3.152 \(\int \frac {a+b x^4}{(c+d x^4)^3} \, dx\)

Optimal. Leaf size=273 \[ -\frac {3 (7 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}+\frac {3 (7 a d+b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}-\frac {3 (7 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}+\frac {3 (7 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}+\frac {x (7 a d+b c)}{32 c^2 d \left (c+d x^4\right )}-\frac {x (b c-a d)}{8 c d \left (c+d x^4\right )^2} \]

[Out]

-1/8*(-a*d+b*c)*x/c/d/(d*x^4+c)^2+1/32*(7*a*d+b*c)*x/c^2/d/(d*x^4+c)+3/128*(7*a*d+b*c)*arctan(-1+d^(1/4)*x*2^(
1/2)/c^(1/4))/c^(11/4)/d^(5/4)*2^(1/2)+3/128*(7*a*d+b*c)*arctan(1+d^(1/4)*x*2^(1/2)/c^(1/4))/c^(11/4)/d^(5/4)*
2^(1/2)-3/256*(7*a*d+b*c)*ln(-c^(1/4)*d^(1/4)*x*2^(1/2)+c^(1/2)+x^2*d^(1/2))/c^(11/4)/d^(5/4)*2^(1/2)+3/256*(7
*a*d+b*c)*ln(c^(1/4)*d^(1/4)*x*2^(1/2)+c^(1/2)+x^2*d^(1/2))/c^(11/4)/d^(5/4)*2^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {385, 199, 211, 1165, 628, 1162, 617, 204} \[ -\frac {3 (7 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}+\frac {3 (7 a d+b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}-\frac {3 (7 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}+\frac {3 (7 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}+\frac {x (7 a d+b c)}{32 c^2 d \left (c+d x^4\right )}-\frac {x (b c-a d)}{8 c d \left (c+d x^4\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)/(c + d*x^4)^3,x]

[Out]

-((b*c - a*d)*x)/(8*c*d*(c + d*x^4)^2) + ((b*c + 7*a*d)*x)/(32*c^2*d*(c + d*x^4)) - (3*(b*c + 7*a*d)*ArcTan[1
- (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(64*Sqrt[2]*c^(11/4)*d^(5/4)) + (3*(b*c + 7*a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x
)/c^(1/4)])/(64*Sqrt[2]*c^(11/4)*d^(5/4)) - (3*(b*c + 7*a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]
*x^2])/(128*Sqrt[2]*c^(11/4)*d^(5/4)) + (3*(b*c + 7*a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2
])/(128*Sqrt[2]*c^(11/4)*d^(5/4))

Rule 199

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

Rule 204

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 385

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

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

Rubi steps

\begin {align*} \int \frac {a+b x^4}{\left (c+d x^4\right )^3} \, dx &=-\frac {(b c-a d) x}{8 c d \left (c+d x^4\right )^2}+\frac {(b c+7 a d) \int \frac {1}{\left (c+d x^4\right )^2} \, dx}{8 c d}\\ &=-\frac {(b c-a d) x}{8 c d \left (c+d x^4\right )^2}+\frac {(b c+7 a d) x}{32 c^2 d \left (c+d x^4\right )}+\frac {(3 (b c+7 a d)) \int \frac {1}{c+d x^4} \, dx}{32 c^2 d}\\ &=-\frac {(b c-a d) x}{8 c d \left (c+d x^4\right )^2}+\frac {(b c+7 a d) x}{32 c^2 d \left (c+d x^4\right )}+\frac {(3 (b c+7 a d)) \int \frac {\sqrt {c}-\sqrt {d} x^2}{c+d x^4} \, dx}{64 c^{5/2} d}+\frac {(3 (b c+7 a d)) \int \frac {\sqrt {c}+\sqrt {d} x^2}{c+d x^4} \, dx}{64 c^{5/2} d}\\ &=-\frac {(b c-a d) x}{8 c d \left (c+d x^4\right )^2}+\frac {(b c+7 a d) x}{32 c^2 d \left (c+d x^4\right )}+\frac {(3 (b c+7 a d)) \int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx}{128 c^{5/2} d^{3/2}}+\frac {(3 (b c+7 a d)) \int \frac {1}{\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}+x^2} \, dx}{128 c^{5/2} d^{3/2}}-\frac {(3 (b c+7 a d)) \int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}+2 x}{-\frac {\sqrt {c}}{\sqrt {d}}-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx}{128 \sqrt {2} c^{11/4} d^{5/4}}-\frac {(3 (b c+7 a d)) \int \frac {\frac {\sqrt {2} \sqrt [4]{c}}{\sqrt [4]{d}}-2 x}{-\frac {\sqrt {c}}{\sqrt {d}}+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{d}}-x^2} \, dx}{128 \sqrt {2} c^{11/4} d^{5/4}}\\ &=-\frac {(b c-a d) x}{8 c d \left (c+d x^4\right )^2}+\frac {(b c+7 a d) x}{32 c^2 d \left (c+d x^4\right )}-\frac {3 (b c+7 a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}+\frac {3 (b c+7 a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}+\frac {(3 (b c+7 a d)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}-\frac {(3 (b c+7 a d)) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}\\ &=-\frac {(b c-a d) x}{8 c d \left (c+d x^4\right )^2}+\frac {(b c+7 a d) x}{32 c^2 d \left (c+d x^4\right )}-\frac {3 (b c+7 a d) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}+\frac {3 (b c+7 a d) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{64 \sqrt {2} c^{11/4} d^{5/4}}-\frac {3 (b c+7 a d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}+\frac {3 (b c+7 a d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{128 \sqrt {2} c^{11/4} d^{5/4}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 243, normalized size = 0.89 \[ \frac {-\frac {32 c^{7/4} \sqrt [4]{d} x (b c-a d)}{\left (c+d x^4\right )^2}+\frac {8 c^{3/4} \sqrt [4]{d} x (7 a d+b c)}{c+d x^4}-3 \sqrt {2} (7 a d+b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )+3 \sqrt {2} (7 a d+b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )-6 \sqrt {2} (7 a d+b c) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )+6 \sqrt {2} (7 a d+b c) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{256 c^{11/4} d^{5/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)/(c + d*x^4)^3,x]

[Out]

((-32*c^(7/4)*d^(1/4)*(b*c - a*d)*x)/(c + d*x^4)^2 + (8*c^(3/4)*d^(1/4)*(b*c + 7*a*d)*x)/(c + d*x^4) - 6*Sqrt[
2]*(b*c + 7*a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)] + 6*Sqrt[2]*(b*c + 7*a*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)
*x)/c^(1/4)] - 3*Sqrt[2]*(b*c + 7*a*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2] + 3*Sqrt[2]*(b*c
+ 7*a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(256*c^(11/4)*d^(5/4))

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fricas [B]  time = 1.28, size = 787, normalized size = 2.88 \[ \frac {4 \, {\left (b c d + 7 \, a d^{2}\right )} x^{5} + 12 \, {\left (c^{2} d^{3} x^{8} + 2 \, c^{3} d^{2} x^{4} + c^{4} d\right )} \left (-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}\right )^{\frac {1}{4}} \arctan \left (-\frac {c^{8} d^{4} x \left (-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}\right )^{\frac {3}{4}} - c^{8} d^{4} \sqrt {\frac {c^{6} d^{2} \sqrt {-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}} + {\left (b^{2} c^{2} + 14 \, a b c d + 49 \, a^{2} d^{2}\right )} x^{2}}{b^{2} c^{2} + 14 \, a b c d + 49 \, a^{2} d^{2}}} \left (-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}\right )^{\frac {3}{4}}}{b^{3} c^{3} + 21 \, a b^{2} c^{2} d + 147 \, a^{2} b c d^{2} + 343 \, a^{3} d^{3}}\right ) + 3 \, {\left (c^{2} d^{3} x^{8} + 2 \, c^{3} d^{2} x^{4} + c^{4} d\right )} \left (-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}\right )^{\frac {1}{4}} \log \left (3 \, c^{3} d \left (-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}\right )^{\frac {1}{4}} + 3 \, {\left (b c + 7 \, a d\right )} x\right ) - 3 \, {\left (c^{2} d^{3} x^{8} + 2 \, c^{3} d^{2} x^{4} + c^{4} d\right )} \left (-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}\right )^{\frac {1}{4}} \log \left (-3 \, c^{3} d \left (-\frac {b^{4} c^{4} + 28 \, a b^{3} c^{3} d + 294 \, a^{2} b^{2} c^{2} d^{2} + 1372 \, a^{3} b c d^{3} + 2401 \, a^{4} d^{4}}{c^{11} d^{5}}\right )^{\frac {1}{4}} + 3 \, {\left (b c + 7 \, a d\right )} x\right ) - 4 \, {\left (3 \, b c^{2} - 11 \, a c d\right )} x}{128 \, {\left (c^{2} d^{3} x^{8} + 2 \, c^{3} d^{2} x^{4} + c^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)/(d*x^4+c)^3,x, algorithm="fricas")

[Out]

1/128*(4*(b*c*d + 7*a*d^2)*x^5 + 12*(c^2*d^3*x^8 + 2*c^3*d^2*x^4 + c^4*d)*(-(b^4*c^4 + 28*a*b^3*c^3*d + 294*a^
2*b^2*c^2*d^2 + 1372*a^3*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5))^(1/4)*arctan(-(c^8*d^4*x*(-(b^4*c^4 + 28*a*b^3*c^
3*d + 294*a^2*b^2*c^2*d^2 + 1372*a^3*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5))^(3/4) - c^8*d^4*sqrt((c^6*d^2*sqrt(-(
b^4*c^4 + 28*a*b^3*c^3*d + 294*a^2*b^2*c^2*d^2 + 1372*a^3*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5)) + (b^2*c^2 + 14*
a*b*c*d + 49*a^2*d^2)*x^2)/(b^2*c^2 + 14*a*b*c*d + 49*a^2*d^2))*(-(b^4*c^4 + 28*a*b^3*c^3*d + 294*a^2*b^2*c^2*
d^2 + 1372*a^3*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5))^(3/4))/(b^3*c^3 + 21*a*b^2*c^2*d + 147*a^2*b*c*d^2 + 343*a^
3*d^3)) + 3*(c^2*d^3*x^8 + 2*c^3*d^2*x^4 + c^4*d)*(-(b^4*c^4 + 28*a*b^3*c^3*d + 294*a^2*b^2*c^2*d^2 + 1372*a^3
*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5))^(1/4)*log(3*c^3*d*(-(b^4*c^4 + 28*a*b^3*c^3*d + 294*a^2*b^2*c^2*d^2 + 137
2*a^3*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5))^(1/4) + 3*(b*c + 7*a*d)*x) - 3*(c^2*d^3*x^8 + 2*c^3*d^2*x^4 + c^4*d)
*(-(b^4*c^4 + 28*a*b^3*c^3*d + 294*a^2*b^2*c^2*d^2 + 1372*a^3*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5))^(1/4)*log(-3
*c^3*d*(-(b^4*c^4 + 28*a*b^3*c^3*d + 294*a^2*b^2*c^2*d^2 + 1372*a^3*b*c*d^3 + 2401*a^4*d^4)/(c^11*d^5))^(1/4)
+ 3*(b*c + 7*a*d)*x) - 4*(3*b*c^2 - 11*a*c*d)*x)/(c^2*d^3*x^8 + 2*c^3*d^2*x^4 + c^4*d)

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giac [A]  time = 0.19, size = 286, normalized size = 1.05 \[ \frac {3 \, \sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c + 7 \, \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{128 \, c^{3} d^{2}} + \frac {3 \, \sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c + 7 \, \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {c}{d}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{128 \, c^{3} d^{2}} + \frac {3 \, \sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c + 7 \, \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{256 \, c^{3} d^{2}} - \frac {3 \, \sqrt {2} {\left (\left (c d^{3}\right )^{\frac {1}{4}} b c + 7 \, \left (c d^{3}\right )^{\frac {1}{4}} a d\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {c}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {c}{d}}\right )}{256 \, c^{3} d^{2}} + \frac {b c d x^{5} + 7 \, a d^{2} x^{5} - 3 \, b c^{2} x + 11 \, a c d x}{32 \, {\left (d x^{4} + c\right )}^{2} c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)/(d*x^4+c)^3,x, algorithm="giac")

[Out]

3/128*sqrt(2)*((c*d^3)^(1/4)*b*c + 7*(c*d^3)^(1/4)*a*d)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(c/d)^(1/4))/(c/d)^(
1/4))/(c^3*d^2) + 3/128*sqrt(2)*((c*d^3)^(1/4)*b*c + 7*(c*d^3)^(1/4)*a*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(c
/d)^(1/4))/(c/d)^(1/4))/(c^3*d^2) + 3/256*sqrt(2)*((c*d^3)^(1/4)*b*c + 7*(c*d^3)^(1/4)*a*d)*log(x^2 + sqrt(2)*
x*(c/d)^(1/4) + sqrt(c/d))/(c^3*d^2) - 3/256*sqrt(2)*((c*d^3)^(1/4)*b*c + 7*(c*d^3)^(1/4)*a*d)*log(x^2 - sqrt(
2)*x*(c/d)^(1/4) + sqrt(c/d))/(c^3*d^2) + 1/32*(b*c*d*x^5 + 7*a*d^2*x^5 - 3*b*c^2*x + 11*a*c*d*x)/((d*x^4 + c)
^2*c^2*d)

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maple [A]  time = 0.05, size = 314, normalized size = 1.15 \[ \frac {21 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{128 c^{3}}+\frac {21 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{128 c^{3}}+\frac {21 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, a \ln \left (\frac {x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}{x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}\right )}{256 c^{3}}+\frac {3 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )}{128 c^{2} d}+\frac {3 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )}{128 c^{2} d}+\frac {3 \left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, b \ln \left (\frac {x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}{x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {\frac {c}{d}}}\right )}{256 c^{2} d}+\frac {\frac {\left (7 a d +b c \right ) x^{5}}{32 c^{2}}+\frac {\left (11 a d -3 b c \right ) x}{32 c d}}{\left (d \,x^{4}+c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)/(d*x^4+c)^3,x)

[Out]

(1/32*(7*a*d+b*c)/c^2*x^5+1/32*(11*a*d-3*b*c)/c/d*x)/(d*x^4+c)^2+21/128/c^3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)
/(c/d)^(1/4)*x-1)*a+3/128/c^2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x-1)*b+21/256/c^3*(c/d)^(1/4)*2
^(1/2)*ln((x^2+(c/d)^(1/4)*2^(1/2)*x+(c/d)^(1/2))/(x^2-(c/d)^(1/4)*2^(1/2)*x+(c/d)^(1/2)))*a+3/256/c^2/d*(c/d)
^(1/4)*2^(1/2)*ln((x^2+(c/d)^(1/4)*2^(1/2)*x+(c/d)^(1/2))/(x^2-(c/d)^(1/4)*2^(1/2)*x+(c/d)^(1/2)))*b+21/128/c^
3*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(1/4)*x+1)*a+3/128/c^2/d*(c/d)^(1/4)*2^(1/2)*arctan(2^(1/2)/(c/d)^(
1/4)*x+1)*b

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maxima [A]  time = 1.21, size = 271, normalized size = 0.99 \[ \frac {{\left (b c d + 7 \, a d^{2}\right )} x^{5} - {\left (3 \, b c^{2} - 11 \, a c d\right )} x}{32 \, {\left (c^{2} d^{3} x^{8} + 2 \, c^{3} d^{2} x^{4} + c^{4} d\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} {\left (b c + 7 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {2 \, \sqrt {2} {\left (b c + 7 \, a d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {c} \sqrt {\sqrt {c} \sqrt {d}}} + \frac {\sqrt {2} {\left (b c + 7 \, a d\right )} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (b c + 7 \, a d\right )} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {3}{4}} d^{\frac {1}{4}}}\right )}}{256 \, c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)/(d*x^4+c)^3,x, algorithm="maxima")

[Out]

1/32*((b*c*d + 7*a*d^2)*x^5 - (3*b*c^2 - 11*a*c*d)*x)/(c^2*d^3*x^8 + 2*c^3*d^2*x^4 + c^4*d) + 3/256*(2*sqrt(2)
*(b*c + 7*a*d)*arctan(1/2*sqrt(2)*(2*sqrt(d)*x + sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(c)*sqrt
(sqrt(c)*sqrt(d))) + 2*sqrt(2)*(b*c + 7*a*d)*arctan(1/2*sqrt(2)*(2*sqrt(d)*x - sqrt(2)*c^(1/4)*d^(1/4))/sqrt(s
qrt(c)*sqrt(d)))/(sqrt(c)*sqrt(sqrt(c)*sqrt(d))) + sqrt(2)*(b*c + 7*a*d)*log(sqrt(d)*x^2 + sqrt(2)*c^(1/4)*d^(
1/4)*x + sqrt(c))/(c^(3/4)*d^(1/4)) - sqrt(2)*(b*c + 7*a*d)*log(sqrt(d)*x^2 - sqrt(2)*c^(1/4)*d^(1/4)*x + sqrt
(c))/(c^(3/4)*d^(1/4)))/(c^2*d)

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mupad [B]  time = 1.58, size = 762, normalized size = 2.79 \[ \frac {\frac {x^5\,\left (7\,a\,d+b\,c\right )}{32\,c^2}+\frac {x\,\left (11\,a\,d-3\,b\,c\right )}{32\,c\,d}}{c^2+2\,c\,d\,x^4+d^2\,x^8}-\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}-\frac {9\,\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )\,3{}\mathrm {i}}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}+\frac {\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}+\frac {9\,\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )\,3{}\mathrm {i}}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}}{\frac {3\,\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}-\frac {9\,\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}-\frac {3\,\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}+\frac {9\,\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}}\right )\,\left (7\,a\,d+b\,c\right )\,3{}\mathrm {i}}{64\,{\left (-c\right )}^{11/4}\,d^{5/4}}-\frac {3\,\mathrm {atan}\left (\frac {\frac {3\,\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}-\frac {\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )\,9{}\mathrm {i}}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}+\frac {3\,\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}+\frac {\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )\,9{}\mathrm {i}}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}}{\frac {\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}-\frac {\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )\,9{}\mathrm {i}}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )\,3{}\mathrm {i}}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}-\frac {\left (\frac {9\,x\,\left (49\,a^2\,d^3+14\,a\,b\,c\,d^2+b^2\,c^2\,d\right )}{256\,c^4}+\frac {\left (7\,a\,d+b\,c\right )\,\left (7\,a\,d^3+b\,c\,d^2\right )\,9{}\mathrm {i}}{256\,{\left (-c\right )}^{15/4}\,d^{5/4}}\right )\,\left (7\,a\,d+b\,c\right )\,3{}\mathrm {i}}{128\,{\left (-c\right )}^{11/4}\,d^{5/4}}}\right )\,\left (7\,a\,d+b\,c\right )}{64\,{\left (-c\right )}^{11/4}\,d^{5/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)/(c + d*x^4)^3,x)

[Out]

((x^5*(7*a*d + b*c))/(32*c^2) + (x*(11*a*d - 3*b*c))/(32*c*d))/(c^2 + d^2*x^8 + 2*c*d*x^4) - (atan(((((9*x*(49
*a^2*d^3 + b^2*c^2*d + 14*a*b*c*d^2))/(256*c^4) - (9*(7*a*d + b*c)*(7*a*d^3 + b*c*d^2))/(256*(-c)^(15/4)*d^(5/
4)))*(7*a*d + b*c)*3i)/(128*(-c)^(11/4)*d^(5/4)) + (((9*x*(49*a^2*d^3 + b^2*c^2*d + 14*a*b*c*d^2))/(256*c^4) +
(9*(7*a*d + b*c)*(7*a*d^3 + b*c*d^2))/(256*(-c)^(15/4)*d^(5/4)))*(7*a*d + b*c)*3i)/(128*(-c)^(11/4)*d^(5/4)))
/((3*((9*x*(49*a^2*d^3 + b^2*c^2*d + 14*a*b*c*d^2))/(256*c^4) - (9*(7*a*d + b*c)*(7*a*d^3 + b*c*d^2))/(256*(-c
)^(15/4)*d^(5/4)))*(7*a*d + b*c))/(128*(-c)^(11/4)*d^(5/4)) - (3*((9*x*(49*a^2*d^3 + b^2*c^2*d + 14*a*b*c*d^2)
)/(256*c^4) + (9*(7*a*d + b*c)*(7*a*d^3 + b*c*d^2))/(256*(-c)^(15/4)*d^(5/4)))*(7*a*d + b*c))/(128*(-c)^(11/4)
*d^(5/4))))*(7*a*d + b*c)*3i)/(64*(-c)^(11/4)*d^(5/4)) - (3*atan(((3*((9*x*(49*a^2*d^3 + b^2*c^2*d + 14*a*b*c*
d^2))/(256*c^4) - ((7*a*d + b*c)*(7*a*d^3 + b*c*d^2)*9i)/(256*(-c)^(15/4)*d^(5/4)))*(7*a*d + b*c))/(128*(-c)^(
11/4)*d^(5/4)) + (3*((9*x*(49*a^2*d^3 + b^2*c^2*d + 14*a*b*c*d^2))/(256*c^4) + ((7*a*d + b*c)*(7*a*d^3 + b*c*d
^2)*9i)/(256*(-c)^(15/4)*d^(5/4)))*(7*a*d + b*c))/(128*(-c)^(11/4)*d^(5/4)))/((((9*x*(49*a^2*d^3 + b^2*c^2*d +
14*a*b*c*d^2))/(256*c^4) - ((7*a*d + b*c)*(7*a*d^3 + b*c*d^2)*9i)/(256*(-c)^(15/4)*d^(5/4)))*(7*a*d + b*c)*3i
)/(128*(-c)^(11/4)*d^(5/4)) - (((9*x*(49*a^2*d^3 + b^2*c^2*d + 14*a*b*c*d^2))/(256*c^4) + ((7*a*d + b*c)*(7*a*
d^3 + b*c*d^2)*9i)/(256*(-c)^(15/4)*d^(5/4)))*(7*a*d + b*c)*3i)/(128*(-c)^(11/4)*d^(5/4))))*(7*a*d + b*c))/(64
*(-c)^(11/4)*d^(5/4))

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sympy [A]  time = 1.03, size = 151, normalized size = 0.55 \[ \frac {x^{5} \left (7 a d^{2} + b c d\right ) + x \left (11 a c d - 3 b c^{2}\right )}{32 c^{4} d + 64 c^{3} d^{2} x^{4} + 32 c^{2} d^{3} x^{8}} + \operatorname {RootSum} {\left (268435456 t^{4} c^{11} d^{5} + 194481 a^{4} d^{4} + 111132 a^{3} b c d^{3} + 23814 a^{2} b^{2} c^{2} d^{2} + 2268 a b^{3} c^{3} d + 81 b^{4} c^{4}, \left (t \mapsto t \log {\left (\frac {128 t c^{3} d}{21 a d + 3 b c} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)/(d*x**4+c)**3,x)

[Out]

(x**5*(7*a*d**2 + b*c*d) + x*(11*a*c*d - 3*b*c**2))/(32*c**4*d + 64*c**3*d**2*x**4 + 32*c**2*d**3*x**8) + Root
Sum(268435456*_t**4*c**11*d**5 + 194481*a**4*d**4 + 111132*a**3*b*c*d**3 + 23814*a**2*b**2*c**2*d**2 + 2268*a*
b**3*c**3*d + 81*b**4*c**4, Lambda(_t, _t*log(128*_t*c**3*d/(21*a*d + 3*b*c) + x)))

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