Integrand size = 43, antiderivative size = 245 \[ \int \frac {\sqrt {b+a x} \left (-g+f x^2\right )}{\left (e+d x^2\right ) \sqrt {c+\sqrt {b+a x}}} \, dx=-\frac {16 c f \sqrt {b+a x} \sqrt {c+\sqrt {b+a x}}}{15 a d}+\frac {4 f \left (3 b+8 c^2+3 a x\right ) \sqrt {c+\sqrt {b+a x}}}{15 a d}-\frac {a (e f+d g) \text {RootSum}\left [b^2 d-2 b c^2 d+c^4 d+a^2 e+4 b c d \text {$\#$1}^2-4 c^3 d \text {$\#$1}^2-2 b d \text {$\#$1}^4+6 c^2 d \text {$\#$1}^4-4 c d \text {$\#$1}^6+d \text {$\#$1}^8\&,\frac {-c \log \left (\sqrt {c+\sqrt {b+a x}}-\text {$\#$1}\right )+\log \left (\sqrt {c+\sqrt {b+a x}}-\text {$\#$1}\right ) \text {$\#$1}^2}{-b \text {$\#$1}+c^2 \text {$\#$1}-2 c \text {$\#$1}^3+\text {$\#$1}^5}\&\right ]}{2 d^2} \]
Time = 0.43 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {b+a x} \left (-g+f x^2\right )}{\left (e+d x^2\right ) \sqrt {c+\sqrt {b+a x}}} \, dx=\frac {4 f \sqrt {c+\sqrt {b+a x}} \left (8 c^2-4 c \sqrt {b+a x}+3 (b+a x)\right )}{15 a d}-\frac {a (e f+d g) \text {RootSum}\left [b^2 d-2 b c^2 d+c^4 d+a^2 e+4 b c d \text {$\#$1}^2-4 c^3 d \text {$\#$1}^2-2 b d \text {$\#$1}^4+6 c^2 d \text {$\#$1}^4-4 c d \text {$\#$1}^6+d \text {$\#$1}^8\&,\frac {-c \log \left (\sqrt {c+\sqrt {b+a x}}-\text {$\#$1}\right )+\log \left (\sqrt {c+\sqrt {b+a x}}-\text {$\#$1}\right ) \text {$\#$1}^2}{-b \text {$\#$1}+c^2 \text {$\#$1}-2 c \text {$\#$1}^3+\text {$\#$1}^5}\&\right ]}{2 d^2} \]
(4*f*Sqrt[c + Sqrt[b + a*x]]*(8*c^2 - 4*c*Sqrt[b + a*x] + 3*(b + a*x)))/(1 5*a*d) - (a*(e*f + d*g)*RootSum[b^2*d - 2*b*c^2*d + c^4*d + a^2*e + 4*b*c* d*#1^2 - 4*c^3*d*#1^2 - 2*b*d*#1^4 + 6*c^2*d*#1^4 - 4*c*d*#1^6 + d*#1^8 & , (-(c*Log[Sqrt[c + Sqrt[b + a*x]] - #1]) + Log[Sqrt[c + Sqrt[b + a*x]] - #1]*#1^2)/(-(b*#1) + c^2*#1 - 2*c*#1^3 + #1^5) & ])/(2*d^2)
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.
| \(\displaystyle \int \frac {\sqrt {a x+b} \left (f x^2-g\right )}{\left (d x^2+e\right ) \sqrt {\sqrt {a x+b}+c}} \, dx\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle \frac {2 \int -\frac {(b+a x) \left (a^2 g-a^2 f x^2\right )}{\left (d x^2+e\right ) \sqrt {c+\sqrt {b+a x}}}d\sqrt {b+a x}}{a^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 \int \frac {(b+a x) \left (a^2 g-a^2 f x^2\right )}{\left (d x^2+e\right ) \sqrt {c+\sqrt {b+a x}}}d\sqrt {b+a x}}{a^3}\) |
| \(\Big \downarrow \) 2091 |
| \(\displaystyle -\frac {2 \int \frac {(b+a x) \left (a^2 g-a^2 f x^2\right )}{\sqrt {c+\sqrt {b+a x}} \left (\frac {d b^2}{a^2}-\frac {2 d (b+a x) b}{a^2}+\frac {d (b+a x)^2}{a^2}+e\right )}d\sqrt {b+a x}}{a^3}\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle -\frac {4 \int \frac {(-b+c-a x)^2 \left (a^2 g-f \left (b-(-b+c-a x)^2\right )^2\right )}{d (-b+c-a x)^4-2 b d (-b+c-a x)^2+b^2 d+a^2 e}d\sqrt {c+\sqrt {b+a x}}}{a}\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle -\frac {4 \int \frac {(-b+c-a x)^2 \left (a^2 g-f \left (b-(-b+c-a x)^2\right )^2\right )}{d (-b+c-a x)^4-2 b d (-b+c-a x)^2+b^2 d \left (\frac {e a^2}{b^2 d}+1\right )}d\sqrt {c+\sqrt {b+a x}}}{a}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {4 \int \left (-\frac {f c^2}{d}+\frac {2 f (b+a x) c}{d}-\frac {f (b+a x)^2}{d}+\frac {(e f+d g) (b+a x)^2 a^2+c^2 (e f+d g) a^2-2 c (e f+d g) (b+a x) a^2}{d \left (d (-b+c-a x)^4-2 b d (-b+c-a x)^2+b^2 d \left (\frac {e a^2}{b^2 d}+1\right )\right )}\right )d\sqrt {c+\sqrt {b+a x}}}{a}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \left (\frac {2 a^2 c (d g+e f) \int \frac {b+a x}{-d (b+a x)^4+4 c d (b+a x)^3+2 b \left (1-\frac {3 c^2}{b}\right ) d (b+a x)^2-4 b c \left (1-\frac {c^2}{b}\right ) d (b+a x)-b^2 d \left (\frac {d c^4-2 b d c^2+a^2 e}{b^2 d}+1\right )}d\sqrt {c+\sqrt {b+a x}}}{d}+\frac {a^2 c^2 (d g+e f) \int \frac {1}{d (b+a x)^4-4 c d (b+a x)^3-2 b \left (1-\frac {3 c^2}{b}\right ) d (b+a x)^2+4 b c \left (1-\frac {c^2}{b}\right ) d (b+a x)+b^2 d \left (\frac {d c^4-2 b d c^2+a^2 e}{b^2 d}+1\right )}d\sqrt {c+\sqrt {b+a x}}}{d}+\frac {a^2 (d g+e f) \int \frac {(b+a x)^2}{d (b+a x)^4-4 c d (b+a x)^3-2 b \left (1-\frac {3 c^2}{b}\right ) d (b+a x)^2+4 b c \left (1-\frac {c^2}{b}\right ) d (b+a x)+b^2 d \left (\frac {d c^4-2 b d c^2+a^2 e}{b^2 d}+1\right )}d\sqrt {c+\sqrt {b+a x}}}{d}-\frac {c^2 f \sqrt {\sqrt {a x+b}+c}}{d}+\frac {2 c f (a x+b)^{3/2}}{3 d}-\frac {f (a x+b)^{5/2}}{5 d}\right )}{a}\) |
3.27.100.3.1 Defintions of rubi rules used
Int[(Px_)*(u_)^(p_.)*(z_)^(q_.), x_Symbol] :> Int[Px*ExpandToSum[z, x]^q*Ex pandToSum[u, x]^p, x] /; FreeQ[{p, q}, x] && PolyQ[Px, x] && BinomialQ[z, x ] && TrinomialQ[u, x] && !(BinomialMatchQ[z, x] && TrinomialMatchQ[u, x])
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.32 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {\frac {4 f \left (\frac {\left (c +\sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+\sqrt {c +\sqrt {a x +b}}\, c^{2}\right )}{d}-\frac {a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{8}-4 c d \,\textit {\_Z}^{6}+\left (6 c^{2} d -2 b d \right ) \textit {\_Z}^{4}+\left (-4 c^{3} d +4 b c d \right ) \textit {\_Z}^{2}+c^{4} d -2 b \,c^{2} d +e \,a^{2}+b^{2} d \right )}{\sum }\frac {\left (\left (d g +e f \right ) \textit {\_R}^{4}+2 c \left (-d g -e f \right ) \textit {\_R}^{2}+c^{2} d g +c^{2} e f \right ) \ln \left (\sqrt {c +\sqrt {a x +b}}-\textit {\_R} \right )}{\textit {\_R}^{7}-3 \textit {\_R}^{5} c +3 \textit {\_R}^{3} c^{2}-\textit {\_R}^{3} b -\textit {\_R} \,c^{3}+\textit {\_R} b c}\right )}{2 d^{2}}}{a}\) | \(226\) |
| default | \(-\frac {2 \left (-\frac {2 f \left (\frac {\left (c +\sqrt {a x +b}\right )^{\frac {5}{2}}}{5}-\frac {2 c \left (c +\sqrt {a x +b}\right )^{\frac {3}{2}}}{3}+\sqrt {c +\sqrt {a x +b}}\, c^{2}\right )}{d}+\frac {a^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,\textit {\_Z}^{8}-4 c d \,\textit {\_Z}^{6}+\left (6 c^{2} d -2 b d \right ) \textit {\_Z}^{4}+\left (-4 c^{3} d +4 b c d \right ) \textit {\_Z}^{2}+c^{4} d -2 b \,c^{2} d +e \,a^{2}+b^{2} d \right )}{\sum }\frac {\left (\left (d g +e f \right ) \textit {\_R}^{4}+2 c \left (-d g -e f \right ) \textit {\_R}^{2}+c^{2} d g +c^{2} e f \right ) \ln \left (\sqrt {c +\sqrt {a x +b}}-\textit {\_R} \right )}{\textit {\_R}^{7}-3 \textit {\_R}^{5} c +3 \textit {\_R}^{3} c^{2}-\textit {\_R}^{3} b -\textit {\_R} \,c^{3}+\textit {\_R} b c}\right )}{4 d^{2}}\right )}{a}\) | \(226\) |
2/a*(2*f/d*(1/5*(c+(a*x+b)^(1/2))^(5/2)-2/3*c*(c+(a*x+b)^(1/2))^(3/2)+(c+( a*x+b)^(1/2))^(1/2)*c^2)-1/4*a^2/d^2*sum(((d*g+e*f)*_R^4+2*c*(-d*g-e*f)*_R ^2+c^2*d*g+c^2*e*f)/(_R^7-3*_R^5*c+3*_R^3*c^2-_R^3*b-_R*c^3+_R*b*c)*ln((c+ (a*x+b)^(1/2))^(1/2)-_R),_R=RootOf(d*_Z^8-4*c*d*_Z^6+(6*c^2*d-2*b*d)*_Z^4+ (-4*c^3*d+4*b*c*d)*_Z^2+c^4*d-2*b*c^2*d+e*a^2+b^2*d)))
Timed out. \[ \int \frac {\sqrt {b+a x} \left (-g+f x^2\right )}{\left (e+d x^2\right ) \sqrt {c+\sqrt {b+a x}}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {b+a x} \left (-g+f x^2\right )}{\left (e+d x^2\right ) \sqrt {c+\sqrt {b+a x}}} \, dx=\text {Timed out} \]
Not integrable
Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {b+a x} \left (-g+f x^2\right )}{\left (e+d x^2\right ) \sqrt {c+\sqrt {b+a x}}} \, dx=\int { \frac {{\left (f x^{2} - g\right )} \sqrt {a x + b}}{{\left (d x^{2} + e\right )} \sqrt {c + \sqrt {a x + b}}} \,d x } \]
Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 76.01 (sec) , antiderivative size = 2635, normalized size of antiderivative = 10.76 \[ \int \frac {\sqrt {b+a x} \left (-g+f x^2\right )}{\left (e+d x^2\right ) \sqrt {c+\sqrt {b+a x}}} \, dx=\text {Too large to display} \]
4/15*(3*a^4*(c + sqrt(a*x + b))^(5/2)*d^4*f - 10*a^4*(c + sqrt(a*x + b))^( 3/2)*c*d^4*f + 15*a^4*sqrt(c + sqrt(a*x + b))*c^2*d^4*f)/(a^5*d^5) + 1/2*( (a^7*(c + sqrt((b*d + sqrt(-d*e)*a)/d))^2*d^4*e*f - 2*a^7*(c + sqrt((b*d + sqrt(-d*e)*a)/d))*c*d^4*e*f + a^7*c^2*d^4*e*f + a^7*(c + sqrt((b*d + sqrt (-d*e)*a)/d))^2*d^5*g - 2*a^7*(c + sqrt((b*d + sqrt(-d*e)*a)/d))*c*d^5*g + a^7*c^2*d^5*g)*log(sqrt(c + sqrt(a*x + b)) + sqrt(c + sqrt((b*d + sqrt(-d *e)*a)/d)))/((c + sqrt((b*d + sqrt(-d*e)*a)/d))^(7/2)*d - 3*(c + sqrt((b*d + sqrt(-d*e)*a)/d))^(5/2)*c*d + (3*c^2*d - b*d)*(c + sqrt((b*d + sqrt(-d* e)*a)/d))^(3/2) - (c^3*d - b*c*d)*sqrt(c + sqrt((b*d + sqrt(-d*e)*a)/d))) - (a^7*(c + sqrt((b*d + sqrt(-d*e)*a)/d))^2*d^4*e*f - 2*a^7*(c + sqrt((b*d + sqrt(-d*e)*a)/d))*c*d^4*e*f + a^7*c^2*d^4*e*f + a^7*(c + sqrt((b*d + sq rt(-d*e)*a)/d))^2*d^5*g - 2*a^7*(c + sqrt((b*d + sqrt(-d*e)*a)/d))*c*d^5*g + a^7*c^2*d^5*g)*log(sqrt(c + sqrt(a*x + b)) - sqrt(c + sqrt((b*d + sqrt( -d*e)*a)/d)))/((c + sqrt((b*d + sqrt(-d*e)*a)/d))^(7/2)*d - 3*(c + sqrt((b *d + sqrt(-d*e)*a)/d))^(5/2)*c*d + (3*c^2*d - b*d)*(c + sqrt((b*d + sqrt(- d*e)*a)/d))^(3/2) - (c^3*d - b*c*d)*sqrt(c + sqrt((b*d + sqrt(-d*e)*a)/d)) ) + (a^7*(c - sqrt((b*d + sqrt(-d*e)*a)/d))^2*d^4*e*f - 2*a^7*(c - sqrt((b *d + sqrt(-d*e)*a)/d))*c*d^4*e*f + a^7*c^2*d^4*e*f + a^7*(c - sqrt((b*d + sqrt(-d*e)*a)/d))^2*d^5*g - 2*a^7*(c - sqrt((b*d + sqrt(-d*e)*a)/d))*c*d^5 *g + a^7*c^2*d^5*g)*log(sqrt(c + sqrt(a*x + b)) + sqrt(c - sqrt((b*d + ...
Not integrable
Time = 7.33 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt {b+a x} \left (-g+f x^2\right )}{\left (e+d x^2\right ) \sqrt {c+\sqrt {b+a x}}} \, dx=\int -\frac {\left (g-f\,x^2\right )\,\sqrt {b+a\,x}}{\sqrt {c+\sqrt {b+a\,x}}\,\left (d\,x^2+e\right )} \,d x \]