Optimal. Leaf size=104 \[ \frac {8 \sqrt [4]{a+b x^2} (8 b c-5 a d)}{15 a^3 e^3 \sqrt {e x}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}} \]
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Rubi [A] time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {453, 273, 264} \begin {gather*} \frac {8 \sqrt [4]{a+b x^2} (8 b c-5 a d)}{15 a^3 e^3 \sqrt {e x}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}} \end {gather*}
Antiderivative was successfully verified.
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Rule 264
Rule 273
Rule 453
Rubi steps
\begin {align*} \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{7/4}} \, dx &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {(8 b c-5 a d) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{7/4}} \, dx}{5 a e^2}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}-\frac {(4 (8 b c-5 a d)) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{3/4}} \, dx}{15 a^2 e^2}\\ &=-\frac {2 c}{5 a e (e x)^{5/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (8 b c-5 a d)}{15 a^2 e^3 \sqrt {e x} \left (a+b x^2\right )^{3/4}}+\frac {8 (8 b c-5 a d) \sqrt [4]{a+b x^2}}{15 a^3 e^3 \sqrt {e x}}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 66, normalized size = 0.63 \begin {gather*} \frac {x \left (-6 a^2 \left (c+5 d x^2\right )+8 a b x^2 \left (6 c-5 d x^2\right )+64 b^2 c x^4\right )}{15 a^3 (e x)^{7/2} \left (a+b x^2\right )^{3/4}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 27.33, size = 100, normalized size = 0.96 \begin {gather*} -\frac {2 \sqrt [4]{a+b x^2} \left (3 a^2 c e^4+15 a^2 d e^4 x^2-24 a b c e^4 x^2+20 a b d e^4 x^4-32 b^2 c e^4 x^4\right )}{15 a^3 e^3 (e x)^{5/2} \left (a e^2+b e^2 x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 2.42, size = 81, normalized size = 0.78 \begin {gather*} \frac {2 \, {\left (4 \, {\left (8 \, b^{2} c - 5 \, a b d\right )} x^{4} - 3 \, a^{2} c + 3 \, {\left (8 \, a b c - 5 \, a^{2} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {e x}}{15 \, {\left (a^{3} b e^{4} x^{5} + a^{4} e^{4} x^{3}\right )}} \end {gather*}
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 \begin {gather*} \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} \left (e x\right )^{\frac {7}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 62, normalized size = 0.60 \begin {gather*} -\frac {2 \left (20 a b d \,x^{4}-32 b^{2} c \,x^{4}+15 a^{2} d \,x^{2}-24 a b c \,x^{2}+3 c \,a^{2}\right ) x}{15 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (e x \right )^{\frac {7}{2}} a^{3}} \end {gather*}
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 \begin {gather*} \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {7}{4}} \left (e x\right )^{\frac {7}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.21, size = 101, normalized size = 0.97 \begin {gather*} -\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,c}{5\,a\,b\,e^3}+\frac {x^2\,\left (30\,a^2\,d-48\,a\,b\,c\right )}{15\,a^3\,b\,e^3}-\frac {x^4\,\left (64\,b^2\,c-40\,a\,b\,d\right )}{15\,a^3\,b\,e^3}\right )}{x^4\,\sqrt {e\,x}+\frac {a\,x^2\,\sqrt {e\,x}}{b}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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