3.2746 \(\int x^m (a+b x^{2+2 m})^{5/2} \, dx\)

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

[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*Sqrt[b]*(1 + m))

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Rubi [A]  time = 0.0539171, 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, Rules used = {345, 195, 217, 206} \[ \frac{5 a^2 x^{m+1} \sqrt{a+b x^{2 (m+1)}}}{16 (m+1)}+\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^{m+1}}{\sqrt{a+b x^{2 (m+1)}}}\right )}{16 \sqrt{b} (m+1)}+\frac{x^{m+1} \left (a+b x^{2 (m+1)}\right )^{5/2}}{6 (m+1)}+\frac{5 a x^{m+1} \left (a+b x^{2 (m+1)}\right )^{3/2}}{24 (m+1)} \]

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*Sqrt[b]*(1 + m))

Rule 345

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +
1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] &&  !IntegerQ[n]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b x^2\right )^{5/2} \, dx,x,x^{1+m}\right )}{1+m}\\ &=\frac{x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac{(5 a) \operatorname{Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,x^{1+m}\right )}{6 (1+m)}\\ &=\frac{5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac{x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac{\left (5 a^2\right ) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,x^{1+m}\right )}{8 (1+m)}\\ &=\frac{5 a^2 x^{1+m} \sqrt{a+b x^{2 (1+m)}}}{16 (1+m)}+\frac{5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac{x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^{1+m}\right )}{16 (1+m)}\\ &=\frac{5 a^2 x^{1+m} \sqrt{a+b x^{2 (1+m)}}}{16 (1+m)}+\frac{5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac{x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac{\left (5 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^{1+m}}{\sqrt{a+b x^{2+2 m}}}\right )}{16 (1+m)}\\ &=\frac{5 a^2 x^{1+m} \sqrt{a+b x^{2 (1+m)}}}{16 (1+m)}+\frac{5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac{x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^{1+m}}{\sqrt{a+b x^{2 (1+m)}}}\right )}{16 \sqrt{b} (1+m)}\\ \end{align*}

Mathematica [C]  time = 0.0667992, size = 88, normalized size = 0.65 \[ \frac{a^2 x^{m+1} \sqrt{a+b x^{2 m+2}} \, _2F_1\left (-\frac{5}{2},\frac{m+1}{2 m+2};\frac{m+1}{2 m+2}+1;-\frac{b x^{2 m+2}}{a}\right )}{(m+1) \sqrt{\frac{b x^{2 m+2}}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x^(1 + m)*Sqrt[a + b*x^(2 + 2*m)]*Hypergeometric2F1[-5/2, (1 + m)/(2 + 2*m), 1 + (1 + m)/(2 + 2*m), -((b*
x^(2 + 2*m))/a)])/((1 + m)*Sqrt[1 + (b*x^(2 + 2*m))/a])

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( a+b{x}^{2+2\,m} \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

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. \begin{align*} \int{\left (b x^{2 \, m + 2} + a\right )}^{\frac{5}{2}} x^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(a+b*x^(2+2*m))^(5/2),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. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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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.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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