3.8.38 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{3/2}}{\sqrt {d x}} \, dx\) [738]

Optimal. Leaf size=193 \[ \frac {2 a^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac {2 a b^2 (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^7 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 193, 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 {2 a b^2 (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 d^7 \left (a+b x^2\right )}+\frac {2 a^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(d*(a + b*x^2)) + (6*a^2*b*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(5*d^3*(a + b*x^2)) + (2*a*b^2*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^5*(a + b*x^2)) +
(2*b^3*(d*x)^(13/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*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 \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt {d x}} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^3}{\sqrt {d x}} \, 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 (\frac {a^3 b^3}{\sqrt {d x}}+\frac {3 a^2 b^4 (d x)^{3/2}}{d^2}+\frac {3 a b^5 (d x)^{7/2}}{d^4}+\frac {b^6 (d x)^{11/2}}{d^6}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {2 a^3 \sqrt {d x} \sqrt {a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )}+\frac {6 a^2 b (d x)^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac {2 a b^2 (d x)^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {2 b^3 (d x)^{13/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 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 \sqrt {\left (a+b x^2\right )^2} \left (195 a^3+117 a^2 b x^2+65 a b^2 x^4+15 b^3 x^6\right )}{195 \sqrt {d x} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*Sqrt[(a + b*x^2)^2]*(195*a^3 + 117*a^2*b*x^2 + 65*a*b^2*x^4 + 15*b^3*x^6))/(195*Sqrt[d*x]*(a + b*x^2))

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Maple [A]
time = 0.01, size = 63, normalized size = 0.33

method	result	size
gosper	\(\frac {2 x \left (15 b^{3} x^{6}+65 a \,b^{2} x^{4}+117 a^{2} b \,x^{2}+195 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{195 \left (b \,x^{2}+a \right )^{3} \sqrt {d x}}\)	\(61\)
risch	\(\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (15 b^{3} x^{6}+65 a \,b^{2} x^{4}+117 a^{2} b \,x^{2}+195 a^{3}\right ) x}{195 \left (b \,x^{2}+a \right ) \sqrt {d x}}\)	\(61\)
default	\(\frac {2 \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} \sqrt {d x}\, \left (15 b^{3} x^{6}+65 a \,b^{2} x^{4}+117 a^{2} b \,x^{2}+195 a^{3}\right )}{195 d \left (b \,x^{2}+a \right )^{3}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/195*((b*x^2+a)^2)^(3/2)*(d*x)^(1/2)/d*(15*b^3*x^6+65*a*b^2*x^4+117*a^2*b*x^2+195*a^3)/(b*x^2+a)^3

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Maxima [A]
time = 0.29, size = 87, normalized size = 0.45 \begin {gather*} \frac {2 \, {\left (5 \, {\left (9 \, b^{3} \sqrt {d} x^{3} + 13 \, a b^{2} \sqrt {d} x\right )} x^{\frac {7}{2}} + 26 \, {\left (5 \, a b^{2} \sqrt {d} x^{3} + 9 \, a^{2} b \sqrt {d} x\right )} x^{\frac {3}{2}} + \frac {117 \, {\left (a^{2} b \sqrt {d} x^{3} + 5 \, a^{3} \sqrt {d} x\right )}}{\sqrt {x}}\right )}}{585 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="maxima")

[Out]

2/585*(5*(9*b^3*sqrt(d)*x^3 + 13*a*b^2*sqrt(d)*x)*x^(7/2) + 26*(5*a*b^2*sqrt(d)*x^3 + 9*a^2*b*sqrt(d)*x)*x^(3/
2) + 117*(a^2*b*sqrt(d)*x^3 + 5*a^3*sqrt(d)*x)/sqrt(x))/d

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Fricas [A]
time = 0.35, size = 42, normalized size = 0.22 \begin {gather*} \frac {2 \, {\left (15 \, b^{3} x^{6} + 65 \, a b^{2} x^{4} + 117 \, a^{2} b x^{2} + 195 \, a^{3}\right )} \sqrt {d x}}{195 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="fricas")

[Out]

2/195*(15*b^3*x^6 + 65*a*b^2*x^4 + 117*a^2*b*x^2 + 195*a^3)*sqrt(d*x)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/(d*x)**(1/2),x)

[Out]

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

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Giac [A]
time = 3.08, size = 89, normalized size = 0.46 \begin {gather*} \frac {2 \, {\left (15 \, \sqrt {d x} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 65 \, \sqrt {d x} a b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 117 \, \sqrt {d x} a^{2} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 195 \, \sqrt {d x} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{195 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="giac")

[Out]

2/195*(15*sqrt(d*x)*b^3*x^6*sgn(b*x^2 + a) + 65*sqrt(d*x)*a*b^2*x^4*sgn(b*x^2 + a) + 117*sqrt(d*x)*a^2*b*x^2*s
gn(b*x^2 + a) + 195*sqrt(d*x)*a^3*sgn(b*x^2 + a))/d

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Mupad [B]
time = 4.50, size = 76, normalized size = 0.39 \begin {gather*} \frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {6\,a^2\,x^3}{5}+\frac {2\,b^2\,x^7}{13}+\frac {2\,a^3\,x}{b}+\frac {2\,a\,b\,x^5}{3}\right )}{x^2\,\sqrt {d\,x}+\frac {a\,\sqrt {d\,x}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)/(d*x)^(1/2),x)

[Out]

((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*((6*a^2*x^3)/5 + (2*b^2*x^7)/13 + (2*a^3*x)/b + (2*a*b*x^5)/3))/(x^2*(d*x)^
(1/2) + (a*(d*x)^(1/2))/b)

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