Integrand size = 26, antiderivative size = 137 \[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=-\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {(b c-a d) (3 b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{3/2} d^{5/2}} \]
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Time = 0.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {457, 81, 52, 65, 223, 212} \[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {(b c-a d) (a d+3 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{3/2} d^{5/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} (a d+3 b c)}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d} \]
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Rule 52
Rule 65
Rule 81
Rule 212
Rule 223
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x \sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right ) \\ & = \frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}-\frac {(3 b c+a d) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{\sqrt {c+d x}} \, dx,x,x^2\right )}{8 b d} \\ & = -\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {((b c-a d) (3 b c+a d)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}} \, dx,x,x^2\right )}{16 b d^2} \\ & = -\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {((b c-a d) (3 b c+a d)) \text {Subst}\left (\int \frac {1}{\sqrt {c-\frac {a d}{b}+\frac {d x^2}{b}}} \, dx,x,\sqrt {a+b x^2}\right )}{8 b^2 d^2} \\ & = -\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {((b c-a d) (3 b c+a d)) \text {Subst}\left (\int \frac {1}{1-\frac {d x^2}{b}} \, dx,x,\frac {\sqrt {a+b x^2}}{\sqrt {c+d x^2}}\right )}{8 b^2 d^2} \\ & = -\frac {(3 b c+a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 b d^2}+\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 b d}+\frac {(b c-a d) (3 b c+a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{3/2} d^{5/2}} \\ \end{align*}
Time = 0.86 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.87 \[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (-3 b c+a d+2 b d x^2\right )}{8 b d^2}+\frac {\left (3 b^2 c^2-2 a b c d-a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )}{8 b^{3/2} d^{5/2}} \]
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Time = 3.06 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\frac {\left (2 b d \,x^{2}+a d -3 b c \right ) \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}{8 b \,d^{2}}-\frac {\left (a^{2} d^{2}+2 a b c d -3 b^{2} c^{2}\right ) \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}}{16 b \,d^{2} \sqrt {b d}\, \sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) | \(159\) |
default | \(-\frac {\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}\, \left (-4 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b d \,x^{2}+\ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} d^{2}+2 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a b c d -3 \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{2} c^{2}-2 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, a d +6 \sqrt {b d}\, \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, b c \right )}{16 \sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, d^{2} b \sqrt {b d}}\) | \(290\) |
elliptic | \(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, \left (\frac {x^{2} \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{4 d}+\frac {\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, a}{8 b d}-\frac {3 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, c}{8 d^{2}}-\frac {\ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a^{2}}{16 b \sqrt {b d}}-\frac {\ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) a c}{8 d \sqrt {b d}}+\frac {3 b \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d \,x^{2}}{\sqrt {b d}}+\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\right ) c^{2}}{16 d^{2} \sqrt {b d}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {d \,x^{2}+c}}\) | \(307\) |
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Time = 0.30 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.44 \[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\left [-\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \, {\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {b d}\right ) - 4 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, b^{2} d^{3}}, -\frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-b d}}{2 \, {\left (b^{2} d^{2} x^{4} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (2 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c d + a b d^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, b^{2} d^{3}}\right ] \]
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\[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\int \frac {x^{3} \sqrt {a + b x^{2}}}{\sqrt {c + d x^{2}}}\, dx \]
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Exception generated. \[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.32 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.16 \[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \sqrt {b x^{2} + a} {\left (\frac {2 \, {\left (b x^{2} + a\right )}}{b^{2} d} - \frac {3 \, b^{3} c d + a b^{2} d^{2}}{b^{4} d^{3}}\right )} - \frac {{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {b x^{2} + a} \sqrt {b d} + \sqrt {b^{2} c + {\left (b x^{2} + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} b}{8 \, {\left | b \right |}} \]
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Time = 30.48 (sec) , antiderivative size = 639, normalized size of antiderivative = 4.66 \[ \int \frac {x^3 \sqrt {a+b x^2}}{\sqrt {c+d x^2}} \, dx=\frac {\frac {\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )\,\left (\frac {a^2\,b^2\,d^2}{4}+\frac {a\,b^3\,c\,d}{2}-\frac {3\,b^4\,c^2}{4}\right )}{d^6\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^3\,\left (\frac {7\,a^2\,b\,d^2}{4}+\frac {23\,a\,b^2\,c\,d}{2}+\frac {11\,b^3\,c^2}{4}\right )}{d^5\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^3}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^5\,\left (\frac {7\,a^2\,d^2}{4}+\frac {23\,a\,b\,c\,d}{2}+\frac {11\,b^2\,c^2}{4}\right )}{d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^5}+\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^7\,\left (\frac {a^2\,d^2}{4}+\frac {a\,b\,c\,d}{2}-\frac {3\,b^2\,c^2}{4}\right )}{b\,d^3\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^7}-\frac {4\,a^{3/2}\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^6}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^6}-\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4\,\left (16\,c\,b^2+8\,a\,d\,b\right )}{d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}-\frac {4\,a^{3/2}\,b^2\,\sqrt {c}\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^4\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}}{\frac {{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^8}{{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^8}+\frac {b^4}{d^4}-\frac {4\,b^3\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^2}{d^3\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^2}+\frac {6\,b^2\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^4}{d^2\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^4}-\frac {4\,b\,{\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}^6}{d\,{\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}^6}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {b\,x^2+a}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {d\,x^2+c}-\sqrt {c}\right )}\right )\,\left (a\,d-b\,c\right )\,\left (a\,d+3\,b\,c\right )}{4\,b^{3/2}\,d^{5/2}} \]
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