3.4.31 \(\int \frac {(a+b x^2)^{7/4}}{(c+d x^2)^2} \, dx\) [331]

Optimal. Leaf size=340 \[ \frac {b (5 b c-a d) x}{2 c d^2 \sqrt [4]{a+b x^2}}-\frac {(b c-a d) x \left (a+b x^2\right )^{3/4}}{2 c d \left (c+d x^2\right )}-\frac {\sqrt {a} \sqrt {b} (5 b c-a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c d^2 \sqrt [4]{a+b x^2}}+\frac {\sqrt [4]{a} \sqrt {-b c+a d} (5 b c+2 a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (-\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c d^{5/2} x}-\frac {\sqrt [4]{a} \sqrt {-b c+a d} (5 b c+2 a d) \sqrt {-\frac {b x^2}{a}} \Pi \left (\frac {\sqrt {a} \sqrt {d}}{\sqrt {-b c+a d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{a+b x^2}}{\sqrt [4]{a}}\right )\right |-1\right )}{4 c d^{5/2} x} \]

[Out]

1/2*b*(-a*d+5*b*c)*x/c/d^2/(b*x^2+a)^(1/4)-1/2*(-a*d+b*c)*x*(b*x^2+a)^(3/4)/c/d/(d*x^2+c)-1/2*(-a*d+5*b*c)*(1+
b*x^2/a)^(1/4)*(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)))*EllipticE(sin(1
/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)*b^(1/2)/c/d^2/(b*x^2+a)^(1/4)+1/4*a^(1/4)*(2*a*d+5*b*c)*Ellipti
cPi((b*x^2+a)^(1/4)/a^(1/4),-a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(a*d-b*c)^(1/2)*(-b*x^2/a)^(1/2)/c/d^(5/2)/x-1
/4*a^(1/4)*(2*a*d+5*b*c)*EllipticPi((b*x^2+a)^(1/4)/a^(1/4),a^(1/2)*d^(1/2)/(a*d-b*c)^(1/2),I)*(a*d-b*c)^(1/2)
*(-b*x^2/a)^(1/2)/c/d^(5/2)/x

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Rubi [A]
time = 0.20, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {424, 544, 235, 233, 202, 408, 504, 1232} \begin {gather*} \frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \sqrt {a d-b c} (2 a d+5 b c) \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 )}{4 c d^{5/2} x}-\frac {\sqrt [4]{a} \sqrt {-\frac {b x^2}{a}} \sqrt {a d-b c} (2 a d+5 b c) \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 )}{4 c d^{5/2} x}-\frac {\sqrt {a} \sqrt {b} \sqrt [4]{\frac {b x^2}{a}+1} (5 b c-a d) E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c d^2 \sqrt [4]{a+b x^2}}+\frac {b x (5 b c-a d)}{2 c d^2 \sqrt [4]{a+b x^2}}-\frac {x \left (a+b x^2\right )^{3/4} (b c-a d)}{2 c d \left (c+d x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(5*b*c - a*d)*x)/(2*c*d^2*(a + b*x^2)^(1/4)) - ((b*c - a*d)*x*(a + b*x^2)^(3/4))/(2*c*d*(c + d*x^2)) - (Sqr
t[a]*Sqrt[b]*(5*b*c - a*d)*(1 + (b*x^2)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*c*d^2*(a + b*
x^2)^(1/4)) + (a^(1/4)*Sqrt[-(b*c) + a*d]*(5*b*c + 2*a*d)*Sqrt[-((b*x^2)/a)]*EllipticPi[-((Sqrt[a]*Sqrt[d])/Sq
rt[-(b*c) + a*d]), ArcSin[(a + b*x^2)^(1/4)/a^(1/4)], -1])/(4*c*d^(5/2)*x) - (a^(1/4)*Sqrt[-(b*c) + a*d]*(5*b*
c + 2*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])/(4*c*d^(5/2)*x)

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 235

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

Rule 408

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

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 504

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

Rule 544

Int[(((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Dist[f/d,
Int[(a + b*x^n)^p, x], x] + Dist[(d*e - c*f)/d, Int[(a + b*x^n)^p/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, p, n}, x]

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 )^{7/4}}{\left (c+d x^2\right )^2} \, dx &=-\frac {(b c-a d) x \left (a+b x^2\right )^{3/4}}{2 c d \left (c+d x^2\right )}+\frac {\int \frac {a (b c+a d)+\frac {1}{2} b (5 b c-a d) x^2}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{2 c d}\\ &=-\frac {(b c-a d) x \left (a+b x^2\right )^{3/4}}{2 c d \left (c+d x^2\right )}+\frac {(b (5 b c-a d)) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{4 c d^2}-\frac {((b c-a d) (5 b c+2 a d)) \int \frac {1}{\sqrt [4]{a+b x^2} \left (c+d x^2\right )} \, dx}{4 c d^2}\\ &=-\frac {(b c-a d) x \left (a+b x^2\right )^{3/4}}{2 c d \left (c+d x^2\right )}-\frac {\left ((b c-a d) (5 b c+2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^4}{a}} \left (b c-a d+d x^4\right )} \, dx,x,\sqrt [4]{a+b x^2}\right )}{2 c d^2 x}+\frac {\left (b (5 b c-a d) \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{4 c d^2 \sqrt [4]{a+b x^2}}\\ &=\frac {b (5 b c-a d) x}{2 c d^2 \sqrt [4]{a+b x^2}}-\frac {(b c-a d) x \left (a+b x^2\right )^{3/4}}{2 c d \left (c+d x^2\right )}+\frac {\left ((b c-a d) (5 b c+2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {-b c+a d}-\sqrt {d} x^2\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{4 c d^{5/2} x}-\frac {\left ((b c-a d) (5 b c+2 a d) \sqrt {-\frac {b x^2}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {-b c+a d}+\sqrt {d} x^2\right ) \sqrt {1-\frac {x^4}{a}}} \, dx,x,\sqrt [4]{a+b x^2}\right )}{4 c d^{5/2} x}-\frac {\left (b (5 b c-a d) \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{4 c d^2 \sqrt [4]{a+b x^2}}\\ &=\frac {b (5 b c-a d) x}{2 c d^2 \sqrt [4]{a+b x^2}}-\frac {(b c-a d) x \left (a+b x^2\right )^{3/4}}{2 c d \left (c+d x^2\right )}-\frac {\sqrt {a} \sqrt {b} (5 b c-a d) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 c d^2 \sqrt [4]{a+b x^2}}+\frac {\sqrt [4]{a} \sqrt {-b c+a d} (5 b c+2 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 )}{4 c d^{5/2} x}-\frac {\sqrt [4]{a} \sqrt {-b c+a d} (5 b c+2 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 )}{4 c d^{5/2} 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 = 10.24, size = 340, normalized size = 1.00 \begin {gather*} \frac {x \left (-b (-5 b c+a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} F_1\left (\frac {3}{2};\frac {1}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {6 c \left (-6 a c \left (2 a^2 d-b^2 c x^2+a b d x^2\right ) F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+(-b c+a d) x^2 \left (a+b x^2\right ) \left (4 a d F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c F_1\left (\frac {3}{2};\frac {5}{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 {1}{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 {1}{4},2;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}\right )}{12 c^2 d \sqrt [4]{a+b x^2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-(b*(-5*b*c + a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, -((b*x^2)/a), -((d*x^2)/c)]) + (6*
c*(-6*a*c*(2*a^2*d - b^2*c*x^2 + a*b*d*x^2)*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + (-(b*c) +
a*d)*x^2*(a + b*x^2)*(4*a*d*AppellF1[3/2, 1/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1
, 5/2, -((b*x^2)/a), -((d*x^2)/c)])))/((c + d*x^2)*(-6*a*c*AppellF1[1/2, 1/4, 1, 3/2, -((b*x^2)/a), -((d*x^2)/
c)] + x^2*(4*a*d*AppellF1[3/2, 1/4, 2, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1, 5/2, -((b*
x^2)/a), -((d*x^2)/c)])))))/(12*c^2*d*(a + b*x^2)^(1/4))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{\frac {7}{4}}}{\left (d \,x^{2}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^2 + a)^(7/4)/(d*x^2 + c)^2, 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)^(7/4)/(d*x^2+c)^2,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 {7}{4}}}{\left (c + d x^{2}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate((b*x^2 + a)^(7/4)/(d*x^2 + c)^2, 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 )}^{7/4}}{{\left (d\,x^2+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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