Integrand size = 22, antiderivative size = 349 \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt [4]{-a} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c-\sqrt [4]{-a} d\right ) (2+n)}-\frac {(c+d x)^{2+n} \operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{4 b^{3/4} \left (\sqrt [4]{b} c+\sqrt [4]{-a} d\right ) (2+n)} \]
-1/4*(d*x+c)^(2+n)*hypergeom([1, 2+n],[3+n],b^(1/4)*(d*x+c)/(b^(1/4)*c-(-a )^(1/4)*d))/b^(3/4)/(b^(1/4)*c-(-a)^(1/4)*d)/(2+n)-1/4*(d*x+c)^(2+n)*hyper geom([1, 2+n],[3+n],b^(1/4)*(d*x+c)/(b^(1/4)*c+(-a)^(1/4)*d))/b^(3/4)/(b^( 1/4)*c+(-a)^(1/4)*d)/(2+n)-1/4*(d*x+c)^(2+n)*hypergeom([1, 2+n],[3+n],b^(1 /4)*(d*x+c)/(b^(1/4)*c-d*(-(-a)^(1/2))^(1/2)))/b^(3/4)/(2+n)/(b^(1/4)*c-d* (-(-a)^(1/2))^(1/2))-1/4*(d*x+c)^(2+n)*hypergeom([1, 2+n],[3+n],b^(1/4)*(d *x+c)/(b^(1/4)*c+d*(-(-a)^(1/2))^(1/2)))/b^(3/4)/(2+n)/(b^(1/4)*c+d*(-(-a) ^(1/2))^(1/2))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\frac {(c+d x)^{2+n} \left (-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c-\sqrt [4]{-a} d}-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-i \sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c-i \sqrt [4]{-a} d}-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+i \sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c+i \sqrt [4]{-a} d}-\frac {\operatorname {Hypergeometric2F1}\left (1,2+n,3+n,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{4 b^{3/4} (2+n)} \]
((c + d*x)^(2 + n)*(-(Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x ))/(b^(1/4)*c - (-a)^(1/4)*d)]/(b^(1/4)*c - (-a)^(1/4)*d)) - Hypergeometri c2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c - I*(-a)^(1/4)*d)]/(b ^(1/4)*c - I*(-a)^(1/4)*d) - Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*( c + d*x))/(b^(1/4)*c + I*(-a)^(1/4)*d)]/(b^(1/4)*c + I*(-a)^(1/4)*d) - Hyp ergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c + (-a)^(1/4 )*d)]/(b^(1/4)*c + (-a)^(1/4)*d)))/(4*b^(3/4)*(2 + n))
Time = 0.86 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {7276, 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.
\(\displaystyle \int \frac {x^3 (c+d x)^{n+1}}{a+b x^4} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {x (c+d x)^{n+1}}{2 \left (b x^2-\sqrt {-a} \sqrt {b}\right )}+\frac {x (c+d x)^{n+1}}{2 \left (\sqrt {-a} \sqrt {b}+b x^2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt [4]{b} c-\sqrt {-\sqrt {-a}} d\right )}-\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt {-\sqrt {-a}} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt {-\sqrt {-a}} d+\sqrt [4]{b} c\right )}-\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c-\sqrt [4]{-a} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt [4]{b} c-\sqrt [4]{-a} d\right )}-\frac {(c+d x)^{n+2} \operatorname {Hypergeometric2F1}\left (1,n+2,n+3,\frac {\sqrt [4]{b} (c+d x)}{\sqrt [4]{b} c+\sqrt [4]{-a} d}\right )}{4 b^{3/4} (n+2) \left (\sqrt [4]{-a} d+\sqrt [4]{b} c\right )}\) |
-1/4*((c + d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d *x))/(b^(1/4)*c - Sqrt[-Sqrt[-a]]*d)])/(b^(3/4)*(b^(1/4)*c - Sqrt[-Sqrt[-a ]]*d)*(2 + n)) - ((c + d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (b^ (1/4)*(c + d*x))/(b^(1/4)*c + Sqrt[-Sqrt[-a]]*d)])/(4*b^(3/4)*(b^(1/4)*c + Sqrt[-Sqrt[-a]]*d)*(2 + n)) - ((c + d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c - (-a)^(1/4)*d)])/(4*b^(3/4)*(b^ (1/4)*c - (-a)^(1/4)*d)*(2 + n)) - ((c + d*x)^(2 + n)*Hypergeometric2F1[1, 2 + n, 3 + n, (b^(1/4)*(c + d*x))/(b^(1/4)*c + (-a)^(1/4)*d)])/(4*b^(3/4) *(b^(1/4)*c + (-a)^(1/4)*d)*(2 + n))
3.3.25.3.1 Defintions of rubi rules used
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
\[\int \frac {x^{3} \left (d x +c \right )^{1+n}}{b \,x^{4}+a}d x\]
\[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int { \frac {{\left (d x + c\right )}^{n + 1} x^{3}}{b x^{4} + a} \,d x } \]
Timed out. \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int { \frac {{\left (d x + c\right )}^{n + 1} x^{3}}{b x^{4} + a} \,d x } \]
\[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int { \frac {{\left (d x + c\right )}^{n + 1} x^{3}}{b x^{4} + a} \,d x } \]
Timed out. \[ \int \frac {x^3 (c+d x)^{1+n}}{a+b x^4} \, dx=\int \frac {x^3\,{\left (c+d\,x\right )}^{n+1}}{b\,x^4+a} \,d x \]