3.2739 \(\int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx\)

Optimal. Leaf size=136 \[ \frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a+b x^{2 (m+1)}}}\right )}{16 \sqrt{b} (m+1)}+\frac{5 a^2 x^{m+1} \sqrt{a+b x^{2 (m+1)}}}{16 (m+1)}+\frac{x^{m+1} \left (a+b x^{2 (m+1)}\right )^{5/2}}{6 (m+1)}+\frac{5 a x^{m+1} \left (a+b x^{2 (m+1)}\right )^{3/2}}{24 (m+1)} \]

[Out]

(5*a^2*x^(1 + m)*Sqrt[a + b*x^(2*(1 + m))])/(16*(1 + m)) + (5*a*x^(1 + m)*(a + b
*x^(2*(1 + m)))^(3/2))/(24*(1 + m)) + (x^(1 + m)*(a + b*x^(2*(1 + m)))^(5/2))/(6
*(1 + m)) + (5*a^3*ArcTanh[(Sqrt[b]*x^(1 + m))/Sqrt[a + b*x^(2*(1 + m))]])/(16*S
qrt[b]*(1 + m))

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Rubi [A]  time = 0.131797, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21 \[ \frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a+b x^{2 (m+1)}}}\right )}{16 \sqrt{b} (m+1)}+\frac{5 a^2 x^{m+1} \sqrt{a+b x^{2 (m+1)}}}{16 (m+1)}+\frac{x^{m+1} \left (a+b x^{2 (m+1)}\right )^{5/2}}{6 (m+1)}+\frac{5 a x^{m+1} \left (a+b x^{2 (m+1)}\right )^{3/2}}{24 (m+1)} \]

Antiderivative was successfully verified.

[In]  Int[x^m*(a + b*x^(2 + 2*m))^(5/2),x]

[Out]

(5*a^2*x^(1 + m)*Sqrt[a + b*x^(2*(1 + m))])/(16*(1 + m)) + (5*a*x^(1 + m)*(a + b
*x^(2*(1 + m)))^(3/2))/(24*(1 + m)) + (x^(1 + m)*(a + b*x^(2*(1 + m)))^(5/2))/(6
*(1 + m)) + (5*a^3*ArcTanh[(Sqrt[b]*x^(1 + m))/Sqrt[a + b*x^(2*(1 + m))]])/(16*S
qrt[b]*(1 + m))

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Rubi in Sympy [A]  time = 7.46824, size = 60, normalized size = 0.44 \[ \frac{a^{2} x^{m + 1} \sqrt{a + b x^{2 m + 2}}{{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{2 m + 2}}{a}} \right )}}{\sqrt{1 + \frac{b x^{2 m + 2}}{a}} \left (m + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**m*(a+b*x**(2+2*m))**(5/2),x)

[Out]

a**2*x**(m + 1)*sqrt(a + b*x**(2*m + 2))*hyper((-5/2, 1/2), (3/2,), -b*x**(2*m +
2)/a)/(sqrt(1 + b*x**(2*m + 2)/a)*(m + 1))

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Mathematica [A]  time = 0.241332, size = 127, normalized size = 0.93 \[ \frac{15 a^{7/2} \sqrt{\frac{b x^{2 m+2}}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a}}\right )+\sqrt{b} x^{m+1} \left (33 a^3+59 a^2 b x^{2 m+2}+34 a b^2 x^{4 m+4}+8 b^3 x^{6 m+6}\right )}{48 \sqrt{b} (m+1) \sqrt{a+b x^{2 m+2}}} \]

Antiderivative was successfully verified.

[In]  Integrate[x^m*(a + b*x^(2 + 2*m))^(5/2),x]

[Out]

(Sqrt[b]*x^(1 + m)*(33*a^3 + 59*a^2*b*x^(2 + 2*m) + 34*a*b^2*x^(4 + 4*m) + 8*b^3
*x^(6 + 6*m)) + 15*a^(7/2)*Sqrt[1 + (b*x^(2 + 2*m))/a]*ArcSinh[(Sqrt[b]*x^(1 + m
))/Sqrt[a]])/(48*Sqrt[b]*(1 + m)*Sqrt[a + b*x^(2 + 2*m)])

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Maple [F]  time = 0.116, size = 0, normalized size = 0. \[ \int{x}^{m} \left ( a+b{x}^{2+2\,m} \right ) ^{{\frac{5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^m*(a+b*x^(2+2*m))^(5/2),x)

[Out]

int(x^m*(a+b*x^(2+2*m))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (b x^{2 \, m + 2} + a\right )}^{\frac{5}{2}} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^(2*m + 2) + a)^(5/2)*x^m,x, algorithm="maxima")

[Out]

integrate((b*x^(2*m + 2) + a)^(5/2)*x^m, x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^(2*m + 2) + a)^(5/2)*x^m,x, algorithm="fricas")

[Out]

Exception raised: TypeError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**m*(a+b*x**(2+2*m))**(5/2),x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (b x^{2 \, m + 2} + a\right )}^{\frac{5}{2}} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^(2*m + 2) + a)^(5/2)*x^m,x, algorithm="giac")

[Out]

integrate((b*x^(2*m + 2) + a)^(5/2)*x^m, x)