3.12.5 \(\int \frac {(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\) [1105]

Optimal. Leaf size=171 \[ -\frac {(4 b c-5 a d) e \sqrt {e x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}+\frac {(4 b c-5 a d) e^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac {(4 b c-5 a d) e^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 171, 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 = {470, 294, 335, 246, 218, 214, 211} \begin {gather*} \frac {e^{3/2} (4 b c-5 a d) \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac {e^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac {e \sqrt {e x} (4 b c-5 a d)}{2 b^2 \sqrt [4]{a+b x^2}}+\frac {d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((4*b*c - 5*a*d)*e*Sqrt[e*x])/(b^2*(a + b*x^2)^(1/4)) + (d*(e*x)^(5/2))/(2*b*e*(a + b*x^2)^(1/4)) + ((4*b
*c - 5*a*d)*e^(3/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(9/4)) + ((4*b*c - 5*a*d)*e^
(3/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(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 470

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

________________________________________________________________________________________

Mathematica [A]
time = 0.77, size = 148, normalized size = 0.87 \begin {gather*} \frac {(e x)^{3/2} \left (2 \sqrt [4]{b} \sqrt {x} \left (-4 b c+5 a d+b d x^2\right )+(4 b c-5 a d) \sqrt [4]{a+b x^2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+(4 b c-5 a d) \sqrt [4]{a+b x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{9/4} x^{3/2} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{\frac {3}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 229, normalized size = 1.34 \begin {gather*} \frac {1}{8} \, {\left (d {\left (\frac {4 \, {\left (4 \, a b - \frac {5 \, {\left (b x^{2} + a\right )} a}{x^{2}}\right )}}{\frac {{\left (b x^{2} + a\right )}^{\frac {1}{4}} b^{3}}{\sqrt {x}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{4}} b^{2}}{x^{\frac {5}{2}}}} + \frac {5 \, a {\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 )} - 4 \, c {\left (\frac {\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}}}}{b} + \frac {4 \, \sqrt {x}}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} b}\right )}\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)^(5/4),x, algorithm="maxima")

[Out]

1/8*(d*(4*(4*a*b - 5*(b*x^2 + a)*a/x^2)/((b*x^2 + a)^(1/4)*b^3/sqrt(x) - (b*x^2 + a)^(5/4)*b^2/x^(5/2)) + 5*a*
(2*arctan((b*x^2 + a)^(1/4)/(b^(1/4)*sqrt(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*((2*arctan((b*x^2 + a)^(1/4)/(b^(1/4)*sqrt(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 + 4*sqrt(x)/((b*x^2
+ a)^(1/4)*b)))*e^(3/2)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 891 vs. \(2 (115) = 230\).
time = 1.68, size = 891, normalized size = 5.21 \begin {gather*} \frac {4 \, {\left (b d x^{2} - 4 \, b c + 5 \, a d\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {x} e^{\frac {3}{2}} + 4 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {{\left ({\left (4 \, b^{8} c - 5 \, a b^{7} d\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {x} \left (\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}\right )^{\frac {3}{4}} e^{6} + {\left (b^{8} x^{2} + a b^{7}\right )} \sqrt {\frac {{\left (16 \, b^{2} c^{2} - 40 \, a b c d + 25 \, a^{2} d^{2}\right )} \sqrt {b x^{2} + a} x e^{3} + {\left (b^{5} x^{2} + a b^{4}\right )} \sqrt {\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}} e^{3}}{b x^{2} + a}} \left (\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}\right )^{\frac {3}{4}} e^{\frac {9}{2}}\right )} e^{\left (-6\right )}}{256 \, a b^{4} c^{4} - 1280 \, a^{2} b^{3} c^{3} d + 2400 \, a^{3} b^{2} c^{2} d^{2} - 2000 \, a^{4} b c d^{3} + 625 \, a^{5} d^{4} + {\left (256 \, b^{5} c^{4} - 1280 \, a b^{4} c^{3} d + 2400 \, a^{2} b^{3} c^{2} d^{2} - 2000 \, a^{3} b^{2} c d^{3} + 625 \, a^{4} b d^{4}\right )} x^{2}}\right ) e^{\frac {3}{2}} + {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}} \log \left (-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (4 \, b c - 5 \, a d\right )} \sqrt {x} e^{\frac {3}{2}} + {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}}}{b x^{2} + a}\right ) - {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}} \log \left (-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (4 \, b c - 5 \, a d\right )} \sqrt {x} e^{\frac {3}{2}} - {\left (b^{3} x^{2} + a b^{2}\right )} \left (\frac {256 \, b^{4} c^{4} - 1280 \, a b^{3} c^{3} d + 2400 \, a^{2} b^{2} c^{2} d^{2} - 2000 \, a^{3} b c d^{3} + 625 \, a^{4} d^{4}}{b^{9}}\right )^{\frac {1}{4}} e^{\frac {3}{2}}}{b x^{2} + a}\right )}{8 \, {\left (b^{3} x^{2} + a 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)^(5/4),x, algorithm="fricas")

[Out]

1/8*(4*(b*d*x^2 - 4*b*c + 5*a*d)*(b*x^2 + a)^(3/4)*sqrt(x)*e^(3/2) + 4*(b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*
a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(1/4)*arctan(((4*b^8*c - 5*a*b^7*d)*
(b*x^2 + a)^(3/4)*sqrt(x)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4
*d^4)/b^9)^(3/4)*e^6 + (b^8*x^2 + a*b^7)*sqrt(((16*b^2*c^2 - 40*a*b*c*d + 25*a^2*d^2)*sqrt(b*x^2 + a)*x*e^3 +
(b^5*x^2 + a*b^4)*sqrt((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4
)/b^9)*e^3)/(b*x^2 + a))*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*
d^4)/b^9)^(3/4)*e^(9/2))*e^(-6)/(256*a*b^4*c^4 - 1280*a^2*b^3*c^3*d + 2400*a^3*b^2*c^2*d^2 - 2000*a^4*b*c*d^3
+ 625*a^5*d^4 + (256*b^5*c^4 - 1280*a*b^4*c^3*d + 2400*a^2*b^3*c^2*d^2 - 2000*a^3*b^2*c*d^3 + 625*a^4*b*d^4)*x
^2))*e^(3/2) + (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 +
625*a^4*d^4)/b^9)^(1/4)*e^(3/2)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(x)*e^(3/2) + (b^3*x^2 + a*b^2)*((
256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(1/4)*e^(3/2))/(b
*x^2 + a)) - (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 62
5*a^4*d^4)/b^9)^(1/4)*e^(3/2)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(x)*e^(3/2) - (b^3*x^2 + a*b^2)*((25
6*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)/b^9)^(1/4)*e^(3/2))/(b*x
^2 + a)))/(b^3*x^2 + a*b^2)

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 13.02, size = 94, normalized size = 0.55 \begin {gather*} \frac {c e^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \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 {5}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{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)**(5/4),x)

[Out]

c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(9/4)) + d
*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(13/4))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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 )}^{5/4}} \,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)^(5/4),x)

[Out]

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

________________________________________________________________________________________