\(\int \frac {(e x)^{7/2} (c+d x^2)}{(a+b x^2)^{7/4}} \, dx\) [1121]

Optimal result

Integrand size = 26, antiderivative size = 192 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {5 (2 b c-3 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^3}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {5 \sqrt {a} (2 b c-3 a d) e^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 b^{5/2} \left (a+b x^2\right )^{3/4}} \]

[Out]

2/3*(-a*d+b*c)*(e*x)^(9/2)/a/b/e/(b*x^2+a)^(3/4)-1/3*(-3*a*d+2*b*c)*e*(e*x)^(5/2)*(b*x^2+a)^(1/4)/a/b^2+5/6*(-
3*a*d+2*b*c)*e^2*(1+a/b/x^2)^(3/4)*(e*x)^(3/2)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b
^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)/b^(5/2)/(b*x^2+a)^(3/4)+5/6*(-3
*a*d+2*b*c)*e^3*(b*x^2+a)^(1/4)*(e*x)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {468, 327, 335, 243, 342, 281, 237} \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {5 \sqrt {a} e^2 (e x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} (2 b c-3 a d) \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 b^{5/2} \left (a+b x^2\right )^{3/4}}+\frac {5 e^3 \sqrt {e x} \sqrt [4]{a+b x^2} (2 b c-3 a d)}{6 b^3}-\frac {e (e x)^{5/2} \sqrt [4]{a+b x^2} (2 b c-3 a d)}{3 a b^2}+\frac {2 (e x)^{9/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \]

[In]

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

[Out]

(2*(b*c - a*d)*(e*x)^(9/2))/(3*a*b*e*(a + b*x^2)^(3/4)) + (5*(2*b*c - 3*a*d)*e^3*Sqrt[e*x]*(a + b*x^2)^(1/4))/
(6*b^3) - ((2*b*c - 3*a*d)*e*(e*x)^(5/2)*(a + b*x^2)^(1/4))/(3*a*b^2) + (5*Sqrt[a]*(2*b*c - 3*a*d)*e^2*(1 + a/
(b*x^2))^(3/4)*(e*x)^(3/2)*EllipticF[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(6*b^(5/2)*(a + b*x^2)^(3/4))

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 327

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 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 342

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

Rule 468

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*} \text {integral}& = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {\left (2 \left (-3 b c+\frac {9 a d}{2}\right )\right ) \int \frac {(e x)^{7/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{3 a b} \\ & = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {\left (5 (2 b c-3 a d) e^2\right ) \int \frac {(e x)^{3/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{6 b^2} \\ & = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {5 (2 b c-3 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^3}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}-\frac {\left (5 a (2 b c-3 a d) e^4\right ) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{3/4}} \, dx}{12 b^3} \\ & = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {5 (2 b c-3 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^3}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}-\frac {\left (5 a (2 b c-3 a d) e^3\right ) \text {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt {e x}\right )}{6 b^3} \\ & = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {5 (2 b c-3 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^3}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}-\frac {\left (5 a (2 b c-3 a d) e^3 \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {e x}\right )}{6 b^3 \left (a+b x^2\right )^{3/4}} \\ & = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {5 (2 b c-3 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^3}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {\left (5 a (2 b c-3 a d) e^3 \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a e^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {e x}}\right )}{6 b^3 \left (a+b x^2\right )^{3/4}} \\ & = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {5 (2 b c-3 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^3}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {\left (5 a (2 b c-3 a d) e^3 \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a e^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{e x}\right )}{12 b^3 \left (a+b x^2\right )^{3/4}} \\ & = \frac {2 (b c-a d) (e x)^{9/2}}{3 a b e \left (a+b x^2\right )^{3/4}}+\frac {5 (2 b c-3 a d) e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}{6 b^3}-\frac {(2 b c-3 a d) e (e x)^{5/2} \sqrt [4]{a+b x^2}}{3 a b^2}+\frac {5 \sqrt {a} (2 b c-3 a d) e^2 \left (1+\frac {a}{b x^2}\right )^{3/4} (e x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 b^{5/2} \left (a+b x^2\right )^{3/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.57 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {e^3 \sqrt {e x} \left (-15 a^2 d+a b \left (10 c-9 d x^2\right )+2 b^2 x^2 \left (3 c+d x^2\right )+5 a (-2 b c+3 a d) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{6 b^3 \left (a+b x^2\right )^{3/4}} \]

[In]

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

[Out]

(e^3*Sqrt[e*x]*(-15*a^2*d + a*b*(10*c - 9*d*x^2) + 2*b^2*x^2*(3*c + d*x^2) + 5*a*(-2*b*c + 3*a*d)*(1 + (b*x^2)
/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, -((b*x^2)/a)]))/(6*b^3*(a + b*x^2)^(3/4))

Maple [F]

\[\int \frac {\left (e x \right )^{\frac {7}{2}} \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \]

[In]

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

[Out]

integral((d*e^3*x^5 + c*e^3*x^3)*(b*x^2 + a)^(1/4)*sqrt(e*x)/(b^2*x^4 + 2*a*b*x^2 + a^2), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 149.19 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.49 \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {c e^{\frac {7}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {13}{4}\right )} + \frac {d e^{\frac {7}{2}} x^{\frac {13}{2}} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {17}{4}\right )} \]

[In]

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

[Out]

c*e**(7/2)*x**(9/2)*gamma(9/4)*hyper((7/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*gamma(13/4)) +
d*e**(7/2)*x**(13/2)*gamma(13/4)*hyper((7/4, 13/4), (17/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*gamma(17/4
))

Maxima [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{2} + a\right )}^{\frac {7}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{7/4}} \, dx=\int \frac {{\left (e\,x\right )}^{7/2}\,\left (d\,x^2+c\right )}{{\left (b\,x^2+a\right )}^{7/4}} \,d x \]

[In]

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

[Out]

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