3.613 \(\int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=156 \[ \frac {5 a^{7/4} c^{7/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a+b x^2}}-\frac {10 a c^3 \sqrt {c x} \sqrt {a+b x^2}}{21 b^2}+\frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b} \]

[Out]

2/7*c*(c*x)^(5/2)*(b*x^2+a)^(1/2)/b-10/21*a*c^3*(c*x)^(1/2)*(b*x^2+a)^(1/2)/b^2+5/21*a^(7/4)*c^(7/2)*(cos(2*ar
ctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*Ellipti
cF(sin(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b
^(1/2))^2)^(1/2)/b^(9/4)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {321, 329, 220} \[ \frac {5 a^{7/4} c^{7/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a+b x^2}}-\frac {10 a c^3 \sqrt {c x} \sqrt {a+b x^2}}{21 b^2}+\frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)/Sqrt[a + b*x^2],x]

[Out]

(-10*a*c^3*Sqrt[c*x]*Sqrt[a + b*x^2])/(21*b^2) + (2*c*(c*x)^(5/2)*Sqrt[a + b*x^2])/(7*b) + (5*a^(7/4)*c^(7/2)*
(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4
)*Sqrt[c])], 1/2])/(21*b^(9/4)*Sqrt[a + b*x^2])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {(c x)^{7/2}}{\sqrt {a+b x^2}} \, dx &=\frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b}-\frac {\left (5 a c^2\right ) \int \frac {(c x)^{3/2}}{\sqrt {a+b x^2}} \, dx}{7 b}\\ &=-\frac {10 a c^3 \sqrt {c x} \sqrt {a+b x^2}}{21 b^2}+\frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b}+\frac {\left (5 a^2 c^4\right ) \int \frac {1}{\sqrt {c x} \sqrt {a+b x^2}} \, dx}{21 b^2}\\ &=-\frac {10 a c^3 \sqrt {c x} \sqrt {a+b x^2}}{21 b^2}+\frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b}+\frac {\left (10 a^2 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{21 b^2}\\ &=-\frac {10 a c^3 \sqrt {c x} \sqrt {a+b x^2}}{21 b^2}+\frac {2 c (c x)^{5/2} \sqrt {a+b x^2}}{7 b}+\frac {5 a^{7/4} c^{7/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{21 b^{9/4} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 87, normalized size = 0.56 \[ \frac {2 c^3 \sqrt {c x} \left (5 a^2 \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )-5 a^2-2 a b x^2+3 b^2 x^4\right )}{21 b^2 \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^3*Sqrt[c*x]*(-5*a^2 - 2*a*b*x^2 + 3*b^2*x^4 + 5*a^2*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4,
-((b*x^2)/a)]))/(21*b^2*Sqrt[a + b*x^2])

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x} c^{3} x^{3}}{\sqrt {b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x)*c^3*x^3/sqrt(b*x^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {7}{2}}}{\sqrt {b x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^(7/2)/sqrt(b*x^2 + a), x)

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maple [A]  time = 0.03, size = 141, normalized size = 0.90 \[ \frac {\sqrt {c x}\, \left (6 b^{3} x^{5}-4 a \,b^{2} x^{3}-10 a^{2} b x +5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, a^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )\right ) c^{3}}{21 \sqrt {b \,x^{2}+a}\, b^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*c^3/x*(c*x)^(1/2)/(b*x^2+a)^(1/2)*(5*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))
/(-a*b)^(1/2))^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2)
)*(-a*b)^(1/2)*a^2+6*b^3*x^5-4*a*b^2*x^3-10*a^2*b*x)/b^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {7}{2}}}{\sqrt {b x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x)^(7/2)/sqrt(b*x^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x\right )}^{7/2}}{\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)/(a + b*x^2)^(1/2),x)

[Out]

int((c*x)^(7/2)/(a + b*x^2)^(1/2), x)

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sympy [C]  time = 19.57, size = 44, normalized size = 0.28 \[ \frac {c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {13}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**(7/2)*x**(9/2)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*gamma(13/4))

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