3.6.70 \(\int \frac {(a^2+2 a b x^2+b^2 x^4)^{5/2}}{(d x)^{7/2}} \, dx\)

Optimal. Leaf size=295 \[ \frac {2 b^5 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {10 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.08, antiderivative size = 295, 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 = {1112, 270} \begin {gather*} \frac {2 b^5 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}-\frac {10 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \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)^(5/2)/(d*x)^(7/2),x]

[Out]

(-2*a^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(d*x)^(5/2)*(a + b*x^2)) - (10*a^4*b*Sqrt[a^2 + 2*a*b*x^2 + b^2*
x^4])/(d^3*Sqrt[d*x]*(a + b*x^2)) + (20*a^3*b^2*(d*x)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^5*(a + b*x^2
)) + (20*a^2*b^3*(d*x)^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*d^7*(a + b*x^2)) + (10*a*b^4*(d*x)^(11/2)*Sqr
t[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*d^9*(a + b*x^2)) + (2*b^5*(d*x)^(15/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(15*
d^11*(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 \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (a b+b^2 x^2\right )^5}{(d x)^{7/2}} \, dx}{b^4 \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^5 b^5}{(d x)^{7/2}}+\frac {5 a^4 b^6}{d^2 (d x)^{3/2}}+\frac {10 a^3 b^7 \sqrt {d x}}{d^4}+\frac {10 a^2 b^8 (d x)^{5/2}}{d^6}+\frac {5 a b^9 (d x)^{9/2}}{d^8}+\frac {b^{10} (d x)^{13/2}}{d^{10}}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )}\\ &=-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {10 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 88, normalized size = 0.30 \begin {gather*} \frac {2 x \sqrt {\left (a+b x^2\right )^2} \left (-231 a^5-5775 a^4 b x^2+3850 a^3 b^2 x^4+1650 a^2 b^3 x^6+525 a b^4 x^8+77 b^5 x^{10}\right )}{1155 (d x)^{7/2} \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)^(5/2)/(d*x)^(7/2),x]

[Out]

(2*x*Sqrt[(a + b*x^2)^2]*(-231*a^5 - 5775*a^4*b*x^2 + 3850*a^3*b^2*x^4 + 1650*a^2*b^3*x^6 + 525*a*b^4*x^8 + 77
*b^5*x^10))/(1155*(d*x)^(7/2)*(a + b*x^2))

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IntegrateAlgebraic [A]  time = 61.31, size = 124, normalized size = 0.42 \begin {gather*} \frac {2 \left (a d^2+b d^2 x^2\right ) \left (-231 a^5 d^{10}-5775 a^4 b d^{10} x^2+3850 a^3 b^2 d^{10} x^4+1650 a^2 b^3 d^{10} x^6+525 a b^4 d^{10} x^8+77 b^5 d^{10} x^{10}\right )}{1155 d^{13} (d x)^{5/2} \sqrt {\frac {\left (a d^2+b d^2 x^2\right )^2}{d^4}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(d*x)^(7/2),x]

[Out]

(2*(a*d^2 + b*d^2*x^2)*(-231*a^5*d^10 - 5775*a^4*b*d^10*x^2 + 3850*a^3*b^2*d^10*x^4 + 1650*a^2*b^3*d^10*x^6 +
525*a*b^4*d^10*x^8 + 77*b^5*d^10*x^10))/(1155*d^13*(d*x)^(5/2)*Sqrt[(a*d^2 + b*d^2*x^2)^2/d^4])

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fricas [A]  time = 0.81, size = 67, normalized size = 0.23 \begin {gather*} \frac {2 \, {\left (77 \, b^{5} x^{10} + 525 \, a b^{4} x^{8} + 1650 \, a^{2} b^{3} x^{6} + 3850 \, a^{3} b^{2} x^{4} - 5775 \, a^{4} b x^{2} - 231 \, a^{5}\right )} \sqrt {d x}}{1155 \, d^{4} x^{3}} \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)^(5/2)/(d*x)^(7/2),x, algorithm="fricas")

[Out]

2/1155*(77*b^5*x^10 + 525*a*b^4*x^8 + 1650*a^2*b^3*x^6 + 3850*a^3*b^2*x^4 - 5775*a^4*b*x^2 - 231*a^5)*sqrt(d*x
)/(d^4*x^3)

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giac [A]  time = 0.20, size = 162, normalized size = 0.55 \begin {gather*} -\frac {2 \, {\left (\frac {231 \, {\left (25 \, a^{4} b d^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} d^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{\sqrt {d x} d^{2} x^{2}} - \frac {77 \, \sqrt {d x} b^{5} d^{105} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 525 \, \sqrt {d x} a b^{4} d^{105} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 1650 \, \sqrt {d x} a^{2} b^{3} d^{105} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 3850 \, \sqrt {d x} a^{3} b^{2} d^{105} x \mathrm {sgn}\left (b x^{2} + a\right )}{d^{105}}\right )}}{1155 \, d^{4}} \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)^(5/2)/(d*x)^(7/2),x, algorithm="giac")

[Out]

-2/1155*(231*(25*a^4*b*d^3*x^2*sgn(b*x^2 + a) + a^5*d^3*sgn(b*x^2 + a))/(sqrt(d*x)*d^2*x^2) - (77*sqrt(d*x)*b^
5*d^105*x^7*sgn(b*x^2 + a) + 525*sqrt(d*x)*a*b^4*d^105*x^5*sgn(b*x^2 + a) + 1650*sqrt(d*x)*a^2*b^3*d^105*x^3*s
gn(b*x^2 + a) + 3850*sqrt(d*x)*a^3*b^2*d^105*x*sgn(b*x^2 + a))/d^105)/d^4

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maple [A]  time = 0.01, size = 83, normalized size = 0.28 \begin {gather*} -\frac {2 \left (-77 b^{5} x^{10}-525 a \,b^{4} x^{8}-1650 a^{2} b^{3} x^{6}-3850 a^{3} b^{2} x^{4}+5775 a^{4} b \,x^{2}+231 a^{5}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} x}{1155 \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/(d*x)^(7/2),x)

[Out]

-2/1155*x*(-77*b^5*x^10-525*a*b^4*x^8-1650*a^2*b^3*x^6-3850*a^3*b^2*x^4+5775*a^4*b*x^2+231*a^5)*((b*x^2+a)^2)^
(5/2)/(b*x^2+a)^5/(d*x)^(7/2)

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maxima [A]  time = 1.55, size = 150, normalized size = 0.51 \begin {gather*} \frac {2 \, {\left (7 \, {\left (11 \, b^{5} \sqrt {d} x^{3} + 15 \, a b^{4} \sqrt {d} x\right )} x^{\frac {9}{2}} + 60 \, {\left (7 \, a b^{4} \sqrt {d} x^{3} + 11 \, a^{2} b^{3} \sqrt {d} x\right )} x^{\frac {5}{2}} + 330 \, {\left (3 \, a^{2} b^{3} \sqrt {d} x^{3} + 7 \, a^{3} b^{2} \sqrt {d} x\right )} \sqrt {x} + \frac {1540 \, {\left (a^{3} b^{2} \sqrt {d} x^{3} - 3 \, a^{4} b \sqrt {d} x\right )}}{x^{\frac {3}{2}}} - \frac {231 \, {\left (5 \, a^{4} b \sqrt {d} x^{3} + a^{5} \sqrt {d} x\right )}}{x^{\frac {7}{2}}}\right )}}{1155 \, d^{4}} \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)^(5/2)/(d*x)^(7/2),x, algorithm="maxima")

[Out]

2/1155*(7*(11*b^5*sqrt(d)*x^3 + 15*a*b^4*sqrt(d)*x)*x^(9/2) + 60*(7*a*b^4*sqrt(d)*x^3 + 11*a^2*b^3*sqrt(d)*x)*
x^(5/2) + 330*(3*a^2*b^3*sqrt(d)*x^3 + 7*a^3*b^2*sqrt(d)*x)*sqrt(x) + 1540*(a^3*b^2*sqrt(d)*x^3 - 3*a^4*b*sqrt
(d)*x)/x^(3/2) - 231*(5*a^4*b*sqrt(d)*x^3 + a^5*sqrt(d)*x)/x^(7/2))/d^4

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mupad [B]  time = 4.72, size = 118, normalized size = 0.40 \begin {gather*} \frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {2\,b^4\,x^{10}}{15\,d^3}-\frac {10\,a^4\,x^2}{d^3}-\frac {2\,a^5}{5\,b\,d^3}+\frac {20\,a^3\,b\,x^4}{3\,d^3}+\frac {10\,a\,b^3\,x^8}{11\,d^3}+\frac {20\,a^2\,b^2\,x^6}{7\,d^3}\right )}{x^4\,\sqrt {d\,x}+\frac {a\,x^2\,\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)^(5/2)/(d*x)^(7/2),x)

[Out]

((a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2)*((2*b^4*x^10)/(15*d^3) - (10*a^4*x^2)/d^3 - (2*a^5)/(5*b*d^3) + (20*a^3*b*x
^4)/(3*d^3) + (10*a*b^3*x^8)/(11*d^3) + (20*a^2*b^2*x^6)/(7*d^3)))/(x^4*(d*x)^(1/2) + (a*x^2*(d*x)^(1/2))/b)

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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 {5}{2}}}{\left (d x\right )^{\frac {7}{2}}}\, 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)**(5/2)/(d*x)**(7/2),x)

[Out]

Integral(((a + b*x**2)**2)**(5/2)/(d*x)**(7/2), x)

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