Integrand size = 26, antiderivative size = 388 \[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=\frac {\left (\frac {b+a x}{d+c x}\right )^{3/4} \left (-32 b^2 d^3-36 b^2 c d^2 x+36 a b d^3 x-48 b d^4 x+3 b^2 c^2 d x^2+42 a b c d^2 x^2-45 a^2 d^3 x^2-60 b c d^3 x^2+60 a d^4 x^2+7 b^2 c^3 x^3+6 a b c^2 d x^3-45 a^2 c d^2 x^3-12 b c^2 d^2 x^3+60 a c d^3 x^3\right )}{96 b^2 d^2 x^3}+\frac {\left (7 b^3 c^3+3 a b^2 c^2 d+5 a^2 b c d^2-12 b^2 c^2 d^2-15 a^3 d^3-8 a b c d^3+20 a^2 d^4\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {b+a x}{d+c x}}}{\sqrt [4]{b}}\right )}{64 b^{9/4} d^{11/4}}+\frac {\left (-7 b^3 c^3-3 a b^2 c^2 d-5 a^2 b c d^2+12 b^2 c^2 d^2+15 a^3 d^3+8 a b c d^3-20 a^2 d^4\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {b+a x}{d+c x}}}{\sqrt [4]{b}}\right )}{64 b^{9/4} d^{11/4}} \]
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Time = 0.38 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1984, 591, 468, 296, 304, 211, 214} \[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=\frac {\left (2 b c d (5 a-6 d)+5 a d^2 (3 a-4 d)+7 b^2 c^2\right ) (b c-a d) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {a x+b}{c x+d}}}{\sqrt [4]{b}}\right )}{64 b^{9/4} d^{11/4}}-\frac {\left (2 b c d (5 a-6 d)+5 a d^2 (3 a-4 d)+7 b^2 c^2\right ) (b c-a d) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {a x+b}{c x+d}}}{\sqrt [4]{b}}\right )}{64 b^{9/4} d^{11/4}}-\frac {\left (2 b c d (5 a-6 d)+5 a d^2 (3 a-4 d)+7 b^2 c^2\right ) (b c-a d) \left (\frac {a x+b}{c x+d}\right )^{3/4}}{32 b^2 d^2 \left (b-\frac {d (a x+b)}{c x+d}\right )}+\frac {(d (9 a-4 d)+7 b c) (b c-a d)^2 \left (\frac {a x+b}{c x+d}\right )^{3/4}}{24 b d^2 \left (b-\frac {d (a x+b)}{c x+d}\right )^2}-\frac {(b+d x) (b c-a d)^3 \left (\frac {a x+b}{c x+d}\right )^{3/4}}{3 b d (c x+d) \left (b-\frac {d (a x+b)}{c x+d}\right )^3} \]
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Rule 211
Rule 214
Rule 296
Rule 304
Rule 468
Rule 591
Rule 1984
Rubi steps \begin{align*} \text {integral}& = -\left ((4 (b c-a d)) \text {Subst}\left (\int \frac {x^2 \left (a-c x^4\right ) \left (b (a-d)+\left (-b c+d^2\right ) x^4\right )}{\left (b-d x^4\right )^4} \, dx,x,\sqrt [4]{\frac {b+a x}{d+c x}}\right )\right ) \\ & = -\frac {(b c-a d)^3 \left (\frac {b+a x}{d+c x}\right )^{3/4} (b+d x)}{3 b d (d+c x) \left (b-\frac {d (b+a x)}{d+c x}\right )^3}-\frac {(b c-a d) \text {Subst}\left (\int \frac {x^2 \left (3 b (a-d) (b c+3 a d)-(7 b c+5 a d) \left (b c-d^2\right ) x^4\right )}{\left (b-d x^4\right )^3} \, dx,x,\sqrt [4]{\frac {b+a x}{d+c x}}\right )}{3 b d} \\ & = -\frac {(b c-a d)^3 \left (\frac {b+a x}{d+c x}\right )^{3/4} (b+d x)}{3 b d (d+c x) \left (b-\frac {d (b+a x)}{d+c x}\right )^3}+\frac {(b c-a d)^2 (7 b c+(9 a-4 d) d) \left (\frac {b+a x}{d+c x}\right )^{3/4}}{24 b d^2 \left (b-\frac {d (b+a x)}{d+c x}\right )^2}-\frac {\left ((b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right )\right ) \text {Subst}\left (\int \frac {x^2}{\left (b-d x^4\right )^2} \, dx,x,\sqrt [4]{\frac {b+a x}{d+c x}}\right )}{8 b d^2} \\ & = -\frac {(b c-a d)^3 \left (\frac {b+a x}{d+c x}\right )^{3/4} (b+d x)}{3 b d (d+c x) \left (b-\frac {d (b+a x)}{d+c x}\right )^3}+\frac {(b c-a d)^2 (7 b c+(9 a-4 d) d) \left (\frac {b+a x}{d+c x}\right )^{3/4}}{24 b d^2 \left (b-\frac {d (b+a x)}{d+c x}\right )^2}-\frac {(b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right ) \left (\frac {b+a x}{d+c x}\right )^{3/4}}{32 b^2 d^2 \left (b-\frac {d (b+a x)}{d+c x}\right )}-\frac {\left ((b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right )\right ) \text {Subst}\left (\int \frac {x^2}{b-d x^4} \, dx,x,\sqrt [4]{\frac {b+a x}{d+c x}}\right )}{32 b^2 d^2} \\ & = -\frac {(b c-a d)^3 \left (\frac {b+a x}{d+c x}\right )^{3/4} (b+d x)}{3 b d (d+c x) \left (b-\frac {d (b+a x)}{d+c x}\right )^3}+\frac {(b c-a d)^2 (7 b c+(9 a-4 d) d) \left (\frac {b+a x}{d+c x}\right )^{3/4}}{24 b d^2 \left (b-\frac {d (b+a x)}{d+c x}\right )^2}-\frac {(b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right ) \left (\frac {b+a x}{d+c x}\right )^{3/4}}{32 b^2 d^2 \left (b-\frac {d (b+a x)}{d+c x}\right )}-\frac {\left ((b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {d} x^2} \, dx,x,\sqrt [4]{\frac {b+a x}{d+c x}}\right )}{64 b^2 d^{5/2}}+\frac {\left ((b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {d} x^2} \, dx,x,\sqrt [4]{\frac {b+a x}{d+c x}}\right )}{64 b^2 d^{5/2}} \\ & = -\frac {(b c-a d)^3 \left (\frac {b+a x}{d+c x}\right )^{3/4} (b+d x)}{3 b d (d+c x) \left (b-\frac {d (b+a x)}{d+c x}\right )^3}+\frac {(b c-a d)^2 (7 b c+(9 a-4 d) d) \left (\frac {b+a x}{d+c x}\right )^{3/4}}{24 b d^2 \left (b-\frac {d (b+a x)}{d+c x}\right )^2}-\frac {(b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right ) \left (\frac {b+a x}{d+c x}\right )^{3/4}}{32 b^2 d^2 \left (b-\frac {d (b+a x)}{d+c x}\right )}+\frac {(b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right ) \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {b+a x}{d+c x}}}{\sqrt [4]{b}}\right )}{64 b^{9/4} d^{11/4}}-\frac {(b c-a d) \left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right ) \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {b+a x}{d+c x}}}{\sqrt [4]{b}}\right )}{64 b^{9/4} d^{11/4}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.40 \[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=\frac {\left (\frac {b+a x}{d+c x}\right )^{3/4} \left (-32 b^3 d (d+c x)^2+x \left (4 b^2 (7 b c+3 (3 a-4 d) d) (d+c x)^2-\left (7 b^2 c^2+2 b c (5 a-6 d) d+5 a (3 a-4 d) d^2\right ) x \left (3 b (d+c x)+(b c-a d) x \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},\frac {d (b+a x)}{b (d+c x)}\right )\right )\right )\right )}{96 b^3 d^2 x^3} \]
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\[\int \frac {d x +b}{x^{4} \left (\frac {a x +b}{c x +d}\right )^{\frac {1}{4}}}d x\]
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Result contains complex when optimal does not.
Time = 0.39 (sec) , antiderivative size = 6825, normalized size of antiderivative = 17.59 \[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=\text {Too large to display} \]
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\[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=\int \frac {b + d x}{x^{4} \sqrt [4]{\frac {a x + b}{c x + d}}}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 535, normalized size of antiderivative = 1.38 \[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=-\frac {3 \, {\left (7 \, b^{3} c^{3} d^{2} + 3 \, a b^{2} c^{2} d^{3} + 20 \, a^{2} d^{6} - {\left (15 \, a^{3} + 8 \, a b c\right )} d^{5} + {\left (5 \, a^{2} b c - 12 \, b^{2} c^{2}\right )} d^{4}\right )} \left (\frac {a x + b}{c x + d}\right )^{\frac {11}{4}} + 6 \, {\left (3 \, b^{4} c^{3} d - 17 \, a b^{3} c^{2} d^{2} - 28 \, a^{2} b d^{5} + 3 \, {\left (7 \, a^{3} b + 8 \, a b^{2} c\right )} d^{4} - {\left (7 \, a^{2} b^{2} c - 4 \, b^{3} c^{2}\right )} d^{3}\right )} \left (\frac {a x + b}{c x + d}\right )^{\frac {7}{4}} - {\left (7 \, b^{5} c^{3} + 3 \, a b^{4} c^{2} d - 108 \, a^{2} b^{2} d^{4} + {\left (113 \, a^{3} b^{2} + 120 \, a b^{3} c\right )} d^{3} - 3 \, {\left (41 \, a^{2} b^{3} c + 4 \, b^{4} c^{2}\right )} d^{2}\right )} \left (\frac {a x + b}{c x + d}\right )^{\frac {3}{4}}}{96 \, {\left (b^{5} d^{2} - \frac {3 \, {\left (a x + b\right )} b^{4} d^{3}}{c x + d} + \frac {3 \, {\left (a x + b\right )}^{2} b^{3} d^{4}}{{\left (c x + d\right )}^{2}} - \frac {{\left (a x + b\right )}^{3} b^{2} d^{5}}{{\left (c x + d\right )}^{3}}\right )}} + \frac {{\left (7 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 20 \, a^{2} d^{4} - {\left (15 \, a^{3} + 8 \, a b c\right )} d^{3} + {\left (5 \, a^{2} b c - 12 \, b^{2} c^{2}\right )} d^{2}\right )} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {d} \left (\frac {a x + b}{c x + d}\right )^{\frac {1}{4}}}{\sqrt {\sqrt {b} \sqrt {d}}}\right )}{\sqrt {\sqrt {b} \sqrt {d}} \sqrt {d}} + \frac {\log \left (\frac {\sqrt {d} \left (\frac {a x + b}{c x + d}\right )^{\frac {1}{4}} - \sqrt {\sqrt {b} \sqrt {d}}}{\sqrt {d} \left (\frac {a x + b}{c x + d}\right )^{\frac {1}{4}} + \sqrt {\sqrt {b} \sqrt {d}}}\right )}{\sqrt {\sqrt {b} \sqrt {d}} \sqrt {d}}\right )}}{128 \, b^{2} d^{2}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 1632 vs. \(2 (360) = 720\).
Time = 1.35 (sec) , antiderivative size = 1632, normalized size of antiderivative = 4.21 \[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=\text {Too large to display} \]
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Time = 8.80 (sec) , antiderivative size = 532, normalized size of antiderivative = 1.37 \[ \int \frac {b+d x}{x^4 \sqrt [4]{\frac {b+a x}{d+c x}}} \, dx=\frac {\mathrm {atanh}\left (\frac {d^{1/4}\,{\left (\frac {b+a\,x}{d+c\,x}\right )}^{1/4}}{b^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (15\,a^2\,d^2+10\,a\,b\,c\,d-20\,a\,d^3+7\,b^2\,c^2-12\,b\,c\,d^2\right )}{64\,b^{9/4}\,d^{11/4}}-\frac {\mathrm {atan}\left (\frac {d^{1/4}\,{\left (\frac {b+a\,x}{d+c\,x}\right )}^{1/4}}{b^{1/4}}\right )\,\left (a\,d-b\,c\right )\,\left (15\,a^2\,d^2+10\,a\,b\,c\,d-20\,a\,d^3+7\,b^2\,c^2-12\,b\,c\,d^2\right )}{64\,b^{9/4}\,d^{11/4}}-\frac {\frac {c^2\,{\left (\frac {b+a\,x}{d+c\,x}\right )}^{7/4}\,\left (\frac {21\,a^3\,c\,d^3}{16}-\frac {7\,a^2\,b\,c^2\,d^2}{16}-\frac {7\,a^2\,c\,d^4}{4}-\frac {17\,a\,b^2\,c^3\,d}{16}+\frac {3\,a\,b\,c^2\,d^3}{2}+\frac {3\,b^3\,c^4}{16}+\frac {b^2\,c^3\,d^2}{4}\right )}{a^3\,b\,d^4}-\frac {c\,{\left (\frac {b+a\,x}{d+c\,x}\right )}^{3/4}\,\left (\frac {113\,a^3\,c^2\,d^3}{96}-\frac {41\,a^2\,b\,c^3\,d^2}{32}-\frac {9\,a^2\,c^2\,d^4}{8}+\frac {a\,b^2\,c^4\,d}{32}+\frac {5\,a\,b\,c^3\,d^3}{4}+\frac {7\,b^3\,c^5}{96}-\frac {b^2\,c^4\,d^2}{8}\right )}{a^3\,d^5}+\frac {c^3\,{\left (\frac {b+a\,x}{d+c\,x}\right )}^{11/4}\,\left (-\frac {15\,a^3\,d^3}{32}+\frac {5\,a^2\,b\,c\,d^2}{32}+\frac {5\,a^2\,d^4}{8}+\frac {3\,a\,b^2\,c^2\,d}{32}-\frac {a\,b\,c\,d^3}{4}+\frac {7\,b^3\,c^3}{32}-\frac {3\,b^2\,c^2\,d^2}{8}\right )}{a^3\,b^2\,d^3}}{\frac {b^3\,c^3}{a^3\,d^3}-\frac {c^3\,{\left (b+a\,x\right )}^3}{a^3\,{\left (d+c\,x\right )}^3}+\frac {3\,b\,c^3\,{\left (b+a\,x\right )}^2}{a^3\,d\,{\left (d+c\,x\right )}^2}-\frac {3\,b^2\,c^3\,\left (b+a\,x\right )}{a^3\,d^2\,\left (d+c\,x\right )}} \]
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