3.1095 \(\int \frac {c+d x^2}{(e x)^{3/2} (a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=113 \[ -\frac {d \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{3/4} e^{3/2}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{3/4} e^{3/2}}-\frac {2 c \sqrt [4]{a+b x^2}}{a e \sqrt {e x}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 113, 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 = {451, 329, 331, 298, 205, 208} \[ -\frac {d \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{3/4} e^{3/2}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{3/4} e^{3/2}}-\frac {2 c \sqrt [4]{a+b x^2}}{a e \sqrt {e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*(a + b*x^2)^(1/4))/(a*e*Sqrt[e*x]) - (d*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(b^(3/4
)*e^(3/2)) + (d*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(b^(3/4)*e^(3/2))

Rule 205

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

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 331

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

Rule 451

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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 100, normalized size = 0.88 \[ \frac {x \left (-2 b^{3/4} c \sqrt [4]{a+b x^2}-a d \sqrt {x} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+a d \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{a b^{3/4} (e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-2*b^(3/4)*c*(a + b*x^2)^(1/4) - a*d*Sqrt[x]*ArcTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + a*d*Sqrt[x]*Arc
Tanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(a*b^(3/4)*(e*x)^(3/2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 10.42, size = 85, normalized size = 0.75 \[ \frac {\sqrt [4]{b} c \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {1}{4}\right )}{2 a e^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right )} + \frac {d x^{\frac {3}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} e^{\frac {3}{2}} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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