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

Optimal. Leaf size=122 \[ \frac {d \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{5/4} \sqrt {e}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {e x}}{\sqrt {e} \sqrt [4]{a+b x^2}}\right )}{b^{5/4} \sqrt {e}}+\frac {2 \sqrt {e x} (b c-a d)}{a b e \sqrt [4]{a+b x^2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {452, 329, 240, 212, 208, 205} \begin {gather*} \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}}+\frac {2 \sqrt {e x} (b c-a d)}{a b e \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[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 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 212

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] &&
!GtQ[a/b, 0]

Rule 240

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 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 452

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 {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 \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) \operatorname {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) \operatorname {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 \operatorname {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 \operatorname {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*}

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Mathematica [C]  time = 0.06, size = 68, normalized size = 0.56 \begin {gather*} \frac {2 \left (d x^3 \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {5}{4},\frac {5}{4};\frac {9}{4};-\frac {b x^2}{a}\right )+5 c x\right )}{5 a \sqrt {e x} \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.36, size = 168, normalized size = 1.38 \begin {gather*} \frac {d \tan ^{-1}\left (\frac {\sqrt [4]{b} e^{3/2} \sqrt {e x} \left (a+b x^2\right )^{3/4}}{a e^2+b e^2 x^2}\right )}{b^{5/4} \sqrt {e}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt [4]{b} e^{3/2} \sqrt {e x} \left (a+b x^2\right )^{3/4}}{a e^2+b e^2 x^2}\right )}{b^{5/4} \sqrt {e}}-\frac {2 e \sqrt {e x} \left (a+b x^2\right )^{3/4} (a d-b c)}{a b \left (a e^2+b e^2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.06, size = 384, normalized size = 3.15 \begin {gather*} \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} {\left (b c - a d\right )} \sqrt {e x} - 4 \, {\left (a b^{2} e x^{2} + a^{2} b e\right )} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {1}{4}} \arctan \left (-\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x} b^{4} d e \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {3}{4}} - {\left (b^{5} e x^{2} + a b^{4} e\right )} \sqrt {\frac {\sqrt {b x^{2} + a} d^{2} e x + {\left (b^{3} e^{2} x^{2} + a b^{2} e^{2}\right )} \sqrt {\frac {d^{4}}{b^{5} e^{2}}}}{b x^{2} + a}} \left (\frac {d^{4}}{b^{5} e^{2}}\right )^{\frac {3}{4}}}{b d^{4} x^{2} + a d^{4}}\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 (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 )}{2 \, {\left (a b^{2} e x^{2} + a^{2} b e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d \,x^{2}+c}{\sqrt {e x}\, \left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [C]  time = 16.06, size = 83, normalized size = 0.68 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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