\(\int \frac {c+d x^2}{\sqrt {e x} (a+b x^2)^{5/4}} \, dx\) [1106]

Optimal result

Integrand size = 26, antiderivative size = 122 \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/4}} \, dx=\frac {2 (b c-a d) \sqrt {e x}}{a b e \sqrt [4]{a+b x^2}}+\frac {d \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{5/4} \sqrt {e}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{5/4} \sqrt {e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {463, 335, 246, 218, 214, 211} \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/4}} \, dx=\frac {d \arctan \left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{5/4} \sqrt {e}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{5/4} \sqrt {e}}+\frac {2 \sqrt {e x} (b c-a d)}{a b e \sqrt [4]{a+b x^2}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/4}} \, dx=\frac {\sqrt {x} \left (\frac {2 \sqrt [4]{b} (b c-a d) \sqrt {x}}{a \sqrt [4]{a+b x^2}}+d \arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+d \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{b^{5/4} \sqrt {e x}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 389, normalized size of antiderivative = 3.19 \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/4}} \, dx=\frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (b c - a d\right )} \sqrt {e x} + {\left (a b^{2} e x^{2} + a^{2} b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d + {\left (b^{2} e x^{2} + a b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) - {\left (a b^{2} e x^{2} + a^{2} b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d - {\left (b^{2} e x^{2} + a b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + {\left (-i \, a b^{2} e x^{2} - i \, a^{2} b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d - {\left (i \, b^{2} e x^{2} + i \, a b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + {\left (i \, a b^{2} e x^{2} + i \, a^{2} b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} d - {\left (-i \, b^{2} e x^{2} - i \, a b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right )}{2 \, {\left (a b^{2} e x^{2} + a^{2} b e\right )}} \]

[In]

integrate((d*x^2+c)/(e*x)^(1/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.93 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/4}} \, dx=\frac {c \Gamma \left (\frac {1}{4}\right )}{2 a \sqrt [4]{b} \sqrt {e} \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \sqrt {e} \Gamma \left (\frac {9}{4}\right )} \]

[In]

integrate((d*x**2+c)/(e*x)**(1/2)/(b*x**2+a)**(5/4),x)

[Out]

c*gamma(1/4)/(2*a*b**(1/4)*sqrt(e)*(a/(b*x**2) + 1)**(1/4)*gamma(5/4)) + d*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)*sqrt(e)*gamma(9/4))

Maxima [F]

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

[In]

integrate((d*x^2+c)/(e*x)^(1/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((d*x^2+c)/(e*x)^(1/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{\sqrt {e x} \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {d\,x^2+c}{\sqrt {e\,x}\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

[In]

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

[Out]

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