3.4.22 \(\int \frac {(a+b x^2)^{5/4}}{c+d x^2} \, dx\) [322]

Optimal. Leaf size=302 \[ \frac {2 b x \sqrt [4]{a+b x^2}}{3 d}+\frac {2 a^{3/2} \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 d \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt {a} \sqrt {b} (b c-a d) \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{d^2 \left (a+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} (b c-a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^2 x}+\frac {\sqrt [4]{a} (b c-a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^2 x} \]

[Out]

2/3*b*x*(b*x^2+a)^(1/4)/d+2/3*a^(3/2)*(1+b*x^2/a)^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*a
rctan(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)/d/(b*x^2+a)^(3/4)-2*(-
a*d+b*c)*(1+b*x^2/a)^(3/4)*(cos(1/2*arctan(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*Ell
ipticF(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)*b^(1/2)/d^2/(b*x^2+a)^(3/4)+a^(1/4)*(-a*d+b*c)*Elli
pticPi((b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/d^2/x+a^(1/4)*(-a*d+b*c)*E
llipticPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(-b*x^2/a)^(1/2)/d^2/x

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Rubi [A]
time = 0.14, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {411, 201, 239, 237, 410, 109, 418, 1232} \begin {gather*} \frac {2 a^{3/2} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 d \left (a+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (b c-a d) \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^2 x}+\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} (b c-a d) \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {a d-b c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{b x^2+a}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^2 x}-\frac {2 \sqrt {a} \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4} (b c-a d) F\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{d^2 \left (a+b x^2\right )^{3/4}}+\frac {2 b x \sqrt [4]{a+b x^2}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/4)/(c + d*x^2),x]

[Out]

(2*b*x*(a + b*x^2)^(1/4))/(3*d) + (2*a^(3/2)*Sqrt[b]*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqrt[b]*x)/Sqrt[a
]]/2, 2])/(3*d*(a + b*x^2)^(3/4)) - (2*Sqrt[a]*Sqrt[b]*(b*c - a*d)*(1 + (b*x^2)/a)^(3/4)*EllipticF[ArcTan[(Sqr
t[b]*x)/Sqrt[a]]/2, 2])/(d^2*(a + b*x^2)^(3/4)) + (a^(1/4)*(b*c - a*d)*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a
]*Sqrt[d])/Sqrt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(d^2*x) + (a^(1/4)*(b*c - a*d)*Sqrt[-(
(b*x^2)/a)]*EllipticPi[(Sqrt[a]*Sqrt[d])/Sqrt[-(b*c) + a*d], ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(d^2*x)

Rule 109

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(3/4)), x_Symbol] :> Dist[-4, Subst[
Int[1/((b*e - a*f - b*x^4)*Sqrt[c - d*(e/f) + d*(x^4/f)]), x], x, (e + f*x)^(1/4)], x] /; FreeQ[{a, b, c, d, e
, f}, x] && GtQ[-f/(d*e - c*f), 0]

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 239

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

Rule 410

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[Sqrt[(-b)*(x^2/a)]/(2*x), Subst[I
nt[1/(Sqrt[(-b)*(x/a)]*(a + b*x)^(3/4)*(c + d*x)), x], x, x^2], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0]

Rule 411

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^4]*((c_) + (d_.)*(x_)^4)), x_Symbol] :> Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1
- Rt[-d/c, 2]*x^2)), x], x] + Dist[1/(2*c), Int[1/(Sqrt[a + b*x^4]*(1 + Rt[-d/c, 2]*x^2)), x], x] /; FreeQ[{a,
b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1232

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[
a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^{5/4}}{c+d x^2} \, dx &=\frac {b \int \sqrt [4]{a+b x^2} \, dx}{d}-\frac {(b c-a d) \int \frac {\sqrt [4]{a+b x^2}}{c+d x^2} \, dx}{d}\\ &=\frac {2 b x \sqrt [4]{a+b x^2}}{3 d}+\frac {(a b) \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{3 d}-\frac {(b (b c-a d)) \int \frac {1}{\left (a+b x^2\right )^{3/4}} \, dx}{d^2}+\frac {(b c-a d)^2 \int \frac {1}{\left (a+b x^2\right )^{3/4} \left (c+d x^2\right )} \, dx}{d^2}\\ &=\frac {2 b x \sqrt [4]{a+b x^2}}{3 d}+\frac {\left ((b c-a d)^2 \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-\frac {b x}{a}} (a+b x)^{3/4} (c+d x)} \, dx,x,x^2\right )}{2 d^2 x}+\frac {\left (a b \left (1+\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{3 d \left (a+b x^2\right )^{3/4}}-\frac {\left (b (b c-a d) \left (1+\frac {b x^2}{a}\right )^{3/4}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \, dx}{d^2 \left (a+b x^2\right )^{3/4}}\\ &=\frac {2 b x \sqrt [4]{a+b x^2}}{3 d}+\frac {2 a^{3/2} \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 d \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt {a} \sqrt {b} (b c-a d) \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{d^2 \left (a+b x^2\right )^{3/4}}-\frac {\left (2 (b c-a d)^2 \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^4}{a}} \left (-b c+a d-d x^4\right )} \, dx,x,\sqrt [4]{a+b x^2}\right )}{d^2 x}\\ &=\frac {2 b x \sqrt [4]{a+b x^2}}{3 d}+\frac {2 a^{3/2} \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 d \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt {a} \sqrt {b} (b c-a d) \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{d^2 \left (a+b x^2\right )^{3/4}}+\frac {\left ((b c-a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {\sqrt {d} x^2}{\sqrt {-b c+a d}}\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{d^2 x}+\frac {\left ((b c-a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {\sqrt {d} x^2}{\sqrt {-b c+a d}}\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{d^2 x}\\ &=\frac {2 b x \sqrt [4]{a+b x^2}}{3 d}+\frac {2 a^{3/2} \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 d \left (a+b x^2\right )^{3/4}}-\frac {2 \sqrt {a} \sqrt {b} (b c-a d) \left (1+\frac {b x^2}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{d^2 \left (a+b x^2\right )^{3/4}}+\frac {\sqrt [4]{a} (b c-a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^2 x}+\frac {\sqrt [4]{a} (b c-a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{d^2 x}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 9.82, size = 348, normalized size = 1.15 \begin {gather*} \frac {x \left (\frac {b (-3 b c+4 a d) x^2 \left (1+\frac {b x^2}{a}\right )^{3/4} F_1\left (\frac {3}{2};\frac {3}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c}+\frac {6 \left (-3 a c \left (3 a^2 d+2 a b d x^2+2 b^2 x^2 \left (c+d x^2\right )\right ) F_1\left (\frac {1}{2};\frac {3}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b x^2 \left (a+b x^2\right ) \left (c+d x^2\right ) \left (4 a d F_1\left (\frac {3}{2};\frac {3}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c F_1\left (\frac {3}{2};\frac {7}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{\left (c+d x^2\right ) \left (-6 a c F_1\left (\frac {1}{2};\frac {3}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (4 a d F_1\left (\frac {3}{2};\frac {3}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c F_1\left (\frac {3}{2};\frac {7}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{9 d \left (a+b x^2\right )^{3/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*((b*(-3*b*c + 4*a*d)*x^2*(1 + (b*x^2)/a)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])/c +
(6*(-3*a*c*(3*a^2*d + 2*a*b*d*x^2 + 2*b^2*x^2*(c + d*x^2))*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/
c)] + b*x^2*(a + b*x^2)*(c + d*x^2)*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c*Appe
llF1[3/2, 7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))/((c + d*x^2)*(-6*a*c*AppellF1[1/2, 3/4, 1, 3/2, -((b*x^2
)/a), -((d*x^2)/c)] + x^2*(4*a*d*AppellF1[3/2, 3/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 3*b*c*AppellF1[3/2,
7/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))))/(9*d*(a + b*x^2)^(3/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {5}{4}}}{d \,x^{2}+c}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**2)**(5/4)/(c + d*x**2), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/4)/(c + d*x^2),x)

[Out]

int((a + b*x^2)^(5/4)/(c + d*x^2), x)

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