Integrand size = 22, antiderivative size = 615 \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac {(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left (b \sqrt {b^2-4 a c} \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)-b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )+6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )-8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^{5/2} \left (b-\sqrt {b^2-4 a c}\right ) d (1+m) n^2}-\frac {c \left (b \sqrt {b^2-4 a c} \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)+b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )\right ) (d x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^{5/2} \left (b+\sqrt {b^2-4 a c}\right ) d (1+m) n^2} \]
[Out]
Time = 6.73 (sec) , antiderivative size = 615, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1398, 1572, 1574, 371} \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=-\frac {c (d x)^{m+1} \left (-8 a^2 c^2 \left (m^2+m (2-6 n)+8 n^2-6 n+1\right )+6 a b^2 c \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b (m-n+1) \sqrt {b^2-4 a c} \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )-\left (b^4 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 a^2 d (m+1) n^2 \left (b^2-4 a c\right )^{5/2} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (d x)^{m+1} \left (8 a^2 c^2 \left (m^2+m (2-6 n)+8 n^2-6 n+1\right )-6 a b^2 c \left (m^2+m (2-4 n)+3 n^2-4 n+1\right )+b (m-n+1) \sqrt {b^2-4 a c} \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )+b^4 \left (m^2+m (2-3 n)+2 n^2-3 n+1\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a^2 d (m+1) n^2 \left (b^2-4 a c\right )^{5/2} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {(d x)^{m+1} \left (4 a^2 c^2 (m-4 n+1)-b c x^n \left (2 a c (2 m-7 n+2)-b^2 (m-2 n+1)\right )-5 a b^2 c (m-3 n+1)+b^4 (m-2 n+1)\right )}{2 a^2 d n^2 \left (b^2-4 a c\right )^2 \left (a+b x^n+c x^{2 n}\right )}+\frac {(d x)^{m+1} \left (-2 a c+b^2+b c x^n\right )}{2 a d n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )^2} \]
[In]
[Out]
Rule 371
Rule 1398
Rule 1572
Rule 1574
Rubi steps \begin{align*} \text {integral}& = \frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac {\int \frac {(d x)^m \left (-2 a c (1+m-4 n)+b^2 (1+m-2 n)+b c (1+m-3 n) x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx}{2 a \left (b^2-4 a c\right ) n} \\ & = \frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac {(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}+\frac {\int \frac {(d x)^m \left (\left (2 a c (1+m-4 n)-b^2 (1+m-2 n)\right ) \left (2 a c (1+m-2 n)-b^2 (1+m-n)\right )-a b^2 c (1+m) (1+m-3 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2} \\ & = \frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac {(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}+\frac {\int \left (\frac {\left (-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)+\frac {c \left (b^4-6 a b^2 c+8 a^2 c^2+2 b^4 m-12 a b^2 c m+16 a^2 c^2 m+b^4 m^2-6 a b^2 c m^2+8 a^2 c^2 m^2-3 b^4 n+24 a b^2 c n-48 a^2 c^2 n-3 b^4 m n+24 a b^2 c m n-48 a^2 c^2 m n+2 b^4 n^2-18 a b^2 c n^2+64 a^2 c^2 n^2\right )}{\sqrt {b^2-4 a c}}\right ) (d x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n}+\frac {\left (-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)-\frac {c \left (b^4-6 a b^2 c+8 a^2 c^2+2 b^4 m-12 a b^2 c m+16 a^2 c^2 m+b^4 m^2-6 a b^2 c m^2+8 a^2 c^2 m^2-3 b^4 n+24 a b^2 c n-48 a^2 c^2 n-3 b^4 m n+24 a b^2 c m n-48 a^2 c^2 m n+2 b^4 n^2-18 a b^2 c n^2+64 a^2 c^2 n^2\right )}{\sqrt {b^2-4 a c}}\right ) (d x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n}\right ) \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2} \\ & = \frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac {(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac {\left (c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)-\frac {b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(d x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}-\frac {\left (c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)+\frac {b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(d x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2} \\ & = \frac {(d x)^{1+m} \left (b^2-2 a c+b c x^n\right )}{2 a \left (b^2-4 a c\right ) d n \left (a+b x^n+c x^{2 n}\right )^2}-\frac {(d x)^{1+m} \left (4 a^2 c^2 (1+m-4 n)-5 a b^2 c (1+m-3 n)+b^4 (1+m-2 n)-b c \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 d n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)-\frac {b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b-\sqrt {b^2-4 a c}\right ) d (1+m) n^2}-\frac {c \left (b \left (2 a c (2+2 m-7 n)-b^2 (1+m-2 n)\right ) (1+m-n)+\frac {b^4 \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )-6 a b^2 c \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )+8 a^2 c^2 \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b+\sqrt {b^2-4 a c}\right ) d (1+m) n^2} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(12289\) vs. \(2(615)=1230\).
Time = 7.24 (sec) , antiderivative size = 12289, normalized size of antiderivative = 19.98 \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\text {Result too large to show} \]
[In]
[Out]
\[\int \frac {\left (d x \right )^{m}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{3}}d x\]
[In]
[Out]
\[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{3}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{3}} \,d x } \]
[In]
[Out]
\[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\int { \frac {\left (d x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{3}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(d x)^m}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx=\int \frac {{\left (d\,x\right )}^m}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^3} \,d x \]
[In]
[Out]