Optimal. Leaf size=372 \[ \frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} (f x)^{m+1} \left (c^4 d^2 (m+2) (m+3) (m+4) (m+5)+e (m+1)^2 \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},c^2 x^2\right )}{c^4 f (m+1)^2 (m+2) (m+3) (m+4) (m+5)}+\frac{d^2 (f x)^{m+1} \left (a+b \text{sech}^{-1}(c x)\right )}{f (m+1)}+\frac{2 d e (f x)^{m+3} \left (a+b \text{sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac{e^2 (f x)^{m+5} \left (a+b \text{sech}^{-1}(c x)\right )}{f^5 (m+5)}-\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 f (m+2) (m+3) (m+4) (m+5)}-\frac{b e^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4) (m+5)} \]
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Rubi [A] time = 0.431709, antiderivative size = 352, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {270, 6301, 12, 1267, 459, 364} \[ \frac{d^2 (f x)^{m+1} \left (a+b \text{sech}^{-1}(c x)\right )}{f (m+1)}+\frac{2 d e (f x)^{m+3} \left (a+b \text{sech}^{-1}(c x)\right )}{f^3 (m+3)}+\frac{e^2 (f x)^{m+5} \left (a+b \text{sech}^{-1}(c x)\right )}{f^5 (m+5)}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} (f x)^{m+1} \left (\frac{e \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 (m+2) (m+3) (m+4) (m+5)}+\frac{d^2}{(m+1)^2}\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};c^2 x^2\right )}{f}-\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} (f x)^{m+1} \left (2 c^2 d \left (m^2+9 m+20\right )+e (m+3)^2\right )}{c^4 f (m+2) (m+3) (m+4) (m+5)}-\frac{b e^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} (f x)^{m+3}}{c^2 f^3 (m+4) (m+5)} \]
Antiderivative was successfully verified.
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Rule 270
Rule 6301
Rule 12
Rule 1267
Rule 459
Rule 364
Rubi steps
\begin{align*} \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=\frac{d^2 (f x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^5 (5+m)}+\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{(1+m) (3+m) (5+m) \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{d^2 (f x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^5 (5+m)}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{(f x)^m \left (d^2 \left (15+8 m+m^2\right )+2 d e \left (5+6 m+m^2\right ) x^2+e^2 \left (3+4 m+m^2\right ) x^4\right )}{\sqrt{1-c^2 x^2}} \, dx}{15+23 m+9 m^2+m^3}\\ &=-\frac{b e^2 (f x)^{3+m} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{c^2 f^3 (4+m) (5+m)}+\frac{d^2 (f x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^5 (5+m)}-\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{(f x)^m \left (-c^2 d^2 (3+m) (4+m) (5+m)-e (1+m) \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) x^2\right )}{\sqrt{1-c^2 x^2}} \, dx}{c^2 (4+m) \left (15+23 m+9 m^2+m^3\right )}\\ &=-\frac{b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) (f x)^{1+m} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{c^4 f (2+m) (4+m) \left (15+8 m+m^2\right )}-\frac{b e^2 (f x)^{3+m} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{c^2 f^3 (4+m) (5+m)}+\frac{d^2 (f x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^5 (5+m)}+-\frac{\left (b \left (-c^4 d^2 (2+m) (3+m) (4+m) (5+m)-e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{(f x)^m}{\sqrt{1-c^2 x^2}} \, dx}{c^4 (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right )}\\ &=-\frac{b e \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right ) (f x)^{1+m} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{c^4 f (2+m) (4+m) \left (15+8 m+m^2\right )}-\frac{b e^2 (f x)^{3+m} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{c^2 f^3 (4+m) (5+m)}+\frac{d^2 (f x)^{1+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f (1+m)}+\frac{2 d e (f x)^{3+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^3 (3+m)}+\frac{e^2 (f x)^{5+m} \left (a+b \text{sech}^{-1}(c x)\right )}{f^5 (5+m)}+\frac{b \left (c^4 d^2 (2+m) (3+m) (4+m) (5+m)+e (1+m)^2 \left (e (3+m)^2+2 c^2 d \left (20+9 m+m^2\right )\right )\right ) (f x)^{1+m} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};c^2 x^2\right )}{c^4 f (1+m) (2+m) (4+m) \left (15+23 m+9 m^2+m^3\right )}\\ \end{align*}
Mathematica [F] time = 0.153183, size = 0, normalized size = 0. \[ \int (f x)^m \left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 2.111, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) ^{2} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} +{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \operatorname{arsech}\left (c x\right )\right )} \left (f x\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}^{2}{\left (b \operatorname{arsech}\left (c x\right ) + a\right )} \left (f x\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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