Optimal. Leaf size=367 \[ \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (-9 a^{3/2} B e^2+15 \sqrt {a} c d (2 A e+B d)-5 a \sqrt {c} e (A e+2 B d)+15 A c^{3/2} d^2\right )}{30 \sqrt [4]{a} c^{7/4} \sqrt {a+c x^4}}+\frac {x \sqrt {a+c x^4} \left (-3 a B e^2+10 A c d e+5 B c d^2\right )}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right ) \left (-3 a B e^2+10 A c d e+5 B c d^2\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {e x \sqrt {a+c x^4} (A e+2 B d)}{3 c}+\frac {B e^2 x^3 \sqrt {a+c x^4}}{5 c} \]
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Rubi [A] time = 0.40, antiderivative size = 706, normalized size of antiderivative = 1.92, number of steps used = 12, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1721, 220, 305, 1196, 321} \[ -\frac {a^{3/4} e \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (A e+2 B d) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 c^{5/4} \sqrt {a+c x^4}}-\frac {3 a^{5/4} B e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 c^{7/4} \sqrt {a+c x^4}}+\frac {3 a^{5/4} B e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (2 A e+B d) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\sqrt [4]{a} d \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} (2 A e+B d) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {e x \sqrt {a+c x^4} (A e+2 B d)}{3 c}+\frac {d x \sqrt {a+c x^4} (2 A e+B d)}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {A d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}-\frac {3 a B e^2 x \sqrt {a+c x^4}}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {B e^2 x^3 \sqrt {a+c x^4}}{5 c} \]
Antiderivative was successfully verified.
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Rule 220
Rule 305
Rule 321
Rule 1196
Rule 1721
Rubi steps
\begin {align*} \int \frac {\left (A+B x^2\right ) \left (d+e x^2\right )^2}{\sqrt {a+c x^4}} \, dx &=\int \left (\frac {A d^2}{\sqrt {a+c x^4}}+\frac {d (B d+2 A e) x^2}{\sqrt {a+c x^4}}+\frac {e (2 B d+A e) x^4}{\sqrt {a+c x^4}}+\frac {B e^2 x^6}{\sqrt {a+c x^4}}\right ) \, dx\\ &=\left (A d^2\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx+\left (B e^2\right ) \int \frac {x^6}{\sqrt {a+c x^4}} \, dx+(e (2 B d+A e)) \int \frac {x^4}{\sqrt {a+c x^4}} \, dx+(d (B d+2 A e)) \int \frac {x^2}{\sqrt {a+c x^4}} \, dx\\ &=\frac {e (2 B d+A e) x \sqrt {a+c x^4}}{3 c}+\frac {B e^2 x^3 \sqrt {a+c x^4}}{5 c}+\frac {A d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}-\frac {\left (3 a B e^2\right ) \int \frac {x^2}{\sqrt {a+c x^4}} \, dx}{5 c}-\frac {(a e (2 B d+A e)) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{3 c}+\frac {\left (\sqrt {a} d (B d+2 A e)\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{\sqrt {c}}-\frac {\left (\sqrt {a} d (B d+2 A e)\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{\sqrt {c}}\\ &=\frac {e (2 B d+A e) x \sqrt {a+c x^4}}{3 c}+\frac {B e^2 x^3 \sqrt {a+c x^4}}{5 c}+\frac {d (B d+2 A e) x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} d (B d+2 A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {A d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}-\frac {a^{3/4} e (2 B d+A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 c^{5/4} \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} d (B d+2 A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}-\frac {\left (3 a^{3/2} B e^2\right ) \int \frac {1}{\sqrt {a+c x^4}} \, dx}{5 c^{3/2}}+\frac {\left (3 a^{3/2} B e^2\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+c x^4}} \, dx}{5 c^{3/2}}\\ &=\frac {e (2 B d+A e) x \sqrt {a+c x^4}}{3 c}+\frac {B e^2 x^3 \sqrt {a+c x^4}}{5 c}-\frac {3 a B e^2 x \sqrt {a+c x^4}}{5 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {d (B d+2 A e) x \sqrt {a+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {3 a^{5/4} B e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 c^{7/4} \sqrt {a+c x^4}}-\frac {\sqrt [4]{a} d (B d+2 A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{c^{3/4} \sqrt {a+c x^4}}+\frac {A d^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}-\frac {3 a^{5/4} B e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 c^{7/4} \sqrt {a+c x^4}}-\frac {a^{3/4} e (2 B d+A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{6 c^{5/4} \sqrt {a+c x^4}}+\frac {\sqrt [4]{a} d (B d+2 A e) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 c^{3/4} \sqrt {a+c x^4}}\\ \end {align*}
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Mathematica [C] time = 0.17, size = 159, normalized size = 0.43 \[ \frac {-5 x \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {c x^4}{a}\right ) \left (a A e^2+2 a B d e-3 A c d^2\right )+x^3 \sqrt {\frac {c x^4}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {c x^4}{a}\right ) \left (-3 a B e^2+10 A c d e+5 B c d^2\right )+e x \left (a+c x^4\right ) \left (5 A e+10 B d+3 B e x^2\right )}{15 c \sqrt {a+c x^4}} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {B e^{2} x^{6} + {\left (2 \, B d e + A e^{2}\right )} x^{4} + A d^{2} + {\left (B d^{2} + 2 \, A d e\right )} x^{2}}{\sqrt {c x^{4} + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{2}}{\sqrt {c x^{4} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.01, size = 403, normalized size = 1.10 \[ \frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, A \,d^{2} \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}+\left (\frac {\sqrt {c \,x^{4}+a}\, x^{3}}{5 c}-\frac {3 i \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) a^{\frac {3}{2}}}{5 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, c^{\frac {3}{2}}}\right ) B \,e^{2}+\frac {i \left (2 A d e +B \,d^{2}\right ) \sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )\right ) \sqrt {a}}{\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}+\left (A \,e^{2}+2 B d e \right ) \left (-\frac {\sqrt {-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}+1}\, a \EllipticF \left (\sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, x , i\right )}{3 \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, c}+\frac {\sqrt {c \,x^{4}+a}\, x}{3 c}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (e x^{2} + d\right )}^{2}}{\sqrt {c x^{4} + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x^2+d\right )}^2}{\sqrt {c\,x^4+a}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 6.61, size = 262, normalized size = 0.71 \[ \frac {A d^{2} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {A d e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {A e^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {B d^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} + \frac {B d e x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} + \frac {B e^{2} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {c x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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