3 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} e^{3} f \,x^{2}+8 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c d \,e^{2} g x -10 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} e g x -6 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d \,e^{2} f x +4 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) b c \,d^{2} e g -5 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{3} g -3 \arctan \left (\frac {\sqrt {-x c e -b e +c d}}{\sqrt {b e -2 c d}}\right ) c^{2} d^{2} e f -4 b \,e^{2} g x \sqrt {-x c e -b e +c d}\, \sqrt {b e -2 c d}+5 c d e g x \sqrt {-x c e -b e +c d}\, \sqrt {b e -2 c d}+3 c \,e^{2} f x \sqrt {-x c e -b e +c d}\, \sqrt {b e -2 c d}-2 b d e g \sqrt {-x c e -b e +c d}\, \sqrt {b e -2 c d}-2 b \,e^{2} f \sqrt {-x c e -b e +c d}\, \sqrt {b e -2 c d}+c \,d^{2} g \sqrt {-x c e -b e +c d}\, \sqrt {b e -2 c d}+7 c d e f \sqrt {-x c e -b e +c d}\, \sqrt {b e -2 c d}\right )}{4 \left (e x +d \right )^{\frac {5}{2}} \left (b e -2 c d \right )^{\frac {5}{2}} e^{2} \sqrt {-x c e -b e +c d}}\) | \(622\) |
-1/4*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)*(4*arctan((-c*e*x-b*e+c*d)^(1/2)/(b* e-2*c*d)^(1/2))*b*c*e^3*g*x^2-5*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^ (1/2))*c^2*d*e^2*g*x^2-3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))* c^2*e^3*f*x^2+8*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*d*e^2 *g*x-10*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d^2*e*g*x-6*a rctan((-c*e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d*e^2*f*x+4*arctan((-c *e*x-b*e+c*d)^(1/2)/(b*e-2*c*d)^(1/2))*b*c*d^2*e*g-5*arctan((-c*e*x-b*e+c* d)^(1/2)/(b*e-2*c*d)^(1/2))*c^2*d^3*g-3*arctan((-c*e*x-b*e+c*d)^(1/2)/(b*e -2*c*d)^(1/2))*c^2*d^2*e*f-4*b*e^2*g*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^ (1/2)+5*c*d*e*g*x*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+3*c*e^2*f*x*(-c *e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)-2*b*d*e*g*(-c*e*x-b*e+c*d)^(1/2)*(b* e-2*c*d)^(1/2)-2*b*e^2*f*(-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+c*d^2*g* (-c*e*x-b*e+c*d)^(1/2)*(b*e-2*c*d)^(1/2)+7*c*d*e*f*(-c*e*x-b*e+c*d)^(1/2)* (b*e-2*c*d)^(1/2))/(e*x+d)^(5/2)/(b*e-2*c*d)^(5/2)/e^2/(-c*e*x-b*e+c*d)^(1 /2)
Leaf count of result is larger than twice the leaf count of optimal. 568 vs. \(2 (211) = 422\).
Time = 0.33 (sec) , antiderivative size = 1168, normalized size of antiderivative = 5.01 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\text {Too large to display} \]
[-1/8*((3*c^2*d^3*e*f + (3*c^2*e^4*f + (5*c^2*d*e^3 - 4*b*c*e^4)*g)*x^3 + 3*(3*c^2*d*e^3*f + (5*c^2*d^2*e^2 - 4*b*c*d*e^3)*g)*x^2 + (5*c^2*d^4 - 4*b *c*d^3*e)*g + 3*(3*c^2*d^2*e^2*f + (5*c^2*d^3*e - 4*b*c*d^2*e^2)*g)*x)*sqr t(2*c*d - b*e)*log(-(c*e^2*x^2 - 3*c*d^2 + 2*b*d*e - 2*(c*d*e - b*e^2)*x - 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(2*c*d - b*e)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) + 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b* d*e)*((14*c^2*d^2*e - 11*b*c*d*e^2 + 2*b^2*e^3)*f + (2*c^2*d^3 - 5*b*c*d^2 *e + 2*b^2*d*e^2)*g + (3*(2*c^2*d*e^2 - b*c*e^3)*f + (10*c^2*d^2*e - 13*b* c*d*e^2 + 4*b^2*e^3)*g)*x)*sqrt(e*x + d))/(8*c^3*d^6*e^2 - 12*b*c^2*d^5*e^ 3 + 6*b^2*c*d^4*e^4 - b^3*d^3*e^5 + (8*c^3*d^3*e^5 - 12*b*c^2*d^2*e^6 + 6* b^2*c*d*e^7 - b^3*e^8)*x^3 + 3*(8*c^3*d^4*e^4 - 12*b*c^2*d^3*e^5 + 6*b^2*c *d^2*e^6 - b^3*d*e^7)*x^2 + 3*(8*c^3*d^5*e^3 - 12*b*c^2*d^4*e^4 + 6*b^2*c* d^3*e^5 - b^3*d^2*e^6)*x), -1/4*((3*c^2*d^3*e*f + (3*c^2*e^4*f + (5*c^2*d* e^3 - 4*b*c*e^4)*g)*x^3 + 3*(3*c^2*d*e^3*f + (5*c^2*d^2*e^2 - 4*b*c*d*e^3) *g)*x^2 + (5*c^2*d^4 - 4*b*c*d^3*e)*g + 3*(3*c^2*d^2*e^2*f + (5*c^2*d^3*e - 4*b*c*d^2*e^2)*g)*x)*sqrt(-2*c*d + b*e)*arctan(sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(-2*c*d + b*e)*sqrt(e*x + d)/(c*e^2*x^2 + b*e^2*x - c*d^2 + b*d*e)) + sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((14*c^2*d^2* e - 11*b*c*d*e^2 + 2*b^2*e^3)*f + (2*c^2*d^3 - 5*b*c*d^2*e + 2*b^2*d*e^2)* g + (3*(2*c^2*d*e^2 - b*c*e^3)*f + (10*c^2*d^2*e - 13*b*c*d*e^2 + 4*b^2...
\[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {f + g x}{\sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )} \left (d + e x\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int { \frac {g x + f}{\sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.48 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.53 \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\frac {\frac {{\left (3 \, c^{3} e f + 5 \, c^{3} d g - 4 \, b c^{2} e g\right )} \arctan \left (\frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e}}{\sqrt {-2 \, c d + b e}}\right )}{{\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} \sqrt {-2 \, c d + b e}} - \frac {10 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{4} d e f - 5 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{3} e^{2} f + 6 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} c^{4} d^{2} g - 11 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b c^{3} d e g + 4 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d - b e} b^{2} c^{2} e^{2} g - 3 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{3} e f - 5 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} c^{3} d g + 4 \, {\left (-{\left (e x + d\right )} c + 2 \, c d - b e\right )}^{\frac {3}{2}} b c^{2} e g}{{\left (4 \, c^{2} d^{2} - 4 \, b c d e + b^{2} e^{2}\right )} {\left (e x + d\right )}^{2} c^{2}}}{4 \, c e^{2}} \]
1/4*((3*c^3*e*f + 5*c^3*d*g - 4*b*c^2*e*g)*arctan(sqrt(-(e*x + d)*c + 2*c* d - b*e)/sqrt(-2*c*d + b*e))/((4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*sqrt(-2*c* d + b*e)) - (10*sqrt(-(e*x + d)*c + 2*c*d - b*e)*c^4*d*e*f - 5*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b*c^3*e^2*f + 6*sqrt(-(e*x + d)*c + 2*c*d - b*e)*c^4 *d^2*g - 11*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b*c^3*d*e*g + 4*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b^2*c^2*e^2*g - 3*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c ^3*e*f - 5*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c^3*d*g + 4*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*b*c^2*e*g)/((4*c^2*d^2 - 4*b*c*d*e + b^2*e^2)*(e*x + d) ^2*c^2))/(c*e^2)
Timed out. \[ \int \frac {f+g x}{(d+e x)^{5/2} \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx=\int \frac {f+g\,x}{{\left (d+e\,x\right )}^{5/2}\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}} \,d x \]