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]

-((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))

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Rubi [A]  time = 0.174277, antiderivative size = 273, normalized size of antiderivative = 1., 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 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 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 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.206871, 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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Maple [A]  time = 0.01, size = 314, normalized size = 1.2 \begin{align*}{\frac{1}{ \left ( d{x}^{4}+c \right ) ^{2}} \left ({\frac{ \left ( 7\,ad+bc \right ){x}^{5}}{32\,{c}^{2}}}+{\frac{ \left ( 11\,ad-3\,bc \right ) x}{32\,cd}} \right ) }+{\frac{21\,\sqrt{2}a}{128\,{c}^{3}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }+{\frac{3\,\sqrt{2}b}{128\,{c}^{2}d}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}+1 \right ) }+{\frac{21\,\sqrt{2}a}{128\,{c}^{3}}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{3\,\sqrt{2}b}{128\,{c}^{2}d}\sqrt [4]{{\frac{c}{d}}}\arctan \left ({x\sqrt{2}{\frac{1}{\sqrt [4]{{\frac{c}{d}}}}}}-1 \right ) }+{\frac{21\,\sqrt{2}a}{256\,{c}^{3}}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) }+{\frac{3\,\sqrt{2}b}{256\,{c}^{2}d}\sqrt [4]{{\frac{c}{d}}}\ln \left ({ \left ({x}^{2}+\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) \left ({x}^{2}-\sqrt [4]{{\frac{c}{d}}}x\sqrt{2}+\sqrt{{\frac{c}{d}}} \right ) ^{-1}} \right ) } \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.39843, size = 1752, normalized size = 6.42 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [A]  time = 1.6505, size = 151, normalized size = 0.55 \begin{align*} \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 )} \end{align*}

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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Giac [A]  time = 1.13543, size = 386, normalized size = 1.41 \begin{align*} \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} \end{align*}

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)