Integrand size = 22, antiderivative size = 226 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\frac {b \left (b c^2+2 a d^2\right ) \sqrt {a+b x^2}}{d^3}-\frac {a^2 \sqrt {a+b x^2}}{c x}-\frac {b^2 c x \sqrt {a+b x^2}}{2 d^2}+\frac {b \left (a+b x^2\right )^{3/2}}{3 d}-\frac {b^{3/2} c \left (2 b c^2+5 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 d^4}-\frac {\left (b c^2+a d^2\right )^{5/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{c^2 d^4}+\frac {a^{5/2} d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{c^2} \] Output:
b*(2*a*d^2+b*c^2)*(b*x^2+a)^(1/2)/d^3-a^2*(b*x^2+a)^(1/2)/c/x-1/2*b^2*c*x* (b*x^2+a)^(1/2)/d^2+1/3*b*(b*x^2+a)^(3/2)/d-1/2*b^(3/2)*c*(5*a*d^2+2*b*c^2 )*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/d^4-(a*d^2+b*c^2)^(5/2)*arctanh((-b*c *x+a*d)/(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/c^2/d^4+a^(5/2)*d*arctanh((b* x^2+a)^(1/2)/a^(1/2))/c^2
Time = 0.81 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\frac {-\frac {c d \sqrt {a+b x^2} \left (6 a^2 d^3-14 a b c d^2 x+b^2 c x \left (-6 c^2+3 c d x-2 d^2 x^2\right )\right )}{x}+12 \left (-b c^2-a d^2\right )^{5/2} \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )-12 a^{5/2} d^5 \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )+3 b^{3/2} c^3 \left (2 b c^2+5 a d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{6 c^2 d^4} \] Input:
Integrate[(a + b*x^2)^(5/2)/(x^2*(c + d*x)),x]
Output:
(-((c*d*Sqrt[a + b*x^2]*(6*a^2*d^3 - 14*a*b*c*d^2*x + b^2*c*x*(-6*c^2 + 3* c*d*x - 2*d^2*x^2)))/x) + 12*(-(b*c^2) - a*d^2)^(5/2)*ArcTan[(Sqrt[b]*(c + d*x) - d*Sqrt[a + b*x^2])/Sqrt[-(b*c^2) - a*d^2]] - 12*a^(5/2)*d^5*ArcTan h[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]] + 3*b^(3/2)*c^3*(2*b*c^2 + 5*a*d^ 2)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(6*c^2*d^4)
Time = 1.13 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.69, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {617, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx\) |
\(\Big \downarrow \) 617 |
\(\displaystyle \int \left (\frac {d^2 \left (a+b x^2\right )^{5/2}}{c^2 (c+d x)}-\frac {d \left (a+b x^2\right )^{5/2}}{c^2 x}+\frac {\left (a+b x^2\right )^{5/2}}{c x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^{5/2} d \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{c^2}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (15 a^2 d^4+20 a b c^2 d^2+8 b^2 c^4\right )}{8 c d^4}+\frac {15 a^2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 c}-\frac {a^2 d \sqrt {a+b x^2}}{c^2}-\frac {\left (a d^2+b c^2\right )^{5/2} \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{c^2 d^4}+\frac {\left (a+b x^2\right )^{3/2} \left (4 \left (a d^2+b c^2\right )-3 b c d x\right )}{12 c^2 d}+\frac {\sqrt {a+b x^2} \left (8 \left (a d^2+b c^2\right )^2-b c d x \left (7 a d^2+4 b c^2\right )\right )}{8 c^2 d^3}-\frac {a d \left (a+b x^2\right )^{3/2}}{3 c^2}+\frac {15 a b x \sqrt {a+b x^2}}{8 c}-\frac {\left (a+b x^2\right )^{5/2}}{c x}+\frac {5 b x \left (a+b x^2\right )^{3/2}}{4 c}\) |
Input:
Int[(a + b*x^2)^(5/2)/(x^2*(c + d*x)),x]
Output:
-((a^2*d*Sqrt[a + b*x^2])/c^2) + (15*a*b*x*Sqrt[a + b*x^2])/(8*c) + ((8*(b *c^2 + a*d^2)^2 - b*c*d*(4*b*c^2 + 7*a*d^2)*x)*Sqrt[a + b*x^2])/(8*c^2*d^3 ) - (a*d*(a + b*x^2)^(3/2))/(3*c^2) + (5*b*x*(a + b*x^2)^(3/2))/(4*c) + (( 4*(b*c^2 + a*d^2) - 3*b*c*d*x)*(a + b*x^2)^(3/2))/(12*c^2*d) - (a + b*x^2) ^(5/2)/(c*x) + (15*a^2*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*c) - (Sqrt[b]*(8*b^2*c^4 + 20*a*b*c^2*d^2 + 15*a^2*d^4)*ArcTanh[(Sqrt[b]*x)/ Sqrt[a + b*x^2]])/(8*c*d^4) - ((b*c^2 + a*d^2)^(5/2)*ArcTanh[(a*d - b*c*x) /(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(c^2*d^4) + (a^(5/2)*d*ArcTanh[Sq rt[a + b*x^2]/Sqrt[a]])/c^2
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Int[ExpandIntegrand[(a + b*x^2)^p, x^m*(c + d*x)^n, x], x] /; FreeQ[{ a, b, c, d, p}, x] && ILtQ[n, 0] && IntegerQ[m] && IntegerQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(393\) vs. \(2(194)=388\).
Time = 0.40 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.74
method | result | size |
risch | \(-\frac {a^{2} \sqrt {b \,x^{2}+a}}{c x}+\frac {\frac {b^{2} c \left (\frac {d \left (3 a \,d^{2}+b \,c^{2}\right ) \sqrt {b \,x^{2}+a}}{b}+b \,d^{3} \left (\frac {x^{2} \sqrt {b \,x^{2}+a}}{3 b}-\frac {2 a \sqrt {b \,x^{2}+a}}{3 b^{2}}\right )-\sqrt {b}\, c^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )-\frac {3 a \,d^{2} c \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}-b c \,d^{2} \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\right )}{d^{4}}-\frac {\left (a^{3} d^{6}+3 a^{2} b \,c^{2} d^{4}+3 a \,b^{2} c^{4} d^{2}+b^{3} c^{6}\right ) \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{5} c \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}+\frac {d \,a^{\frac {5}{2}} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{c}}{c}\) | \(394\) |
default | \(\text {Expression too large to display}\) | \(1001\) |
Input:
int((b*x^2+a)^(5/2)/x^2/(d*x+c),x,method=_RETURNVERBOSE)
Output:
-a^2*(b*x^2+a)^(1/2)/c/x+1/c*(b^2*c/d^4*(d*(3*a*d^2+b*c^2)/b*(b*x^2+a)^(1/ 2)+b*d^3*(1/3*x^2/b*(b*x^2+a)^(1/2)-2/3*a/b^2*(b*x^2+a)^(1/2))-b^(1/2)*c^3 *ln(b^(1/2)*x+(b*x^2+a)^(1/2))-3*a*d^2*c*ln(b^(1/2)*x+(b*x^2+a)^(1/2))/b^( 1/2)-b*c*d^2*(1/2*x/b*(b*x^2+a)^(1/2)-1/2*a/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a) ^(1/2))))-1/d^5*(a^3*d^6+3*a^2*b*c^2*d^4+3*a*b^2*c^4*d^2+b^3*c^6)/c/((a*d^ 2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^ 2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/ d))+d*a^(5/2)/c*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x))
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\text {Timed out} \] Input:
integrate((b*x^2+a)^(5/2)/x^2/(d*x+c),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {5}{2}}}{x^{2} \left (c + d x\right )}\, dx \] Input:
integrate((b*x**2+a)**(5/2)/x**2/(d*x+c),x)
Output:
Integral((a + b*x**2)**(5/2)/(x**2*(c + d*x)), x)
\[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{{\left (d x + c\right )} x^{2}} \,d x } \] Input:
integrate((b*x^2+a)^(5/2)/x^2/(d*x+c),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(5/2)/((d*x + c)*x^2), x)
Exception generated. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((b*x^2+a)^(5/2)/x^2/(d*x+c),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{5/2}}{x^2\,\left (c+d\,x\right )} \,d x \] Input:
int((a + b*x^2)^(5/2)/(x^2*(c + d*x)),x)
Output:
int((a + b*x^2)^(5/2)/(x^2*(c + d*x)), x)
\[ \int \frac {\left (a+b x^2\right )^{5/2}}{x^2 (c+d x)} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{x^{2} \left (d x +c \right )}d x \] Input:
int((b*x^2+a)^(5/2)/x^2/(d*x+c),x)
Output:
int((b*x^2+a)^(5/2)/x^2/(d*x+c),x)