3.1102 \(\int \frac {c+d x^2}{(e x)^{5/2} (a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=107 \[ \frac {2 \sqrt {b} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (2 b c-3 a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 a^{3/2} e^4 \left (a+b x^2\right )^{3/4}}-\frac {2 c \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {453, 329, 237, 335, 275, 231} \[ \frac {2 \sqrt {b} (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (2 b c-3 a d) F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 a^{3/2} e^4 \left (a+b x^2\right )^{3/4}}-\frac {2 c \sqrt [4]{a+b x^2}}{3 a e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 231

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

Rule 237

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 275

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 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]

Rule 335

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 453

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

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Mathematica [C]  time = 0.04, size = 84, normalized size = 0.79 \[ \frac {x \left (2 x^2 \left (\frac {b x^2}{a}+1\right )^{3/4} (3 a d-2 b c) \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};-\frac {b x^2}{a}\right )-2 c \left (a+b x^2\right )\right )}{3 a (e x)^{5/2} \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} {\left (d x^{2} + c\right )} \sqrt {e x}}{b e^{3} x^{5} + a e^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {5}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {d\,x^2+c}{{\left (e\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 24.54, size = 82, normalized size = 0.77 \[ - \frac {d {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{b^{\frac {3}{4}} e^{\frac {5}{2}} x} + \frac {c \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} e^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)/(e*x)**(5/2)/(b*x**2+a)**(3/4),x)

[Out]

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

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