3.7.11 \(\int (d x)^m (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\)

Optimal. Leaf size=205 \[ \frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+5}}{d^5 (m+5) \left (a+b x^2\right )}+\frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+3}}{d^3 (m+3) \left (a+b x^2\right )}+\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+7}}{d^7 (m+7) \left (a+b x^2\right )}+\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+1}}{d (m+1) \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.08, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1112, 270} \begin {gather*} \frac {3 a^2 b \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+3}}{d^3 (m+3) \left (a+b x^2\right )}+\frac {3 a b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+5}}{d^5 (m+5) \left (a+b x^2\right )}+\frac {b^3 \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+7}}{d^7 (m+7) \left (a+b x^2\right )}+\frac {a^3 \sqrt {a^2+2 a b x^2+b^2 x^4} (d x)^{m+1}}{d (m+1) \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 131, normalized size = 0.64 \begin {gather*} \frac {x \sqrt {\left (a+b x^2\right )^2} (d x)^m \left (a^3 \left (m^3+15 m^2+71 m+105\right )+3 a^2 b \left (m^3+13 m^2+47 m+35\right ) x^2+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 )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x*(d*x)^m*Sqrt[(a + b*x^2)^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))

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IntegrateAlgebraic [F]  time = 1.17, size = 0, normalized size = 0.00 \begin {gather*} \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

Defer[IntegrateAlgebraic][(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2), x]

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fricas [A]  time = 0.99, size = 159, normalized size = 0.78 \begin {gather*} \frac {{\left ({\left (b^{3} m^{3} + 9 \, b^{3} m^{2} + 23 \, b^{3} m + 15 \, b^{3}\right )} x^{7} + 3 \, {\left (a b^{2} m^{3} + 11 \, a b^{2} m^{2} + 31 \, a b^{2} m + 21 \, a b^{2}\right )} x^{5} + 3 \, {\left (a^{2} b m^{3} + 13 \, a^{2} b m^{2} + 47 \, a^{2} b m + 35 \, a^{2} b\right )} x^{3} + {\left (a^{3} m^{3} + 15 \, a^{3} m^{2} + 71 \, a^{3} m + 105 \, a^{3}\right )} x\right )} \left (d x\right )^{m}}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.21, size = 384, normalized size = 1.87 \begin {gather*} \frac {\left (d x\right )^{m} b^{3} m^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 9 \, \left (d x\right )^{m} b^{3} m^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, \left (d x\right )^{m} a b^{2} m^{3} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 23 \, \left (d x\right )^{m} b^{3} m x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 33 \, \left (d x\right )^{m} a b^{2} m^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, \left (d x\right )^{m} b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, \left (d x\right )^{m} a^{2} b m^{3} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 93 \, \left (d x\right )^{m} a b^{2} m x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 39 \, \left (d x\right )^{m} a^{2} b m^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 63 \, \left (d x\right )^{m} a b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \left (d x\right )^{m} a^{3} m^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) + 141 \, \left (d x\right )^{m} a^{2} b m x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, \left (d x\right )^{m} a^{3} m^{2} x \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, \left (d x\right )^{m} a^{2} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 71 \, \left (d x\right )^{m} a^{3} m x \mathrm {sgn}\left (b x^{2} + a\right ) + 105 \, \left (d x\right )^{m} a^{3} x \mathrm {sgn}\left (b x^{2} + a\right )}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 199, normalized size = 0.97 \begin {gather*} \frac {\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} b m \,x^{2}+15 a^{3} m^{2}+105 a^{2} b \,x^{2}+71 a^{3} m +105 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} x \left (d x \right )^{m}}{\left (m +7\right ) \left (m +5\right ) \left (m +3\right ) \left (m +1\right ) \left (b \,x^{2}+a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

x*(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)*(d*x)^
m*((b*x^2+a)^2)^(3/2)/(m+7)/(m+5)/(m+3)/(m+1)/(b*x^2+a)^3

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maxima [A]  time = 1.44, size = 119, normalized size = 0.58 \begin {gather*} \frac {{\left ({\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} b^{3} d^{m} x^{7} + 3 \, {\left (m^{3} + 11 \, m^{2} + 31 \, m + 21\right )} a b^{2} d^{m} x^{5} + 3 \, {\left (m^{3} + 13 \, m^{2} + 47 \, m + 35\right )} a^{2} b d^{m} x^{3} + {\left (m^{3} + 15 \, m^{2} + 71 \, m + 105\right )} a^{3} d^{m} x\right )} x^{m}}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (d\,x\right )}^m\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int((d*x)^m*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d x\right )^{m} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral((d*x)**m*((a + b*x**2)**2)**(3/2), x)

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