Integrand size = 19, antiderivative size = 91 \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=-\frac {(b c-a d) x}{5 c d \left (c+d x^2\right )^{5/2}}+\frac {(b c+4 a d) x}{15 c^2 d \left (c+d x^2\right )^{3/2}}+\frac {2 (b c+4 a d) x}{15 c^3 d \sqrt {c+d x^2}} \]
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Time = 0.02 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {393, 198, 197} \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {2 x (4 a d+b c)}{15 c^3 d \sqrt {c+d x^2}}+\frac {x (4 a d+b c)}{15 c^2 d \left (c+d x^2\right )^{3/2}}-\frac {x (b c-a d)}{5 c d \left (c+d x^2\right )^{5/2}} \]
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Rule 197
Rule 198
Rule 393
Rubi steps \begin{align*} \text {integral}& = -\frac {(b c-a d) x}{5 c d \left (c+d x^2\right )^{5/2}}+\frac {(b c+4 a d) \int \frac {1}{\left (c+d x^2\right )^{5/2}} \, dx}{5 c d} \\ & = -\frac {(b c-a d) x}{5 c d \left (c+d x^2\right )^{5/2}}+\frac {(b c+4 a d) x}{15 c^2 d \left (c+d x^2\right )^{3/2}}+\frac {(2 (b c+4 a d)) \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{15 c^2 d} \\ & = -\frac {(b c-a d) x}{5 c d \left (c+d x^2\right )^{5/2}}+\frac {(b c+4 a d) x}{15 c^2 d \left (c+d x^2\right )^{3/2}}+\frac {2 (b c+4 a d) x}{15 c^3 d \sqrt {c+d x^2}} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.66 \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {15 a c^2 x+5 b c^2 x^3+20 a c d x^3+2 b c d x^5+8 a d^2 x^5}{15 c^3 \left (c+d x^2\right )^{5/2}} \]
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Time = 2.42 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57
method | result | size |
pseudoelliptic | \(\frac {x \left (\left (\frac {b \,x^{2}}{3}+a \right ) c^{2}+\frac {4 x^{2} d \left (\frac {b \,x^{2}}{10}+a \right ) c}{3}+\frac {8 a \,d^{2} x^{4}}{15}\right )}{\left (d \,x^{2}+c \right )^{\frac {5}{2}} c^{3}}\) | \(52\) |
gosper | \(\frac {x \left (8 a \,d^{2} x^{4}+2 b c d \,x^{4}+20 a c d \,x^{2}+5 b \,c^{2} x^{2}+15 c^{2} a \right )}{15 \left (d \,x^{2}+c \right )^{\frac {5}{2}} c^{3}}\) | \(57\) |
trager | \(\frac {x \left (8 a \,d^{2} x^{4}+2 b c d \,x^{4}+20 a c d \,x^{2}+5 b \,c^{2} x^{2}+15 c^{2} a \right )}{15 \left (d \,x^{2}+c \right )^{\frac {5}{2}} c^{3}}\) | \(57\) |
default | \(a \left (\frac {x}{5 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}}{c}\right )+b \left (-\frac {x}{4 d \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {c \left (\frac {x}{5 c \left (d \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 c \left (d \,x^{2}+c \right )^{\frac {3}{2}}}+\frac {8 x}{15 c^{2} \sqrt {d \,x^{2}+c}}}{c}\right )}{4 d}\right )\) | \(132\) |
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Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {{\left (2 \, {\left (b c d + 4 \, a d^{2}\right )} x^{5} + 15 \, a c^{2} x + 5 \, {\left (b c^{2} + 4 \, a c d\right )} x^{3}\right )} \sqrt {d x^{2} + c}}{15 \, {\left (c^{3} d^{3} x^{6} + 3 \, c^{4} d^{2} x^{4} + 3 \, c^{5} d x^{2} + c^{6}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 566 vs. \(2 (83) = 166\).
Time = 10.37 (sec) , antiderivative size = 566, normalized size of antiderivative = 6.22 \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=a \left (\frac {15 c^{5} x}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {35 c^{4} d x^{3}}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {28 c^{3} d^{2} x^{5}}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {8 c^{2} d^{3} x^{7}}{15 c^{\frac {17}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {15}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 45 c^{\frac {13}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {11}{2}} d^{3} x^{6} \sqrt {1 + \frac {d x^{2}}{c}}}\right ) + b \left (\frac {5 c x^{3}}{15 c^{\frac {9}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 30 c^{\frac {7}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {5}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {2 d x^{5}}{15 c^{\frac {9}{2}} \sqrt {1 + \frac {d x^{2}}{c}} + 30 c^{\frac {7}{2}} d x^{2} \sqrt {1 + \frac {d x^{2}}{c}} + 15 c^{\frac {5}{2}} d^{2} x^{4} \sqrt {1 + \frac {d x^{2}}{c}}}\right ) \]
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Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.13 \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {8 \, a x}{15 \, \sqrt {d x^{2} + c} c^{3}} + \frac {4 \, a x}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c^{2}} + \frac {a x}{5 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} c} - \frac {b x}{5 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} d} + \frac {2 \, b x}{15 \, \sqrt {d x^{2} + c} c^{2} d} + \frac {b x}{15 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} c d} \]
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Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {2 \, {\left (b c d^{3} + 4 \, a d^{4}\right )} x^{2}}{c^{3} d^{2}} + \frac {5 \, {\left (b c^{2} d^{2} + 4 \, a c d^{3}\right )}}{c^{3} d^{2}}\right )} + \frac {15 \, a}{c}\right )} x}{15 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \]
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Time = 4.90 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int \frac {a+b x^2}{\left (c+d x^2\right )^{7/2}} \, dx=\frac {8\,a\,d\,x\,{\left (d\,x^2+c\right )}^2-3\,b\,c^3\,x+2\,b\,c\,x\,{\left (d\,x^2+c\right )}^2+b\,c^2\,x\,\left (d\,x^2+c\right )+3\,a\,c^2\,d\,x+4\,a\,c\,d\,x\,\left (d\,x^2+c\right )}{15\,c^3\,d\,{\left (d\,x^2+c\right )}^{5/2}} \]
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