3.12.24 \(\int \frac {c+d x^2}{(e x)^{5/2} (a+b x^2)^{7/4}} \, dx\) [1124]

Optimal. Leaf size=144 \[ -\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}+\frac {4 \sqrt {b} (2 b c-a d) \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 a^{5/2} e^4 \left (a+b x^2\right )^{3/4}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 144, 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 = {464, 296, 335, 243, 342, 281, 237} \begin {gather*} \frac {4 \sqrt {b} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (2 b c-a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 a^{5/2} e^4 \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt {e x} (2 b c-a d)}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c)/(3*a*e*(e*x)^(3/2)*(a + b*x^2)^(3/4)) - (2*(2*b*c - a*d)*Sqrt[e*x])/(3*a^2*e^3*(a + b*x^2)^(3/4)) + (4*
Sqrt[b]*(2*b*c - a*d)*(1 + a/(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(3*a^(5/2
)*e^4*(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 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 464

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

Rubi steps

\begin {align*} \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{7/4}} \, dx &=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-a d) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{7/4}} \, dx}{a e^2}\\ &=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}-\frac {(2 (2 b c-a d)) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx}{3 a^2 e^2}\\ &=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}-\frac {(4 (2 b c-a d)) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{3 a^2 e^3}\\ &=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}-\frac {\left (4 (2 b c-a d) \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 )}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}+\frac {\left (4 (2 b c-a d) \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 )}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}+\frac {\left (2 (2 b c-a d) \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 )}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}\\ &=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 (2 b c-a d) \sqrt {e x}}{3 a^2 e^3 \left (a+b x^2\right )^{3/4}}+\frac {4 \sqrt {b} (2 b c-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 )}{3 a^{5/2} e^4 \left (a+b x^2\right )^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 91, normalized size = 0.63 \begin {gather*} \frac {x \left (-2 a c-4 b c x^2+2 a d x^2+4 (-2 b c+a d) x^2 \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 )}{3 a^2 (e x)^{5/2} \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {7}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

e^(-5/2)*integrate((d*x^2 + c)/((b*x^2 + a)^(7/4)*x^(5/2)), x)

________________________________________________________________________________________

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((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*(d*x^2 + c)*sqrt(x)*e^(-5/2)/(b^2*x^7 + 2*a*b*x^5 + a^2*x^3), x)

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 59.63, size = 97, normalized size = 0.67 \begin {gather*} \frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {7}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} + \frac {d \sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} e^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)/(e*x)**(5/2)/(b*x**2+a)**(7/4),x)

[Out]

c*gamma(-3/4)*hyper((-3/4, 7/4), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*e**(5/2)*x**(3/2)*gamma(1/4)) +
d*sqrt(x)*gamma(1/4)*hyper((1/4, 7/4), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*e**(5/2)*gamma(5/4))

________________________________________________________________________________________

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((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*e^(-5/2)/((b*x^2 + a)^(7/4)*x^(5/2)), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {d\,x^2+c}{{\left (e\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{7/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________