3.227 \(\int x^m (c (a+b x^2)^2)^{3/2} \, dx\)

Optimal. Leaf size=161 \[ \frac{a^3 c x^{m+1} \sqrt{c \left (a+b x^2\right )^2}}{(m+1) \left (a+b x^2\right )}+\frac{3 a^2 b c x^{m+3} \sqrt{c \left (a+b x^2\right )^2}}{(m+3) \left (a+b x^2\right )}+\frac{3 a b^2 c x^{m+5} \sqrt{c \left (a+b x^2\right )^2}}{(m+5) \left (a+b x^2\right )}+\frac{b^3 c x^{m+7} \sqrt{c \left (a+b x^2\right )^2}}{(m+7) \left (a+b x^2\right )} \]

[Out]

(a^3*c*x^(1 + m)*Sqrt[c*(a + b*x^2)^2])/((1 + m)*(a + b*x^2)) + (3*a^2*b*c*x^(3 + m)*Sqrt[c*(a + b*x^2)^2])/((
3 + m)*(a + b*x^2)) + (3*a*b^2*c*x^(5 + m)*Sqrt[c*(a + b*x^2)^2])/((5 + m)*(a + b*x^2)) + (b^3*c*x^(7 + m)*Sqr
t[c*(a + b*x^2)^2])/((7 + m)*(a + b*x^2))

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Rubi [A]  time = 0.127775, antiderivative size = 205, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1989, 1112, 270} \[ \frac{a^3 c x^{m+1} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(m+1) \left (a+b x^2\right )}+\frac{3 a^2 b c x^{m+3} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(m+3) \left (a+b x^2\right )}+\frac{3 a b^2 c x^{m+5} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(m+5) \left (a+b x^2\right )}+\frac{b^3 c x^{m+7} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(m+7) \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(a^3*c*x^(1 + m)*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/((1 + m)*(a + b*x^2)) + (3*a^2*b*c*x^(3 + m)*Sqrt[a^2*
c + 2*a*b*c*x^2 + b^2*c*x^4])/((3 + m)*(a + b*x^2)) + (3*a*b^2*c*x^(5 + m)*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^
4])/((5 + m)*(a + b*x^2)) + (b^3*c*x^(7 + m)*Sqrt[a^2*c + 2*a*b*c*x^2 + b^2*c*x^4])/((7 + m)*(a + b*x^2))

Rule 1989

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int x^m \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx &=\int x^m \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \, dx\\ &=\frac{\sqrt{a^2 c+2 a b c x^2+b^2 c x^4} \int x^m \left (a b c+b^2 c x^2\right )^3 \, dx}{b^2 c \left (a b c+b^2 c x^2\right )}\\ &=\frac{\sqrt{a^2 c+2 a b c x^2+b^2 c x^4} \int \left (a^3 b^3 c^3 x^m+3 a^2 b^4 c^3 x^{2+m}+3 a b^5 c^3 x^{4+m}+b^6 c^3 x^{6+m}\right ) \, dx}{b^2 c \left (a b c+b^2 c x^2\right )}\\ &=\frac{a^3 c x^{1+m} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(1+m) \left (a+b x^2\right )}+\frac{3 a^2 b c x^{3+m} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(3+m) \left (a+b x^2\right )}+\frac{3 a b^2 c x^{5+m} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(5+m) \left (a+b x^2\right )}+\frac{b^3 c x^{7+m} \sqrt{a^2 c+2 a b c x^2+b^2 c x^4}}{(7+m) \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0933486, size = 132, normalized size = 0.82 \[ \frac{x^{m+1} \left (c \left (a+b x^2\right )^2\right )^{3/2} \left (3 a^2 b \left (m^3+13 m^2+47 m+35\right ) x^2+a^3 \left (m^3+15 m^2+71 m+105\right )+3 a b^2 \left (m^3+11 m^2+31 m+21\right ) x^4+b^3 \left (m^3+9 m^2+23 m+15\right ) x^6\right )}{(m+1) (m+3) (m+5) (m+7) \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(x^(1 + m)*(c*(a + b*x^2)^2)^(3/2)*(a^3*(105 + 71*m + 15*m^2 + m^3) + 3*a^2*b*(35 + 47*m + 13*m^2 + m^3)*x^2 +
3*a*b^2*(21 + 31*m + 11*m^2 + m^3)*x^4 + b^3*(15 + 23*m + 9*m^2 + m^3)*x^6))/((1 + m)*(3 + m)*(5 + m)*(7 + m)
*(a + b*x^2)^3)

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Maple [A]  time = 0.007, size = 200, normalized size = 1.2 \begin{align*}{\frac{{x}^{1+m} \left ({b}^{3}{m}^{3}{x}^{6}+9\,{b}^{3}{m}^{2}{x}^{6}+3\,a{b}^{2}{m}^{3}{x}^{4}+23\,{b}^{3}m{x}^{6}+33\,a{b}^{2}{m}^{2}{x}^{4}+15\,{b}^{3}{x}^{6}+3\,{a}^{2}b{m}^{3}{x}^{2}+93\,a{b}^{2}m{x}^{4}+39\,{a}^{2}b{m}^{2}{x}^{2}+63\,a{b}^{2}{x}^{4}+{a}^{3}{m}^{3}+141\,{a}^{2}bm{x}^{2}+15\,{a}^{3}{m}^{2}+105\,{a}^{2}b{x}^{2}+71\,m{a}^{3}+105\,{a}^{3} \right ) }{ \left ( 7+m \right ) \left ( 5+m \right ) \left ( 3+m \right ) \left ( 1+m \right ) \left ( b{x}^{2}+a \right ) ^{3}} \left ( c \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(c*(b*x^2+a)^2)^(3/2),x)

[Out]

x^(1+m)*(b^3*m^3*x^6+9*b^3*m^2*x^6+3*a*b^2*m^3*x^4+23*b^3*m*x^6+33*a*b^2*m^2*x^4+15*b^3*x^6+3*a^2*b*m^3*x^2+93
*a*b^2*m*x^4+39*a^2*b*m^2*x^2+63*a*b^2*x^4+a^3*m^3+141*a^2*b*m*x^2+15*a^3*m^2+105*a^2*b*x^2+71*a^3*m+105*a^3)*
(c*(b*x^2+a)^2)^(3/2)/(7+m)/(5+m)/(3+m)/(1+m)/(b*x^2+a)^3

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Maxima [A]  time = 1.08889, size = 161, normalized size = 1. \begin{align*} \frac{{\left ({\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} b^{3} c^{\frac{3}{2}} x^{7} + 3 \,{\left (m^{3} + 11 \, m^{2} + 31 \, m + 21\right )} a b^{2} c^{\frac{3}{2}} x^{5} + 3 \,{\left (m^{3} + 13 \, m^{2} + 47 \, m + 35\right )} a^{2} b c^{\frac{3}{2}} x^{3} +{\left (m^{3} + 15 \, m^{2} + 71 \, m + 105\right )} a^{3} c^{\frac{3}{2}} x\right )} x^{m}}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(c*(b*x^2+a)^2)^(3/2),x, algorithm="maxima")

[Out]

((m^3 + 9*m^2 + 23*m + 15)*b^3*c^(3/2)*x^7 + 3*(m^3 + 11*m^2 + 31*m + 21)*a*b^2*c^(3/2)*x^5 + 3*(m^3 + 13*m^2
+ 47*m + 35)*a^2*b*c^(3/2)*x^3 + (m^3 + 15*m^2 + 71*m + 105)*a^3*c^(3/2)*x)*x^m/(m^4 + 16*m^3 + 86*m^2 + 176*m
+ 105)

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Fricas [A]  time = 1.54038, size = 528, normalized size = 3.28 \begin{align*} \frac{{\left ({\left (b^{3} c m^{3} + 9 \, b^{3} c m^{2} + 23 \, b^{3} c m + 15 \, b^{3} c\right )} x^{7} + 3 \,{\left (a b^{2} c m^{3} + 11 \, a b^{2} c m^{2} + 31 \, a b^{2} c m + 21 \, a b^{2} c\right )} x^{5} + 3 \,{\left (a^{2} b c m^{3} + 13 \, a^{2} b c m^{2} + 47 \, a^{2} b c m + 35 \, a^{2} b c\right )} x^{3} +{\left (a^{3} c m^{3} + 15 \, a^{3} c m^{2} + 71 \, a^{3} c m + 105 \, a^{3} c\right )} x\right )} \sqrt{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c} x^{m}}{a m^{4} + 16 \, a m^{3} + 86 \, a m^{2} +{\left (b m^{4} + 16 \, b m^{3} + 86 \, b m^{2} + 176 \, b m + 105 \, b\right )} x^{2} + 176 \, a m + 105 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(c*(b*x^2+a)^2)^(3/2),x, algorithm="fricas")

[Out]

((b^3*c*m^3 + 9*b^3*c*m^2 + 23*b^3*c*m + 15*b^3*c)*x^7 + 3*(a*b^2*c*m^3 + 11*a*b^2*c*m^2 + 31*a*b^2*c*m + 21*a
*b^2*c)*x^5 + 3*(a^2*b*c*m^3 + 13*a^2*b*c*m^2 + 47*a^2*b*c*m + 35*a^2*b*c)*x^3 + (a^3*c*m^3 + 15*a^3*c*m^2 + 7
1*a^3*c*m + 105*a^3*c)*x)*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)*x^m/(a*m^4 + 16*a*m^3 + 86*a*m^2 + (b*m^4 + 16
*b*m^3 + 86*b*m^2 + 176*b*m + 105*b)*x^2 + 176*a*m + 105*a)

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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*(c*(b*x**2+a)**2)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.23075, size = 479, normalized size = 2.98 \begin{align*} \frac{{\left (b^{3} m^{3} x^{7} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 9 \, b^{3} m^{2} x^{7} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 3 \, a b^{2} m^{3} x^{5} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 23 \, b^{3} m x^{7} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 33 \, a b^{2} m^{2} x^{5} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 15 \, b^{3} x^{7} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 3 \, a^{2} b m^{3} x^{3} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 93 \, a b^{2} m x^{5} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 39 \, a^{2} b m^{2} x^{3} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 63 \, a b^{2} x^{5} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + a^{3} m^{3} x x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 141 \, a^{2} b m x^{3} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 15 \, a^{3} m^{2} x x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 105 \, a^{2} b x^{3} x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 71 \, a^{3} m x x^{m} \mathrm{sgn}\left (b x^{2} + a\right ) + 105 \, a^{3} x x^{m} \mathrm{sgn}\left (b x^{2} + a\right )\right )} c^{\frac{3}{2}}}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(c*(b*x^2+a)^2)^(3/2),x, algorithm="giac")

[Out]

(b^3*m^3*x^7*x^m*sgn(b*x^2 + a) + 9*b^3*m^2*x^7*x^m*sgn(b*x^2 + a) + 3*a*b^2*m^3*x^5*x^m*sgn(b*x^2 + a) + 23*b
^3*m*x^7*x^m*sgn(b*x^2 + a) + 33*a*b^2*m^2*x^5*x^m*sgn(b*x^2 + a) + 15*b^3*x^7*x^m*sgn(b*x^2 + a) + 3*a^2*b*m^
3*x^3*x^m*sgn(b*x^2 + a) + 93*a*b^2*m*x^5*x^m*sgn(b*x^2 + a) + 39*a^2*b*m^2*x^3*x^m*sgn(b*x^2 + a) + 63*a*b^2*
x^5*x^m*sgn(b*x^2 + a) + a^3*m^3*x*x^m*sgn(b*x^2 + a) + 141*a^2*b*m*x^3*x^m*sgn(b*x^2 + a) + 15*a^3*m^2*x*x^m*
sgn(b*x^2 + a) + 105*a^2*b*x^3*x^m*sgn(b*x^2 + a) + 71*a^3*m*x*x^m*sgn(b*x^2 + a) + 105*a^3*x*x^m*sgn(b*x^2 +
a))*c^(3/2)/(m^4 + 16*m^3 + 86*m^2 + 176*m + 105)