3.8.36 \(\int (d x)^{3/2} (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\) [736]

Optimal. Leaf size=195 \[ \frac {2 a^3 (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}+\frac {2 a^2 b (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}+\frac {6 a b^2 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{17/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 d^7 \left (a+b x^2\right )} \]

[Out]

2/5*a^3*(d*x)^(5/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+2/3*a^2*b*(d*x)^(9/2)*((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+6/
13*a*b^2*(d*x)^(13/2)*((b*x^2+a)^2)^(1/2)/d^5/(b*x^2+a)+2/17*b^3*(d*x)^(17/2)*((b*x^2+a)^2)^(1/2)/d^7/(b*x^2+a
)

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Rubi [A]
time = 0.04, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1126, 276} \begin {gather*} \frac {6 a b^2 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^5 \left (a+b x^2\right )}+\frac {2 a^2 b (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{17/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 d^7 \left (a+b x^2\right )}+\frac {2 a^3 (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(a + b*x^2)) + (2*a^2*b*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4])/(3*d^3*(a + b*x^2)) + (6*a*b^2*(d*x)^(13/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*d^5*(a + b*x^
2)) + (2*b^3*(d*x)^(17/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(17*d^7*(a + b*x^2))

Rule 276

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 1126

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)^{3/2} \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)^{3/2} \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)^{3/2}+\frac {3 a^2 b^4 (d x)^{7/2}}{d^2}+\frac {3 a b^5 (d x)^{11/2}}{d^4}+\frac {b^6 (d x)^{15/2}}{d^6}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {2 a^3 (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}+\frac {2 a^2 b (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}+\frac {6 a b^2 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{17/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{17 d^7 \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 66, normalized size = 0.34 \begin {gather*} \frac {2 x (d x)^{3/2} \sqrt {\left (a+b x^2\right )^2} \left (663 a^3+1105 a^2 b x^2+765 a b^2 x^4+195 b^3 x^6\right )}{3315 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(d*x)^(3/2)*Sqrt[(a + b*x^2)^2]*(663*a^3 + 1105*a^2*b*x^2 + 765*a*b^2*x^4 + 195*b^3*x^6))/(3315*(a + b*x^
2))

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Maple [A]
time = 0.04, size = 63, normalized size = 0.32

method	result	size
gosper	\(\frac {2 x \left (195 b^{3} x^{6}+765 a \,b^{2} x^{4}+1105 a^{2} b \,x^{2}+663 a^{3}\right ) \left (d x \right )^{\frac {3}{2}} \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{3315 \left (b \,x^{2}+a \right )^{3}}\)	\(61\)
default	\(\frac {2 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} \left (d x \right )^{\frac {5}{2}} \left (195 b^{3} x^{6}+765 a \,b^{2} x^{4}+1105 a^{2} b \,x^{2}+663 a^{3}\right )}{3315 d \left (b \,x^{2}+a \right )^{3}}\)	\(63\)
risch	\(\frac {2 d^{2} \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{3} \left (195 b^{3} x^{6}+765 a \,b^{2} x^{4}+1105 a^{2} b \,x^{2}+663 a^{3}\right )}{3315 \left (b \,x^{2}+a \right ) \sqrt {d x}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3315*((b*x^2+a)^2)^(3/2)*(d*x)^(5/2)/d*(195*b^3*x^6+765*a*b^2*x^4+1105*a^2*b*x^2+663*a^3)/(b*x^2+a)^3

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Maxima [A]
time = 0.30, size = 83, normalized size = 0.43 \begin {gather*} \frac {2}{221} \, {\left (13 \, b^{3} d^{\frac {3}{2}} x^{3} + 17 \, a b^{2} d^{\frac {3}{2}} x\right )} x^{\frac {11}{2}} + \frac {4}{117} \, {\left (9 \, a b^{2} d^{\frac {3}{2}} x^{3} + 13 \, a^{2} b d^{\frac {3}{2}} x\right )} x^{\frac {7}{2}} + \frac {2}{45} \, {\left (5 \, a^{2} b d^{\frac {3}{2}} x^{3} + 9 \, a^{3} d^{\frac {3}{2}} x\right )} x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/221*(13*b^3*d^(3/2)*x^3 + 17*a*b^2*d^(3/2)*x)*x^(11/2) + 4/117*(9*a*b^2*d^(3/2)*x^3 + 13*a^2*b*d^(3/2)*x)*x^
(7/2) + 2/45*(5*a^2*b*d^(3/2)*x^3 + 9*a^3*d^(3/2)*x)*x^(3/2)

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Fricas [A]
time = 0.35, size = 46, normalized size = 0.24 \begin {gather*} \frac {2}{3315} \, {\left (195 \, b^{3} d x^{8} + 765 \, a b^{2} d x^{6} + 1105 \, a^{2} b d x^{4} + 663 \, a^{3} d x^{2}\right )} \sqrt {d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3315*(195*b^3*d*x^8 + 765*a*b^2*d*x^6 + 1105*a^2*b*d*x^4 + 663*a^3*d*x^2)*sqrt(d*x)

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

[Out]

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

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Giac [A]
time = 2.95, size = 90, normalized size = 0.46 \begin {gather*} \frac {2}{3315} \, {\left (195 \, \sqrt {d x} b^{3} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 765 \, \sqrt {d x} a b^{2} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 1105 \, \sqrt {d x} a^{2} b x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 663 \, \sqrt {d x} a^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right )\right )} d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3315*(195*sqrt(d*x)*b^3*x^8*sgn(b*x^2 + a) + 765*sqrt(d*x)*a*b^2*x^6*sgn(b*x^2 + a) + 1105*sqrt(d*x)*a^2*b*x
^4*sgn(b*x^2 + a) + 663*sqrt(d*x)*a^3*x^2*sgn(b*x^2 + a))*d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d\,x\right )}^{3/2}\,{\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)^(3/2)*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

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

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