3.10.19 \(\int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx\) [919]

Optimal. Leaf size=152 \[ -\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac {a^{5/2} c^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{12 b^{3/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

1/30*a*c*(c*x)^(5/2)*(b*x^2+a)^(1/4)/b+1/5*(c*x)^(9/2)*(b*x^2+a)^(1/4)/c-1/12*a^(5/2)*c^2*(1+a/b/x^2)^(3/4)*(c
*x)^(3/2)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*ar
ccot(x*b^(1/2)/a^(1/2))),2^(1/2))/b^(3/2)/(b*x^2+a)^(3/4)-1/12*a^2*c^3*(b*x^2+a)^(1/4)*(c*x)^(1/2)/b^2

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Rubi [A]
time = 0.08, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {285, 327, 335, 243, 342, 281, 237} \begin {gather*} -\frac {a^{5/2} c^2 (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{12 b^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)*(a + b*x^2)^(1/4),x]

[Out]

-1/12*(a^2*c^3*Sqrt[c*x]*(a + b*x^2)^(1/4))/b^2 + (a*c*(c*x)^(5/2)*(a + b*x^2)^(1/4))/(30*b) + ((c*x)^(9/2)*(a
+ b*x^2)^(1/4))/(5*c) - (a^(5/2)*c^2*(1 + a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/
2, 2])/(12*b^(3/2)*(a + b*x^2)^(3/4))

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 281

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

Rule 285

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

Rule 327

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 335

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]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int (c x)^{7/2} \sqrt [4]{a+b x^2} \, dx &=\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac {1}{10} a \int \frac {(c x)^{7/2}}{\left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac {\left (a^2 c^2\right ) \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{12 b}\\ &=-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac {\left (a^3 c^4\right ) \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx}{24 b^2}\\ &=-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac {\left (a^3 c^3\right ) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{c^2}\right )^{3/4}} \, dx,x,\sqrt {c x}\right )}{12 b^2}\\ &=-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}+\frac {\left (a^3 c^3 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {c x}\right )}{12 b^2 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac {\left (a^3 c^3 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a c^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {c x}}\right )}{12 b^2 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac {\left (a^3 c^3 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{c x}\right )}{24 b^2 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}{12 b^2}+\frac {a c (c x)^{5/2} \sqrt [4]{a+b x^2}}{30 b}+\frac {(c x)^{9/2} \sqrt [4]{a+b x^2}}{5 c}-\frac {a^{5/2} c^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{12 b^{3/2} \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 102, normalized size = 0.67 \begin {gather*} \frac {c^3 \sqrt {c x} \sqrt [4]{a+b x^2} \left (\sqrt [4]{1+\frac {b x^2}{a}} \left (-5 a^2+a b x^2+6 b^2 x^4\right )+5 a^2 \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )\right )}{30 b^2 \sqrt [4]{1+\frac {b x^2}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(7/2)*(a + b*x^2)^(1/4),x]

[Out]

(c^3*Sqrt[c*x]*(a + b*x^2)^(1/4)*((1 + (b*x^2)/a)^(1/4)*(-5*a^2 + a*b*x^2 + 6*b^2*x^4) + 5*a^2*Hypergeometric2
F1[-1/4, 1/4, 5/4, -((b*x^2)/a)]))/(30*b^2*(1 + (b*x^2)/a)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (c x \right )^{\frac {7}{2}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)*(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(7/2)*(b*x^2+a)^(1/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)*(c*x)^(7/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*sqrt(c*x)*c^3*x^3, x)

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Sympy [C] Result contains complex when optimal does not.
time = 16.98, size = 46, normalized size = 0.30 \begin {gather*} \frac {\sqrt [4]{a} c^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(7/2)*(b*x**2+a)**(1/4),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(7/2)*(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/4)*(c*x)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (c\,x\right )}^{7/2}\,{\left (b\,x^2+a\right )}^{1/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(7/2)*(a + b*x^2)^(1/4),x)

[Out]

int((c*x)^(7/2)*(a + b*x^2)^(1/4), x)

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