3.1093 \(\int \frac {(e x)^{5/2} (c+d x^2)}{(a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=173 \[ \frac {3 a e^{5/2} (8 b c-7 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac {3 a e^{5/2} (8 b c-7 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}+\frac {e (e x)^{3/2} \sqrt [4]{a+b x^2} (8 b c-7 a d)}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e} \]

[Out]

1/16*(-7*a*d+8*b*c)*e*(e*x)^(3/2)*(b*x^2+a)^(1/4)/b^2+1/4*d*(e*x)^(7/2)*(b*x^2+a)^(1/4)/b/e+3/32*a*(-7*a*d+8*b
*c)*e^(5/2)*arctan(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))/b^(11/4)-3/32*a*(-7*a*d+8*b*c)*e^(5/2)*arctanh
(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))/b^(11/4)

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Rubi [A]  time = 0.13, antiderivative size = 173, 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 = {459, 321, 329, 331, 298, 205, 208} \[ \frac {3 a e^{5/2} (8 b c-7 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac {3 a e^{5/2} (8 b c-7 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}+\frac {e (e x)^{3/2} \sqrt [4]{a+b x^2} (8 b c-7 a d)}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*b*c - 7*a*d)*e*(e*x)^(3/2)*(a + b*x^2)^(1/4))/(16*b^2) + (d*(e*x)^(7/2)*(a + b*x^2)^(1/4))/(4*b*e) + (3*a*
(8*b*c - 7*a*d)*e^(5/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(32*b^(11/4)) - (3*a*(8*b*c -
7*a*d)*e^(5/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(32*b^(11/4))

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 321

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 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 459

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)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx &=\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {\left (-4 b c+\frac {7 a d}{2}\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{4 b}\\ &=\frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {\left (3 a (8 b c-7 a d) e^2\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{3/4}} \, dx}{32 b^2}\\ &=\frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {(3 a (8 b c-7 a d) e) \operatorname {Subst}\left (\int \frac {x^2}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{16 b^2}\\ &=\frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {(3 a (8 b c-7 a d) e) \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 )}{16 b^2}\\ &=\frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac {\left (3 a (8 b c-7 a d) e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e-\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{5/2}}+\frac {\left (3 a (8 b c-7 a d) e^3\right ) \operatorname {Subst}\left (\int \frac {1}{e+\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{5/2}}\\ &=\frac {(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac {d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}+\frac {3 a (8 b c-7 a d) e^{5/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac {3 a (8 b c-7 a d) e^{5/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 131, normalized size = 0.76 \[ \frac {(e x)^{5/2} \left (2 b^{3/4} x^{3/2} \sqrt [4]{a+b x^2} \left (-7 a d+8 b c+4 b d x^2\right )-3 a (7 a d-8 b c) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+3 a (7 a d-8 b c) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{32 b^{11/4} x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x)^(5/2)*(2*b^(3/4)*x^(3/2)*(a + b*x^2)^(1/4)*(8*b*c - 7*a*d + 4*b*d*x^2) - 3*a*(-8*b*c + 7*a*d)*ArcTan[(b
^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + 3*a*(-8*b*c + 7*a*d)*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(32*b
^(11/4)*x^(5/2))

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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((e*x)^(5/2)*(d*x^2+c)/(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 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{\frac {5}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x)^(5/2)*(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 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 41.58, size = 94, normalized size = 0.54 \[ \frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{4}} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(11/4)) +
d*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(15/4
))

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