Integrand size = 26, antiderivative size = 273 \[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\frac {2 d^2 \sqrt {e x}}{3 c \left (b c^2+a d^2\right ) e (c+d x)^{3/2}}+\frac {4 d^2 \left (4 b c^2+a d^2\right ) \sqrt {e x}}{3 c^2 \left (b c^2+a d^2\right )^2 e \sqrt {c+d x}}-\frac {b \arctan \left (\frac {\sqrt {\sqrt {b} c-\sqrt {-a} d} \sqrt {e x}}{\sqrt [4]{-a} \sqrt {e} \sqrt {c+d x}}\right )}{(-a)^{3/4} \left (\sqrt {b} c-\sqrt {-a} d\right )^{5/2} \sqrt {e}}-\frac {b \text {arctanh}\left (\frac {\sqrt {\sqrt {b} c+\sqrt {-a} d} \sqrt {e x}}{\sqrt [4]{-a} \sqrt {e} \sqrt {c+d x}}\right )}{(-a)^{3/4} \left (\sqrt {b} c+\sqrt {-a} d\right )^{5/2} \sqrt {e}} \] Output:
2/3*d^2*(e*x)^(1/2)/c/(a*d^2+b*c^2)/e/(d*x+c)^(3/2)+4/3*d^2*(a*d^2+4*b*c^2 )*(e*x)^(1/2)/c^2/(a*d^2+b*c^2)^2/e/(d*x+c)^(1/2)-b*arctan((b^(1/2)*c-(-a) ^(1/2)*d)^(1/2)*(e*x)^(1/2)/(-a)^(1/4)/e^(1/2)/(d*x+c)^(1/2))/(-a)^(3/4)/( b^(1/2)*c-(-a)^(1/2)*d)^(5/2)/e^(1/2)-b*arctanh((b^(1/2)*c+(-a)^(1/2)*d)^( 1/2)*(e*x)^(1/2)/(-a)^(1/4)/e^(1/2)/(d*x+c)^(1/2))/(-a)^(3/4)/(b^(1/2)*c+( -a)^(1/2)*d)^(5/2)/e^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.42 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\frac {2 \left (a d^4 x (3 c+2 d x)+b c^2 d^2 x (9 c+8 d x)-3 b c^2 d^{3/2} \sqrt {x} (c+d x)^{3/2} \text {RootSum}\left [b c^4-4 b c^3 \text {$\#$1}+6 b c^2 \text {$\#$1}^2+16 a d^2 \text {$\#$1}^2-4 b c \text {$\#$1}^3+b \text {$\#$1}^4\&,\frac {b c^3 \log \left (c+2 d x-2 \sqrt {d} \sqrt {x} \sqrt {c+d x}-\text {$\#$1}\right )-4 b c^2 \log \left (c+2 d x-2 \sqrt {d} \sqrt {x} \sqrt {c+d x}-\text {$\#$1}\right ) \text {$\#$1}+2 a d^2 \log \left (c+2 d x-2 \sqrt {d} \sqrt {x} \sqrt {c+d x}-\text {$\#$1}\right ) \text {$\#$1}+b c \log \left (c+2 d x-2 \sqrt {d} \sqrt {x} \sqrt {c+d x}-\text {$\#$1}\right ) \text {$\#$1}^2}{b c^3-3 b c^2 \text {$\#$1}-8 a d^2 \text {$\#$1}+3 b c \text {$\#$1}^2-b \text {$\#$1}^3}\&\right ]\right )}{3 \left (b c^3+a c d^2\right )^2 \sqrt {e x} (c+d x)^{3/2}} \] Input:
Integrate[1/(Sqrt[e*x]*(c + d*x)^(5/2)*(a + b*x^2)),x]
Output:
(2*(a*d^4*x*(3*c + 2*d*x) + b*c^2*d^2*x*(9*c + 8*d*x) - 3*b*c^2*d^(3/2)*Sq rt[x]*(c + d*x)^(3/2)*RootSum[b*c^4 - 4*b*c^3*#1 + 6*b*c^2*#1^2 + 16*a*d^2 *#1^2 - 4*b*c*#1^3 + b*#1^4 & , (b*c^3*Log[c + 2*d*x - 2*Sqrt[d]*Sqrt[x]*S qrt[c + d*x] - #1] - 4*b*c^2*Log[c + 2*d*x - 2*Sqrt[d]*Sqrt[x]*Sqrt[c + d* x] - #1]*#1 + 2*a*d^2*Log[c + 2*d*x - 2*Sqrt[d]*Sqrt[x]*Sqrt[c + d*x] - #1 ]*#1 + b*c*Log[c + 2*d*x - 2*Sqrt[d]*Sqrt[x]*Sqrt[c + d*x] - #1]*#1^2)/(b* c^3 - 3*b*c^2*#1 - 8*a*d^2*#1 + 3*b*c*#1^2 - b*#1^3) & ]))/(3*(b*c^3 + a*c *d^2)^2*Sqrt[e*x]*(c + d*x)^(3/2))
Time = 1.56 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.51, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {615, 2009}
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.
\(\displaystyle \int \frac {1}{\sqrt {e x} \left (a+b x^2\right ) (c+d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 615 |
\(\displaystyle \int \left (\frac {\sqrt {-a}}{2 a \sqrt {e x} \left (\sqrt {-a}-\sqrt {b} x\right ) (c+d x)^{5/2}}+\frac {\sqrt {-a}}{2 a \sqrt {e x} \left (\sqrt {-a}+\sqrt {b} x\right ) (c+d x)^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b \arctan \left (\frac {\sqrt {e x} \sqrt {\sqrt {b} c-\sqrt {-a} d}}{\sqrt [4]{-a} \sqrt {e} \sqrt {c+d x}}\right )}{(-a)^{3/4} \sqrt {e} \left (\sqrt {b} c-\sqrt {-a} d\right )^{5/2}}-\frac {b \text {arctanh}\left (\frac {\sqrt {e x} \sqrt {\sqrt {-a} d+\sqrt {b} c}}{\sqrt [4]{-a} \sqrt {e} \sqrt {c+d x}}\right )}{(-a)^{3/4} \sqrt {e} \left (\sqrt {-a} d+\sqrt {b} c\right )^{5/2}}+\frac {d \sqrt {e x} \left (5 \sqrt {-a} \sqrt {b} c-2 a d\right )}{3 a c^2 e \sqrt {c+d x} \left (\sqrt {-a} d+\sqrt {b} c\right )^2}-\frac {d \sqrt {e x} \left (5 \sqrt {-a} \sqrt {b} c+2 a d\right )}{3 a c^2 e \sqrt {c+d x} \left (\sqrt {b} c-\sqrt {-a} d\right )^2}-\frac {d \sqrt {e x}}{3 c e (c+d x)^{3/2} \left (\sqrt {-a} \sqrt {b} c-a d\right )}+\frac {d \sqrt {e x}}{3 c e (c+d x)^{3/2} \left (\sqrt {-a} \sqrt {b} c+a d\right )}\) |
Input:
Int[1/(Sqrt[e*x]*(c + d*x)^(5/2)*(a + b*x^2)),x]
Output:
-1/3*(d*Sqrt[e*x])/(c*(Sqrt[-a]*Sqrt[b]*c - a*d)*e*(c + d*x)^(3/2)) + (d*S qrt[e*x])/(3*c*(Sqrt[-a]*Sqrt[b]*c + a*d)*e*(c + d*x)^(3/2)) + (d*(5*Sqrt[ -a]*Sqrt[b]*c - 2*a*d)*Sqrt[e*x])/(3*a*c^2*(Sqrt[b]*c + Sqrt[-a]*d)^2*e*Sq rt[c + d*x]) - (d*(5*Sqrt[-a]*Sqrt[b]*c + 2*a*d)*Sqrt[e*x])/(3*a*c^2*(Sqrt [b]*c - Sqrt[-a]*d)^2*e*Sqrt[c + d*x]) - (b*ArcTan[(Sqrt[Sqrt[b]*c - Sqrt[ -a]*d]*Sqrt[e*x])/((-a)^(1/4)*Sqrt[e]*Sqrt[c + d*x])])/((-a)^(3/4)*(Sqrt[b ]*c - Sqrt[-a]*d)^(5/2)*Sqrt[e]) - (b*ArcTanh[(Sqrt[Sqrt[b]*c + Sqrt[-a]*d ]*Sqrt[e*x])/((-a)^(1/4)*Sqrt[e]*Sqrt[c + d*x])])/((-a)^(3/4)*(Sqrt[b]*c + Sqrt[-a]*d)^(5/2)*Sqrt[e])
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(3130\) vs. \(2(213)=426\).
Time = 0.51 (sec) , antiderivative size = 3131, normalized size of antiderivative = 11.47
Input:
int(1/(e*x)^(1/2)/(d*x+c)^(5/2)/(b*x^2+a),x,method=_RETURNVERBOSE)
Output:
1/6*(12*ln((2*(-a*b)^(1/2)*d*e*x+b*c*e*x+2*((d*x+c)*e*x)^(1/2)*(e*(-a*d+c* (-a*b)^(1/2))/b)^(1/2)*b+c*e*(-a*b)^(1/2))/(b*x-(-a*b)^(1/2)))*a*b^2*c^6*d ^2*e*x*(-e*(a*d+c*(-a*b)^(1/2))/b)^(1/2)+6*ln((2*(-a*b)^(1/2)*d*e*x+b*c*e* x+2*((d*x+c)*e*x)^(1/2)*(e*(-a*d+c*(-a*b)^(1/2))/b)^(1/2)*b+c*e*(-a*b)^(1/ 2))/(b*x-(-a*b)^(1/2)))*b^2*c^7*d*e*x*(-e*(a*d+c*(-a*b)^(1/2))/b)^(1/2)*(- a*b)^(1/2)+12*ln((-2*(-a*b)^(1/2)*d*e*x+b*c*e*x+2*((d*x+c)*e*x)^(1/2)*(-e* (a*d+c*(-a*b)^(1/2))/b)^(1/2)*b-c*e*(-a*b)^(1/2))/(b*x+(-a*b)^(1/2)))*a^2* b*c^4*d^4*e*x*(e*(-a*d+c*(-a*b)^(1/2))/b)^(1/2)+6*ln((-2*(-a*b)^(1/2)*d*e* x+b*c*e*x+2*((d*x+c)*e*x)^(1/2)*(-e*(a*d+c*(-a*b)^(1/2))/b)^(1/2)*b-c*e*(- a*b)^(1/2))/(b*x+(-a*b)^(1/2)))*a^2*c^3*d^5*e*x*(-a*b)^(1/2)*(e*(-a*d+c*(- a*b)^(1/2))/b)^(1/2)+12*ln((-2*(-a*b)^(1/2)*d*e*x+b*c*e*x+2*((d*x+c)*e*x)^ (1/2)*(-e*(a*d+c*(-a*b)^(1/2))/b)^(1/2)*b-c*e*(-a*b)^(1/2))/(b*x+(-a*b)^(1 /2)))*a*b^2*c^6*d^2*e*x*(e*(-a*d+c*(-a*b)^(1/2))/b)^(1/2)-6*ln((-2*(-a*b)^ (1/2)*d*e*x+b*c*e*x+2*((d*x+c)*e*x)^(1/2)*(-e*(a*d+c*(-a*b)^(1/2))/b)^(1/2 )*b-c*e*(-a*b)^(1/2))/(b*x+(-a*b)^(1/2)))*b^2*c^7*d*e*x*(-a*b)^(1/2)*(e*(- a*d+c*(-a*b)^(1/2))/b)^(1/2)+40*a^2*b*c^2*d^5*x*(-e*(a*d+c*(-a*b)^(1/2))/b )^(1/2)*((d*x+c)*e*x)^(1/2)*(e*(-a*d+c*(-a*b)^(1/2))/b)^(1/2)+32*a*b^2*c^4 *d^3*x*(-e*(a*d+c*(-a*b)^(1/2))/b)^(1/2)*((d*x+c)*e*x)^(1/2)*(e*(-a*d+c*(- a*b)^(1/2))/b)^(1/2)+6*ln((2*(-a*b)^(1/2)*d*e*x+b*c*e*x+2*((d*x+c)*e*x)^(1 /2)*(e*(-a*d+c*(-a*b)^(1/2))/b)^(1/2)*b+c*e*(-a*b)^(1/2))/(b*x-(-a*b)^(...
Leaf count of result is larger than twice the leaf count of optimal. 5370 vs. \(2 (213) = 426\).
Time = 1.27 (sec) , antiderivative size = 5370, normalized size of antiderivative = 19.67 \[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:
integrate(1/(e*x)^(1/2)/(d*x+c)^(5/2)/(b*x^2+a),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\int \frac {1}{\sqrt {e x} \left (a + b x^{2}\right ) \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(e*x)**(1/2)/(d*x+c)**(5/2)/(b*x**2+a),x)
Output:
Integral(1/(sqrt(e*x)*(a + b*x**2)*(c + d*x)**(5/2)), x)
\[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )} {\left (d x + c\right )}^{\frac {5}{2}} \sqrt {e x}} \,d x } \] Input:
integrate(1/(e*x)^(1/2)/(d*x+c)^(5/2)/(b*x^2+a),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)*(d*x + c)^(5/2)*sqrt(e*x)), x)
Timed out. \[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x)^(1/2)/(d*x+c)^(5/2)/(b*x^2+a),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\int \frac {1}{\sqrt {e\,x}\,\left (b\,x^2+a\right )\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int(1/((e*x)^(1/2)*(a + b*x^2)*(c + d*x)^(5/2)),x)
Output:
int(1/((e*x)^(1/2)*(a + b*x^2)*(c + d*x)^(5/2)), x)
\[ \int \frac {1}{\sqrt {e x} (c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\frac {\int \frac {1}{\sqrt {x}\, \sqrt {d x +c}\, a \,c^{2}+2 \sqrt {x}\, \sqrt {d x +c}\, a c d x +\sqrt {x}\, \sqrt {d x +c}\, a \,d^{2} x^{2}+\sqrt {x}\, \sqrt {d x +c}\, b \,c^{2} x^{2}+2 \sqrt {x}\, \sqrt {d x +c}\, b c d \,x^{3}+\sqrt {x}\, \sqrt {d x +c}\, b \,d^{2} x^{4}}d x}{\sqrt {e}} \] Input:
int(1/(e*x)^(1/2)/(d*x+c)^(5/2)/(b*x^2+a),x)
Output:
int(1/(sqrt(x)*sqrt(c + d*x)*a*c**2 + 2*sqrt(x)*sqrt(c + d*x)*a*c*d*x + sq rt(x)*sqrt(c + d*x)*a*d**2*x**2 + sqrt(x)*sqrt(c + d*x)*b*c**2*x**2 + 2*sq rt(x)*sqrt(c + d*x)*b*c*d*x**3 + sqrt(x)*sqrt(c + d*x)*b*d**2*x**4),x)/sqr t(e)