Integrand size = 24, antiderivative size = 180 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x^2}}{2 a^2 b \left (a+b x^2\right )}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )}+\frac {c^{3/2} (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3}-\frac {(b c-a d)^{3/2} (4 b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^3 b^{3/2}} \]
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Time = 0.19 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 100, 154, 162, 65, 214} \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {(b c-a d)^{3/2} (a d+4 b c) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^3 b^{3/2}}+\frac {c^{3/2} (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3}-\frac {\sqrt {c+d x^2} (b c-a d) (2 b c-a d)}{2 a^2 b \left (a+b x^2\right )}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )} \]
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Rule 65
Rule 100
Rule 154
Rule 162
Rule 214
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(c+d x)^{5/2}}{x^2 (a+b x)^2} \, dx,x,x^2\right ) \\ & = -\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )}-\frac {\text {Subst}\left (\int \frac {\sqrt {c+d x} \left (\frac {1}{2} c (4 b c-5 a d)+\frac {1}{2} d (b c-2 a d) x\right )}{x (a+b x)^2} \, dx,x,x^2\right )}{2 a} \\ & = -\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x^2}}{2 a^2 b \left (a+b x^2\right )}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{2} b c^2 (4 b c-5 a d)-\frac {1}{2} d \left (2 b^2 c^2-2 a b c d-a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{2 a^2 b} \\ & = -\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x^2}}{2 a^2 b \left (a+b x^2\right )}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )}-\frac {\left (c^2 (4 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^2\right )}{4 a^3}+\frac {\left ((b c-a d)^2 (4 b c+a d)\right ) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^2\right )}{4 a^3 b} \\ & = -\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x^2}}{2 a^2 b \left (a+b x^2\right )}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )}-\frac {\left (c^2 (4 b c-5 a d)\right ) \text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 a^3 d}+\frac {\left ((b c-a d)^2 (4 b c+a d)\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^2}\right )}{2 a^3 b d} \\ & = -\frac {(b c-a d) (2 b c-a d) \sqrt {c+d x^2}}{2 a^2 b \left (a+b x^2\right )}-\frac {c \left (c+d x^2\right )^{3/2}}{2 a x^2 \left (a+b x^2\right )}+\frac {c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3}-\frac {(b c-a d)^{3/2} (4 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {b c-a d}}\right )}{2 a^3 b^{3/2}} \\ \end{align*}
Time = 0.48 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.89 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {-\frac {a \sqrt {c+d x^2} \left (2 b^2 c^2 x^2+a^2 d^2 x^2+a b c \left (c-2 d x^2\right )\right )}{b x^2 \left (a+b x^2\right )}+\frac {(-b c+a d)^{3/2} (4 b c+a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^2}}{\sqrt {-b c+a d}}\right )}{b^{3/2}}+c^{3/2} (4 b c-5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{2 a^3} \]
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Time = 3.23 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.04
method | result | size |
pseudoelliptic | \(\frac {\frac {x^{2} \left (b \,x^{2}+a \right ) \left (a d +4 b c \right ) \left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{2}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2}+2 \sqrt {\left (a d -b c \right ) b}\, \left (\left (b \,x^{2}+a \right ) x^{2} b \left (c^{\frac {5}{2}} b -\frac {5 a d \,c^{\frac {3}{2}}}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )-\frac {\left (2 b^{2} c^{2} x^{2}+a c \left (-2 d \,x^{2}+c \right ) b +a^{2} d^{2} x^{2}\right ) \sqrt {d \,x^{2}+c}\, a}{4}\right )}{x^{2} \sqrt {\left (a d -b c \right ) b}\, \left (b \,x^{2}+a \right ) a^{3} b}\) | \(187\) |
risch | \(\text {Expression too large to display}\) | \(1012\) |
default | \(\text {Expression too large to display}\) | \(5452\) |
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Time = 1.06 (sec) , antiderivative size = 1266, normalized size of antiderivative = 7.03 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \]
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\[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {5}{2}}}{x^{3} \left (a + b x^{2}\right )^{2}}\, dx \]
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\[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{2} x^{3}} \,d x } \]
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Time = 0.30 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.57 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{2 \, a^{3} \sqrt {-c}} + \frac {{\left (4 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{2 \, \sqrt {-b^{2} c + a b d} a^{3} b} - \frac {2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} d - 2 \, \sqrt {d x^{2} + c} b^{2} c^{3} d - 2 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d^{2} + 3 \, \sqrt {d x^{2} + c} a b c^{2} d^{2} + {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3} - \sqrt {d x^{2} + c} a^{2} c d^{3}}{2 \, {\left ({\left (d x^{2} + c\right )}^{2} b - 2 \, {\left (d x^{2} + c\right )} b c + b c^{2} + {\left (d x^{2} + c\right )} a d - a c d\right )} a^{2} b} \]
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Time = 6.75 (sec) , antiderivative size = 1152, normalized size of antiderivative = 6.40 \[ \int \frac {\left (c+d x^2\right )^{5/2}}{x^3 \left (a+b x^2\right )^2} \, dx=\frac {\frac {\sqrt {d\,x^2+c}\,\left (a^2\,c\,d^3-3\,a\,b\,c^2\,d^2+2\,b^2\,c^3\,d\right )}{2\,a^2\,b}-\frac {d\,{\left (d\,x^2+c\right )}^{3/2}\,\left (a^2\,d^2-2\,a\,b\,c\,d+2\,b^2\,c^2\right )}{2\,a^2\,b}}{\left (d\,x^2+c\right )\,\left (a\,d-2\,b\,c\right )+b\,{\left (d\,x^2+c\right )}^2+b\,c^2-a\,c\,d}-\frac {\mathrm {atanh}\left (\frac {5\,d^9\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,\left (\frac {5\,c^2\,d^9}{4}+\frac {4\,b\,c^3\,d^8}{a}-\frac {33\,b^2\,c^4\,d^7}{2\,a^2}+\frac {65\,b^3\,c^5\,d^6}{4\,a^3}-\frac {5\,b^4\,c^6\,d^5}{a^4}\right )}+\frac {4\,c\,d^8\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,c^3\,d^8+\frac {5\,a\,c^2\,d^9}{4\,b}-\frac {33\,b\,c^4\,d^7}{2\,a}+\frac {65\,b^2\,c^5\,d^6}{4\,a^2}-\frac {5\,b^3\,c^6\,d^5}{a^3}}+\frac {65\,b^2\,c^3\,d^6\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,\left (4\,a^2\,c^3\,d^8+\frac {65\,b^2\,c^5\,d^6}{4}-\frac {5\,b^3\,c^6\,d^5}{a}+\frac {5\,a^3\,c^2\,d^9}{4\,b}-\frac {33\,a\,b\,c^4\,d^7}{2}\right )}-\frac {5\,b^3\,c^4\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{4\,a^3\,c^3\,d^8-5\,b^3\,c^6\,d^5+\frac {65\,a\,b^2\,c^5\,d^6}{4}-\frac {33\,a^2\,b\,c^4\,d^7}{2}+\frac {5\,a^4\,c^2\,d^9}{4\,b}}-\frac {33\,b\,c^2\,d^7\,\sqrt {d\,x^2+c}\,\sqrt {c^3}}{2\,\left (4\,a\,c^3\,d^8-\frac {33\,b\,c^4\,d^7}{2}+\frac {65\,b^2\,c^5\,d^6}{4\,a}+\frac {5\,a^2\,c^2\,d^9}{4\,b}-\frac {5\,b^3\,c^6\,d^5}{a^2}\right )}\right )\,\left (5\,a\,d-4\,b\,c\right )\,\sqrt {c^3}}{2\,a^3}-\frac {\mathrm {atanh}\left (\frac {15\,c^3\,d^6\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,\left (\frac {7\,a^3\,c^2\,d^9}{4}+\frac {55\,b^3\,c^5\,d^6}{4}-\frac {41\,a\,b^2\,c^4\,d^7}{4}-\frac {a^2\,b\,c^3\,d^8}{2}+\frac {a^4\,c\,d^{10}}{4\,b}-\frac {5\,b^4\,c^6\,d^5}{a}\right )}+\frac {9\,c^2\,d^7\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,\left (\frac {a^3\,c\,d^{10}}{4}-\frac {41\,b^3\,c^4\,d^7}{4}-\frac {a\,b^2\,c^3\,d^8}{2}+\frac {7\,a^2\,b\,c^2\,d^9}{4}+\frac {55\,b^4\,c^5\,d^6}{4\,a}-\frac {5\,b^5\,c^6\,d^5}{a^2}\right )}+\frac {5\,c^4\,d^5\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{\frac {a^3\,c^3\,d^8}{2}+5\,b^3\,c^6\,d^5-\frac {55\,a\,b^2\,c^5\,d^6}{4}+\frac {41\,a^2\,b\,c^4\,d^7}{4}-\frac {a^5\,c\,d^{10}}{4\,b^2}-\frac {7\,a^4\,c^2\,d^9}{4\,b}}-\frac {c\,d^8\,\sqrt {d\,x^2+c}\,\sqrt {-a^3\,b^3\,d^3+3\,a^2\,b^4\,c\,d^2-3\,a\,b^5\,c^2\,d+b^6\,c^3}}{4\,\left (\frac {b^3\,c^3\,d^8}{2}-\frac {7\,a\,b^2\,c^2\,d^9}{4}+\frac {41\,b^4\,c^4\,d^7}{4\,a}-\frac {55\,b^5\,c^5\,d^6}{4\,a^2}+\frac {5\,b^6\,c^6\,d^5}{a^3}-\frac {a^2\,b\,c\,d^{10}}{4}\right )}\right )\,\sqrt {-b^3\,{\left (a\,d-b\,c\right )}^3}\,\left (a\,d+4\,b\,c\right )}{2\,a^3\,b^3} \]
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