3.11.100 \(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{3/4}} \, dx\) [1100]

Optimal. Leaf size=139 \[ \frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\sqrt {a} (6 b c-5 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 b^{3/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

1/3*d*(e*x)^(5/2)*(b*x^2+a)^(1/4)/b/e+1/6*(-5*a*d+6*b*c)*(1+a/b/x^2)^(3/4)*(e*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*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))
*a^(1/2)/b^(3/2)/(b*x^2+a)^(3/4)+1/6*(-5*a*d+6*b*c)*e*(b*x^2+a)^(1/4)*(e*x)^(1/2)/b^2

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Rubi [A]
time = 0.07, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {470, 327, 335, 243, 342, 281, 237} \begin {gather*} \frac {\sqrt {a} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (6 b c-5 a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 b^{3/2} \left (a+b x^2\right )^{3/4}}+\frac {e \sqrt {e x} \sqrt [4]{a+b x^2} (6 b c-5 a d)}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((6*b*c - 5*a*d)*e*Sqrt[e*x]*(a + b*x^2)^(1/4))/(6*b^2) + (d*(e*x)^(5/2)*(a + b*x^2)^(1/4))/(3*b*e) + (Sqrt[a]
*(6*b*c - 5*a*d)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(6*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 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]

Rule 470

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

Rubi steps

\begin {align*} \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx &=\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {\left (-3 b c+\frac {5 a d}{2}\right ) \int \frac {(e x)^{3/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{3 b}\\ &=\frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {\left (a (6 b c-5 a d) e^2\right ) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx}{12 b^2}\\ &=\frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {(a (6 b c-5 a d) e) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{6 b^2}\\ &=\frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}-\frac {\left (a (6 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {e x}\right )}{6 b^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\left (a (6 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a e^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{6 b^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\left (a (6 b c-5 a d) e \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{e x}\right )}{12 b^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {(6 b c-5 a d) e \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^2}+\frac {d (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 b e}+\frac {\sqrt {a} (6 b c-5 a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 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 = 10.09, size = 97, normalized size = 0.70 \begin {gather*} \frac {e \sqrt {e x} \left (-\left (\left (a+b x^2\right ) \left (5 a d-2 b \left (3 c+d x^2\right )\right )\right )+a (-6 b c+5 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};-\frac {b x^2}{a}\right )\right )}{6 b^2 \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[e*x]*(-((a + b*x^2)*(5*a*d - 2*b*(3*c + d*x^2))) + a*(-6*b*c + 5*a*d)*(1 + (b*x^2)/a)^(3/4)*Hypergeome
tric2F1[1/4, 3/4, 5/4, -((b*x^2)/a)]))/(6*b^2*(a + b*x^2)^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(3/4),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((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 4.94, size = 94, normalized size = 0.68 \begin {gather*} \frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {9}{4}\right )} + \frac {d e^{\frac {3}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {13}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((3/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(9/4)) + d
*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((3/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*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((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*x^(3/2)*e^(3/2)/(b*x^2 + a)^(3/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^{3/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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