3.3.49 \(\int x^7 (c \sqrt {a+b x^2})^{3/2} \, dx\) [249]

Optimal. Leaf size=138 \[ -\frac {2 a^3 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )}{7 b^4}+\frac {6 a^2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^2}{11 b^4}-\frac {2 a \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^3}{5 b^4}+\frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right )^4}{19 b^4} \]

[Out]

-2/7*a^3*(b*x^2+a)*(c*(b*x^2+a)^(1/2))^(3/2)/b^4+6/11*a^2*(b*x^2+a)^2*(c*(b*x^2+a)^(1/2))^(3/2)/b^4-2/5*a*(b*x
^2+a)^3*(c*(b*x^2+a)^(1/2))^(3/2)/b^4+2/19*(b*x^2+a)^4*(c*(b*x^2+a)^(1/2))^(3/2)/b^4

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Rubi [A]
time = 0.07, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1973, 272, 45} \begin {gather*} -\frac {2 a^3 \left (a+b x^2\right ) \left (c \sqrt {a+b x^2}\right )^{3/2}}{7 b^4}+\frac {6 a^2 \left (a+b x^2\right )^2 \left (c \sqrt {a+b x^2}\right )^{3/2}}{11 b^4}+\frac {2 \left (a+b x^2\right )^4 \left (c \sqrt {a+b x^2}\right )^{3/2}}{19 b^4}-\frac {2 a \left (a+b x^2\right )^3 \left (c \sqrt {a+b x^2}\right )^{3/2}}{5 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(-2*a^3*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2))/(7*b^4) + (6*a^2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)^2)/(11*b
^4) - (2*a*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)^3)/(5*b^4) + (2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)^4)/(19*
b^4)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps

\begin {align*} \int x^7 \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \int x^7 \left (a+b x^2\right )^{3/4} \, dx}{\sqrt [4]{a+b x^2}}\\ &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \text {Subst}\left (\int x^3 (a+b x)^{3/4} \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \text {Subst}\left (\int \left (-\frac {a^3 (a+b x)^{3/4}}{b^3}+\frac {3 a^2 (a+b x)^{7/4}}{b^3}-\frac {3 a (a+b x)^{11/4}}{b^3}+\frac {(a+b x)^{15/4}}{b^3}\right ) \, dx,x,x^2\right )}{2 \sqrt [4]{a+b x^2}}\\ &=-\frac {2 a^3 c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{3/2}}{7 b^4}+\frac {6 a^2 c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{5/2}}{11 b^4}-\frac {2 a c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{7/2}}{5 b^4}+\frac {2 c \sqrt {c \sqrt {a+b x^2}} \left (a+b x^2\right )^{9/2}}{19 b^4}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 63, normalized size = 0.46 \begin {gather*} \frac {2 \left (c \sqrt {a+b x^2}\right )^{3/2} \left (a+b x^2\right ) \left (-128 a^3+224 a^2 b x^2-308 a b^2 x^4+385 b^3 x^6\right )}{7315 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^7*(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(2*(c*Sqrt[a + b*x^2])^(3/2)*(a + b*x^2)*(-128*a^3 + 224*a^2*b*x^2 - 308*a*b^2*x^4 + 385*b^3*x^6))/(7315*b^4)

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Maple [A]
time = 0.02, size = 58, normalized size = 0.42

method	result	size
gosper	\(-\frac {2 \left (b \,x^{2}+a \right ) \left (-385 b^{3} x^{6}+308 a \,b^{2} x^{4}-224 a^{2} b \,x^{2}+128 a^{3}\right ) \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}}{7315 b^{4}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(c*(b*x^2+a)^(1/2))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/7315*(b*x^2+a)*(-385*b^3*x^6+308*a*b^2*x^4-224*a^2*b*x^2+128*a^3)*(c*(b*x^2+a)^(1/2))^(3/2)/b^4

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Maxima [A]
time = 0.27, size = 85, normalized size = 0.62 \begin {gather*} -\frac {2 \, {\left (1045 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {7}{2}} a^{3} c^{6} - 1995 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {11}{2}} a^{2} c^{4} + 1463 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {15}{2}} a c^{2} - 385 \, \left (\sqrt {b x^{2} + a} c\right )^{\frac {19}{2}}\right )}}{7315 \, b^{4} c^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="maxima")

[Out]

-2/7315*(1045*(sqrt(b*x^2 + a)*c)^(7/2)*a^3*c^6 - 1995*(sqrt(b*x^2 + a)*c)^(11/2)*a^2*c^4 + 1463*(sqrt(b*x^2 +
a)*c)^(15/2)*a*c^2 - 385*(sqrt(b*x^2 + a)*c)^(19/2))/(b^4*c^8)

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Fricas [A]
time = 0.38, size = 75, normalized size = 0.54 \begin {gather*} \frac {2 \, {\left (385 \, b^{4} c x^{8} + 77 \, a b^{3} c x^{6} - 84 \, a^{2} b^{2} c x^{4} + 96 \, a^{3} b c x^{2} - 128 \, a^{4} c\right )} \sqrt {b x^{2} + a} \sqrt {\sqrt {b x^{2} + a} c}}{7315 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="fricas")

[Out]

2/7315*(385*b^4*c*x^8 + 77*a*b^3*c*x^6 - 84*a^2*b^2*c*x^4 + 96*a^3*b*c*x^2 - 128*a^4*c)*sqrt(b*x^2 + a)*sqrt(s
qrt(b*x^2 + a)*c)/b^4

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Sympy [A]
time = 12.49, size = 144, normalized size = 1.04 \begin {gather*} \begin {cases} - \frac {256 a^{4} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{7315 b^{4}} + \frac {192 a^{3} x^{2} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{7315 b^{3}} - \frac {24 a^{2} x^{4} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{1045 b^{2}} + \frac {2 a x^{6} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{95 b} + \frac {2 x^{8} \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{19} & \text {for}\: b \neq 0 \\\frac {x^{8} \left (\sqrt {a} c\right )^{\frac {3}{2}}}{8} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(c*(b*x**2+a)**(1/2))**(3/2),x)

[Out]

Piecewise((-256*a**4*(c*sqrt(a + b*x**2))**(3/2)/(7315*b**4) + 192*a**3*x**2*(c*sqrt(a + b*x**2))**(3/2)/(7315
*b**3) - 24*a**2*x**4*(c*sqrt(a + b*x**2))**(3/2)/(1045*b**2) + 2*a*x**6*(c*sqrt(a + b*x**2))**(3/2)/(95*b) +
2*x**8*(c*sqrt(a + b*x**2))**(3/2)/19, Ne(b, 0)), (x**8*(sqrt(a)*c)**(3/2)/8, True))

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Giac [A]
time = 4.80, size = 137, normalized size = 0.99 \begin {gather*} \frac {2 \, c^{\frac {3}{2}} {\left (\frac {19 \, {\left (77 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{3}\right )} a}{b^{3}} + \frac {1155 \, {\left (b x^{2} + a\right )}^{\frac {19}{4}} - 5852 \, {\left (b x^{2} + a\right )}^{\frac {15}{4}} a + 11970 \, {\left (b x^{2} + a\right )}^{\frac {11}{4}} a^{2} - 12540 \, {\left (b x^{2} + a\right )}^{\frac {7}{4}} a^{3} + 7315 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} a^{4}}{b^{3}}\right )}}{21945 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="giac")

[Out]

2/21945*c^(3/2)*(19*(77*(b*x^2 + a)^(15/4) - 315*(b*x^2 + a)^(11/4)*a + 495*(b*x^2 + a)^(7/4)*a^2 - 385*(b*x^2
+ a)^(3/4)*a^3)*a/b^3 + (1155*(b*x^2 + a)^(19/4) - 5852*(b*x^2 + a)^(15/4)*a + 11970*(b*x^2 + a)^(11/4)*a^2 -
12540*(b*x^2 + a)^(7/4)*a^3 + 7315*(b*x^2 + a)^(3/4)*a^4)/b^3)/b

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Mupad [B]
time = 2.96, size = 109, normalized size = 0.79 \begin {gather*} \sqrt {c\,\sqrt {b\,x^2+a}}\,\left (\frac {2\,c\,x^8\,\sqrt {b\,x^2+a}}{19}-\frac {256\,a^4\,c\,\sqrt {b\,x^2+a}}{7315\,b^4}+\frac {2\,a\,c\,x^6\,\sqrt {b\,x^2+a}}{95\,b}-\frac {24\,a^2\,c\,x^4\,\sqrt {b\,x^2+a}}{1045\,b^2}+\frac {192\,a^3\,c\,x^2\,\sqrt {b\,x^2+a}}{7315\,b^3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(c*(a + b*x^2)^(1/2))^(3/2),x)

[Out]

(c*(a + b*x^2)^(1/2))^(1/2)*((2*c*x^8*(a + b*x^2)^(1/2))/19 - (256*a^4*c*(a + b*x^2)^(1/2))/(7315*b^4) + (2*a*
c*x^6*(a + b*x^2)^(1/2))/(95*b) - (24*a^2*c*x^4*(a + b*x^2)^(1/2))/(1045*b^2) + (192*a^3*c*x^2*(a + b*x^2)^(1/
2))/(7315*b^3))

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