3.1116 \(\int \frac {(e x)^{5/2} (c+d x^2)}{(a+b x^2)^{7/4}} \, dx\)

Optimal. Leaf size=184 \[ -\frac {e^{5/2} (4 b c-7 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{11/4}}+\frac {e^{5/2} (4 b c-7 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{11/4}}-\frac {e (e x)^{3/2} \sqrt [4]{a+b x^2} (4 b c-7 a d)}{6 a b^2}+\frac {2 (e x)^{7/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \]

[Out]

2/3*(-a*d+b*c)*(e*x)^(7/2)/a/b/e/(b*x^2+a)^(3/4)-1/6*(-7*a*d+4*b*c)*e*(e*x)^(3/2)*(b*x^2+a)^(1/4)/a/b^2-1/4*(-
7*a*d+4*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)+1/4*(-7*a*d+4*b*c)*e^(5/2)*a
rctanh(b^(1/4)*(e*x)^(1/2)/(b*x^2+a)^(1/4)/e^(1/2))/b^(11/4)

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 184, 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 = {457, 321, 329, 331, 298, 205, 208} \[ -\frac {e^{5/2} (4 b c-7 a d) \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{11/4}}+\frac {e^{5/2} (4 b c-7 a d) \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{4 b^{11/4}}-\frac {e (e x)^{3/2} \sqrt [4]{a+b x^2} (4 b c-7 a d)}{6 a b^2}+\frac {2 (e x)^{7/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x]

[Out]

(2*(b*c - a*d)*(e*x)^(7/2))/(3*a*b*e*(a + b*x^2)^(3/4)) - ((4*b*c - 7*a*d)*e*(e*x)^(3/2)*(a + b*x^2)^(1/4))/(6
*a*b^2) - ((4*b*c - 7*a*d)*e^(5/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(11/4)) + ((4
*b*c - 7*a*d)*e^(5/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*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 457

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*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rubi steps

\begin {align*} \int \frac {(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx &=\frac {2 (b c-a d) (e x)^{7/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {\left (2 \left (-2 b c+\frac {7 a d}{2}\right )\right ) \int \frac {(e x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{3 a b}\\ &=\frac {2 (b c-a d) (e x)^{7/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(4 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{6 a b^2}+\frac {\left ((4 b c-7 a d) e^2\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{3/4}} \, dx}{4 b^2}\\ &=\frac {2 (b c-a d) (e x)^{7/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(4 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{6 a b^2}+\frac {((4 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 )}{2 b^2}\\ &=\frac {2 (b c-a d) (e x)^{7/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(4 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{6 a b^2}+\frac {((4 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 )}{2 b^2}\\ &=\frac {2 (b c-a d) (e x)^{7/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(4 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{6 a b^2}+\frac {\left ((4 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 )}{4 b^{5/2}}-\frac {\left ((4 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 )}{4 b^{5/2}}\\ &=\frac {2 (b c-a d) (e x)^{7/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(4 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{6 a b^2}-\frac {(4 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 )}{4 b^{11/4}}+\frac {(4 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 )}{4 b^{11/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.13, size = 77, normalized size = 0.42 \[ \frac {x (e x)^{5/2} \left (\left (\frac {b x^2}{a}+1\right )^{3/4} (4 b c-7 a d) \, _2F_1\left (\frac {7}{4},\frac {7}{4};\frac {11}{4};-\frac {b x^2}{a}\right )+7 a d\right )}{14 a b \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x]

[Out]

(x*(e*x)^(5/2)*(7*a*d + (4*b*c - 7*a*d)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[7/4, 7/4, 11/4, -((b*x^2)/a)])
)/(14*a*b*(a + b*x^2)^(3/4))

________________________________________________________________________________________

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)^(7/4),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(7/4), x)

________________________________________________________________________________________

maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{\frac {5}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

[Out]

int((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*(e*x)^(5/2)/(b*x^2 + a)^(7/4), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,x\right )}^{5/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{7/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4),x)

[Out]

int(((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(7/4), x)

________________________________________________________________________________________

sympy [C]  time = 166.39, size = 94, normalized size = 0.51 \[ \frac {c e^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {11}{4}\right )} + \frac {d e^{\frac {5}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(5/2)*(d*x**2+c)/(b*x**2+a)**(7/4),x)

[Out]

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

________________________________________________________________________________________