Integrand size = 19, antiderivative size = 203 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx=-\frac {5 a^2 c^2 (a e-c d x) \sqrt {a+c x^2}}{16 \left (c d^2+a e^2\right )^3 (d+e x)^2}-\frac {5 a c (a e-c d x) \left (a+c x^2\right )^{3/2}}{24 \left (c d^2+a e^2\right )^2 (d+e x)^4}-\frac {(a e-c d x) \left (a+c x^2\right )^{5/2}}{6 \left (c d^2+a e^2\right ) (d+e x)^6}-\frac {5 a^3 c^3 \text {arctanh}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{16 \left (c d^2+a e^2\right )^{7/2}} \]
-5/24*a*c*(-c*d*x+a*e)*(c*x^2+a)^(3/2)/(a*e^2+c*d^2)^2/(e*x+d)^4-1/6*(-c*d *x+a*e)*(c*x^2+a)^(5/2)/(a*e^2+c*d^2)/(e*x+d)^6-5/16*a^3*c^3*arctanh((-c*d *x+a*e)/(a*e^2+c*d^2)^(1/2)/(c*x^2+a)^(1/2))/(a*e^2+c*d^2)^(7/2)-5/16*a^2* c^2*(-c*d*x+a*e)*(c*x^2+a)^(1/2)/(a*e^2+c*d^2)^3/(e*x+d)^2
Time = 10.50 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx=\frac {1}{48} \left (\frac {\sqrt {a+c x^2} \left (-8 a^5 e^5+8 c^5 d^5 x^5-2 a^4 c e^3 \left (13 d^2+6 d e x+13 e^2 x^2\right )+2 a c^4 d^3 x^3 \left (13 d^2+6 d e x+13 e^2 x^2\right )-a^3 c^2 e \left (33 d^4+54 d^3 e x+122 d^2 e^2 x^2+54 d e^3 x^3+33 e^4 x^4\right )+a^2 c^3 d x \left (33 d^4+54 d^3 e x+122 d^2 e^2 x^2+54 d e^3 x^3+33 e^4 x^4\right )\right )}{\left (c d^2+a e^2\right )^3 (d+e x)^6}+\frac {15 a^3 c^3 \log (d+e x)}{\left (c d^2+a e^2\right )^{7/2}}-\frac {15 a^3 c^3 \log \left (a e-c d x+\sqrt {c d^2+a e^2} \sqrt {a+c x^2}\right )}{\left (c d^2+a e^2\right )^{7/2}}\right ) \]
((Sqrt[a + c*x^2]*(-8*a^5*e^5 + 8*c^5*d^5*x^5 - 2*a^4*c*e^3*(13*d^2 + 6*d* e*x + 13*e^2*x^2) + 2*a*c^4*d^3*x^3*(13*d^2 + 6*d*e*x + 13*e^2*x^2) - a^3* c^2*e*(33*d^4 + 54*d^3*e*x + 122*d^2*e^2*x^2 + 54*d*e^3*x^3 + 33*e^4*x^4) + a^2*c^3*d*x*(33*d^4 + 54*d^3*e*x + 122*d^2*e^2*x^2 + 54*d*e^3*x^3 + 33*e ^4*x^4)))/((c*d^2 + a*e^2)^3*(d + e*x)^6) + (15*a^3*c^3*Log[d + e*x])/(c*d ^2 + a*e^2)^(7/2) - (15*a^3*c^3*Log[a*e - c*d*x + Sqrt[c*d^2 + a*e^2]*Sqrt [a + c*x^2]])/(c*d^2 + a*e^2)^(7/2))/48
Time = 0.32 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {486, 486, 486, 488, 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.
\(\displaystyle \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx\) |
\(\Big \downarrow \) 486 |
\(\displaystyle \frac {5 a c \int \frac {\left (c x^2+a\right )^{3/2}}{(d+e x)^5}dx}{6 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{5/2} (a e-c d x)}{6 (d+e x)^6 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 486 |
\(\displaystyle \frac {5 a c \left (\frac {3 a c \int \frac {\sqrt {c x^2+a}}{(d+e x)^3}dx}{4 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{3/2} (a e-c d x)}{4 (d+e x)^4 \left (a e^2+c d^2\right )}\right )}{6 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{5/2} (a e-c d x)}{6 (d+e x)^6 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 486 |
\(\displaystyle \frac {5 a c \left (\frac {3 a c \left (\frac {a c \int \frac {1}{(d+e x) \sqrt {c x^2+a}}dx}{2 \left (a e^2+c d^2\right )}-\frac {\sqrt {a+c x^2} (a e-c d x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\right )}{4 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{3/2} (a e-c d x)}{4 (d+e x)^4 \left (a e^2+c d^2\right )}\right )}{6 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{5/2} (a e-c d x)}{6 (d+e x)^6 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {5 a c \left (\frac {3 a c \left (-\frac {a c \int \frac {1}{c d^2+a e^2-\frac {(a e-c d x)^2}{c x^2+a}}d\frac {a e-c d x}{\sqrt {c x^2+a}}}{2 \left (a e^2+c d^2\right )}-\frac {\sqrt {a+c x^2} (a e-c d x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\right )}{4 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{3/2} (a e-c d x)}{4 (d+e x)^4 \left (a e^2+c d^2\right )}\right )}{6 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{5/2} (a e-c d x)}{6 (d+e x)^6 \left (a e^2+c d^2\right )}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 a c \left (\frac {3 a c \left (-\frac {a c \text {arctanh}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{2 \left (a e^2+c d^2\right )^{3/2}}-\frac {\sqrt {a+c x^2} (a e-c d x)}{2 (d+e x)^2 \left (a e^2+c d^2\right )}\right )}{4 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{3/2} (a e-c d x)}{4 (d+e x)^4 \left (a e^2+c d^2\right )}\right )}{6 \left (a e^2+c d^2\right )}-\frac {\left (a+c x^2\right )^{5/2} (a e-c d x)}{6 (d+e x)^6 \left (a e^2+c d^2\right )}\) |
-1/6*((a*e - c*d*x)*(a + c*x^2)^(5/2))/((c*d^2 + a*e^2)*(d + e*x)^6) + (5* a*c*(-1/4*((a*e - c*d*x)*(a + c*x^2)^(3/2))/((c*d^2 + a*e^2)*(d + e*x)^4) + (3*a*c*(-1/2*((a*e - c*d*x)*Sqrt[a + c*x^2])/((c*d^2 + a*e^2)*(d + e*x)^ 2) - (a*c*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(2 *(c*d^2 + a*e^2)^(3/2))))/(4*(c*d^2 + a*e^2))))/(6*(c*d^2 + a*e^2))
3.6.55.3.1 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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n + 1)*(a*d - b*c*x)*((a + b*x^2)^p/((n + 1)*(b*c^2 + a*d^2))), x] - Simp[2*a*b*(p/((n + 1)*(b*c^2 + a*d^2))) Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[n + 2*p + 2, 0] && GtQ[p, 0]
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(15867\) vs. \(2(183)=366\).
Time = 2.79 (sec) , antiderivative size = 15868, normalized size of antiderivative = 78.17
Leaf count of result is larger than twice the leaf count of optimal. 951 vs. \(2 (184) = 368\).
Time = 8.67 (sec) , antiderivative size = 1929, normalized size of antiderivative = 9.50 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx=\text {Too large to display} \]
[1/96*(15*(a^3*c^3*e^6*x^6 + 6*a^3*c^3*d*e^5*x^5 + 15*a^3*c^3*d^2*e^4*x^4 + 20*a^3*c^3*d^3*e^3*x^3 + 15*a^3*c^3*d^4*e^2*x^2 + 6*a^3*c^3*d^5*e*x + a^ 3*c^3*d^6)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2 *c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*(33*a^3*c^3*d^6*e + 59*a^4*c^2*d^4*e^3 + 34*a^5*c*d^2*e^5 + 8*a^6*e^7 - (8*c^6*d^7 + 34*a*c^5*d^5*e^2 + 59*a^2*c ^4*d^3*e^4 + 33*a^3*c^3*d*e^6)*x^5 - 3*(4*a*c^5*d^6*e + 22*a^2*c^4*d^4*e^3 + 7*a^3*c^3*d^2*e^5 - 11*a^4*c^2*e^7)*x^4 - 2*(13*a*c^5*d^7 + 74*a^2*c^4* d^5*e^2 + 34*a^3*c^3*d^3*e^4 - 27*a^4*c^2*d*e^6)*x^3 - 2*(27*a^2*c^4*d^6*e - 34*a^3*c^3*d^4*e^3 - 74*a^4*c^2*d^2*e^5 - 13*a^5*c*e^7)*x^2 - 3*(11*a^2 *c^4*d^7 - 7*a^3*c^3*d^5*e^2 - 22*a^4*c^2*d^3*e^4 - 4*a^5*c*d*e^6)*x)*sqrt (c*x^2 + a))/(c^4*d^14 + 4*a*c^3*d^12*e^2 + 6*a^2*c^2*d^10*e^4 + 4*a^3*c*d ^8*e^6 + a^4*d^6*e^8 + (c^4*d^8*e^6 + 4*a*c^3*d^6*e^8 + 6*a^2*c^2*d^4*e^10 + 4*a^3*c*d^2*e^12 + a^4*e^14)*x^6 + 6*(c^4*d^9*e^5 + 4*a*c^3*d^7*e^7 + 6 *a^2*c^2*d^5*e^9 + 4*a^3*c*d^3*e^11 + a^4*d*e^13)*x^5 + 15*(c^4*d^10*e^4 + 4*a*c^3*d^8*e^6 + 6*a^2*c^2*d^6*e^8 + 4*a^3*c*d^4*e^10 + a^4*d^2*e^12)*x^ 4 + 20*(c^4*d^11*e^3 + 4*a*c^3*d^9*e^5 + 6*a^2*c^2*d^7*e^7 + 4*a^3*c*d^5*e ^9 + a^4*d^3*e^11)*x^3 + 15*(c^4*d^12*e^2 + 4*a*c^3*d^10*e^4 + 6*a^2*c^2*d ^8*e^6 + 4*a^3*c*d^6*e^8 + a^4*d^4*e^10)*x^2 + 6*(c^4*d^13*e + 4*a*c^3*d^1 1*e^3 + 6*a^2*c^2*d^9*e^5 + 4*a^3*c*d^7*e^7 + a^4*d^5*e^9)*x), -1/48*(1...
\[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx=\int \frac {\left (a + c x^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{7}}\, dx \]
Exception generated. \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Leaf count of result is larger than twice the leaf count of optimal. 1953 vs. \(2 (184) = 368\).
Time = 0.34 (sec) , antiderivative size = 1953, normalized size of antiderivative = 9.62 \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx=\text {Too large to display} \]
5/8*a^3*c^3*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c* d^2 - a*e^2))/((c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6)*sqr t(-c*d^2 - a*e^2)) + 1/24*(48*(sqrt(c)*x - sqrt(c*x^2 + a))^11*c^6*d^6*e^5 + 144*(sqrt(c)*x - sqrt(c*x^2 + a))^11*a*c^5*d^4*e^7 + 144*(sqrt(c)*x - s qrt(c*x^2 + a))^11*a^2*c^4*d^2*e^9 + 33*(sqrt(c)*x - sqrt(c*x^2 + a))^11*a ^3*c^3*e^11 + 240*(sqrt(c)*x - sqrt(c*x^2 + a))^10*c^(13/2)*d^7*e^4 + 720* (sqrt(c)*x - sqrt(c*x^2 + a))^10*a*c^(11/2)*d^5*e^6 + 720*(sqrt(c)*x - sqr t(c*x^2 + a))^10*a^2*c^(9/2)*d^3*e^8 + 75*(sqrt(c)*x - sqrt(c*x^2 + a))^10 *a^3*c^(7/2)*d*e^10 + 640*(sqrt(c)*x - sqrt(c*x^2 + a))^9*c^7*d^8*e^3 + 18 40*(sqrt(c)*x - sqrt(c*x^2 + a))^9*a*c^6*d^6*e^5 + 1680*(sqrt(c)*x - sqrt( c*x^2 + a))^9*a^2*c^5*d^4*e^7 - 340*(sqrt(c)*x - sqrt(c*x^2 + a))^9*a^3*c^ 4*d^2*e^9 + 5*(sqrt(c)*x - sqrt(c*x^2 + a))^9*a^4*c^3*e^11 + 960*(sqrt(c)* x - sqrt(c*x^2 + a))^8*c^(15/2)*d^9*e^2 + 2160*(sqrt(c)*x - sqrt(c*x^2 + a ))^8*a*c^(13/2)*d^7*e^4 + 720*(sqrt(c)*x - sqrt(c*x^2 + a))^8*a^2*c^(11/2) *d^5*e^6 - 2910*(sqrt(c)*x - sqrt(c*x^2 + a))^8*a^3*c^(9/2)*d^3*e^8 + 45*( sqrt(c)*x - sqrt(c*x^2 + a))^8*a^4*c^(7/2)*d*e^10 + 768*(sqrt(c)*x - sqrt( c*x^2 + a))^7*c^8*d^10*e + 576*(sqrt(c)*x - sqrt(c*x^2 + a))^7*a*c^7*d^8*e ^3 - 2592*(sqrt(c)*x - sqrt(c*x^2 + a))^7*a^2*c^6*d^6*e^5 - 5640*(sqrt(c)* x - sqrt(c*x^2 + a))^7*a^3*c^5*d^4*e^7 + 1800*(sqrt(c)*x - sqrt(c*x^2 + a) )^7*a^4*c^4*d^2*e^9 + 90*(sqrt(c)*x - sqrt(c*x^2 + a))^7*a^5*c^3*e^11 +...
Timed out. \[ \int \frac {\left (a+c x^2\right )^{5/2}}{(d+e x)^7} \, dx=\int \frac {{\left (c\,x^2+a\right )}^{5/2}}{{\left (d+e\,x\right )}^7} \,d x \]