3.12.27 \(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{9/4}} \, dx\) [1127]

Optimal. Leaf size=149 \[ \frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {d e^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}+\frac {d e^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}} \]

[Out]

2/5*(-a*d+b*c)*(e*x)^(5/2)/a/b/e/(b*x^2+a)^(5/4)+d*e^(3/2)*arctan(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))
/b^(9/4)+d*e^(3/2)*arctanh(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))/b^(9/4)-2*d*e*(e*x)^(1/2)/b^2/(b*x^2+a
)^(1/4)

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Rubi [A]
time = 0.06, antiderivative size = 149, 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 = {463, 294, 335, 246, 218, 214, 211} \begin {gather*} \frac {d e^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}+\frac {d e^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {2 (e x)^{5/2} (b c-a d)}{5 a b e \left (a+b x^2\right )^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(2*(b*c - a*d)*(e*x)^(5/2))/(5*a*b*e*(a + b*x^2)^(5/4)) - (2*d*e*Sqrt[e*x])/(b^2*(a + b*x^2)^(1/4)) + (d*e^(3/
2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/b^(9/4) + (d*e^(3/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(
Sqrt[e]*(a + b*x^2)^(1/4))])/b^(9/4)

Rule 211

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

Rule 214

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

Rule 218

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 463

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

Rubi steps

\begin {align*} \int \frac {(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{9/4}} \, dx &=\frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}+\frac {d \int \frac {(e x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{b}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {\left (d e^2\right ) \int \frac {1}{\sqrt {e x} \sqrt [4]{a+b x^2}} \, dx}{b^2}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {(2 d e) \text {Subst}\left (\int \frac {1}{\sqrt [4]{a+\frac {b x^4}{e^2}}} \, dx,x,\sqrt {e x}\right )}{b^2}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {(2 d e) \text {Subst}\left (\int \frac {1}{1-\frac {b x^4}{e^2}} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{b^2}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {\left (d e^2\right ) \text {Subst}\left (\int \frac {1}{e-\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{b^2}+\frac {\left (d e^2\right ) \text {Subst}\left (\int \frac {1}{e+\sqrt {b} x^2} \, dx,x,\frac {\sqrt {e x}}{\sqrt [4]{a+b x^2}}\right )}{b^2}\\ &=\frac {2 (b c-a d) (e x)^{5/2}}{5 a b e \left (a+b x^2\right )^{5/4}}-\frac {2 d e \sqrt {e x}}{b^2 \sqrt [4]{a+b x^2}}+\frac {d e^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}+\frac {d e^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{9/4}}\\ \end {align*}

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Mathematica [A]
time = 1.07, size = 124, normalized size = 0.83 \begin {gather*} \frac {e \sqrt {e x} \left (\frac {2 \sqrt [4]{b} \left (-5 a^2 d+b^2 c x^2-6 a b d x^2\right )}{a \left (a+b x^2\right )^{5/4}}+\frac {5 d \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )}{\sqrt {x}}+\frac {5 d \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )}{\sqrt {x}}\right )}{5 b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4),x]

[Out]

(e*Sqrt[e*x]*((2*b^(1/4)*(-5*a^2*d + b^2*c*x^2 - 6*a*b*d*x^2))/(a*(a + b*x^2)^(5/4)) + (5*d*ArcTan[(b^(1/4)*Sq
rt[x])/(a + b*x^2)^(1/4)])/Sqrt[x] + (5*d*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)])/Sqrt[x]))/(5*b^(9/4))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {9}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

[Out]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(9/4),x)

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Maxima [A]
time = 0.49, size = 129, normalized size = 0.87 \begin {gather*} -\frac {1}{10} \, {\left (d {\left (\frac {4 \, {\left (b + \frac {5 \, {\left (b x^{2} + a\right )}}{x^{2}}\right )} x^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{2}} + \frac {5 \, {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} \sqrt {x}}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {x}}}{b^{\frac {1}{4}} + \frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {x}}}\right )}{b^{\frac {1}{4}}}\right )}}{b^{2}}\right )} - \frac {4 \, c x^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} a}\right )} e^{\frac {3}{2}} \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)^(9/4),x, algorithm="maxima")

[Out]

-1/10*(d*(4*(b + 5*(b*x^2 + a)/x^2)*x^(5/2)/((b*x^2 + a)^(5/4)*b^2) + 5*(2*arctan((b*x^2 + a)^(1/4)/(b^(1/4)*s
qrt(x)))/b^(1/4) + log(-(b^(1/4) - (b*x^2 + a)^(1/4)/sqrt(x))/(b^(1/4) + (b*x^2 + a)^(1/4)/sqrt(x)))/b^(1/4))/
b^2) - 4*c*x^(5/2)/((b*x^2 + a)^(5/4)*a))*e^(3/2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 428 vs. \(2 (99) = 198\).
time = 1.98, size = 428, normalized size = 2.87 \begin {gather*} -\frac {4 \, {\left (5 \, a^{2} d - {\left (b^{2} c - 6 \, a b d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {x} e^{\frac {3}{2}} + 20 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left ({\left (b x^{2} + a\right )}^{\frac {3}{4}} b^{7} d \sqrt {x} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {3}{4}} e^{6} - {\left (b^{8} x^{2} + a b^{7}\right )} \sqrt {\frac {\sqrt {b x^{2} + a} d^{2} x e^{3} + {\left (b^{5} x^{2} + a b^{4}\right )} \sqrt {\frac {d^{4}}{b^{9}}} e^{3}}{b x^{2} + a}} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {3}{4}} e^{\frac {9}{2}}\right )} e^{\left (-6\right )}}{b d^{4} x^{2} + a d^{4}}\right ) e^{\frac {3}{2}} - 5 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} d \sqrt {x} e^{\frac {3}{2}} + {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}}}{b x^{2} + a}\right ) + 5 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} d \sqrt {x} e^{\frac {3}{2}} - {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}}}{b x^{2} + a}\right )}{10 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\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)^(9/4),x, algorithm="fricas")

[Out]

-1/10*(4*(5*a^2*d - (b^2*c - 6*a*b*d)*x^2)*(b*x^2 + a)^(3/4)*sqrt(x)*e^(3/2) + 20*(a*b^4*x^4 + 2*a^2*b^3*x^2 +
a^3*b^2)*(d^4/b^9)^(1/4)*arctan(-((b*x^2 + a)^(3/4)*b^7*d*sqrt(x)*(d^4/b^9)^(3/4)*e^6 - (b^8*x^2 + a*b^7)*sqr
t((sqrt(b*x^2 + a)*d^2*x*e^3 + (b^5*x^2 + a*b^4)*sqrt(d^4/b^9)*e^3)/(b*x^2 + a))*(d^4/b^9)^(3/4)*e^(9/2))*e^(-
6)/(b*d^4*x^2 + a*d^4))*e^(3/2) - 5*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(d^4/b^9)^(1/4)*e^(3/2)*log(((b*x^2
+ a)^(3/4)*d*sqrt(x)*e^(3/2) + (b^3*x^2 + a*b^2)*(d^4/b^9)^(1/4)*e^(3/2))/(b*x^2 + a)) + 5*(a*b^4*x^4 + 2*a^2*
b^3*x^2 + a^3*b^2)*(d^4/b^9)^(1/4)*e^(3/2)*log(((b*x^2 + a)^(3/4)*d*sqrt(x)*e^(3/2) - (b^3*x^2 + a*b^2)*(d^4/b
^9)^(1/4)*e^(3/2))/(b*x^2 + a)))/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)

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Sympy [C] Result contains complex when optimal does not.
time = 70.32, size = 116, normalized size = 0.78 \begin {gather*} \frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right )}{2 a^{\frac {9}{4}} \sqrt [4]{1 + \frac {b x^{2}}{a}} \Gamma \left (\frac {9}{4}\right ) + 2 a^{\frac {5}{4}} b x^{2} \sqrt [4]{1 + \frac {b x^{2}}{a}} \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 {9}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {9}{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)**(9/4),x)

[Out]

c*e**(3/2)*x**(5/2)*gamma(5/4)/(2*a**(9/4)*(1 + b*x**2/a)**(1/4)*gamma(9/4) + 2*a**(5/4)*b*x**2*(1 + b*x**2/a)
**(1/4)*gamma(9/4)) + d*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((9/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a
**(9/4)*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)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*x^(3/2)*e^(3/2)/(b*x^2 + a)^(9/4), 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 )}^{9/4}} \,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)^(9/4),x)

[Out]

int(((e*x)^(3/2)*(c + d*x^2))/(a + b*x^2)^(9/4), x)

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