3.407 \(\int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=71 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d} \]

[Out]

1/2*arctan(b^(1/4)*sin(d*x+c)/a^(1/4))/a^(3/4)/b^(1/4)/d+1/2*arctanh(b^(1/4)*sin(d*x+c)/a^(1/4))/a^(3/4)/b^(1/
4)/d

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Rubi [A]  time = 0.07, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3223, 212, 208, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a - b*Sin[c + d*x]^4),x]

[Out]

ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)]/(2*a^(3/4)*b^(1/4)*d) + ArcTanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)]/(2*a^(3
/4)*b^(1/4)*d)

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 3223

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x]
, x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (EqQ[n, 4] || GtQ[m, 0
] || IGtQ[p, 0] || IntegersQ[m, p])

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{a-b \sin ^4(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{a-b x^4} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {a}-\sqrt {b} x^2} \, dx,x,\sin (c+d x)\right )}{2 \sqrt {a} d}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {a}+\sqrt {b} x^2} \, dx,x,\sin (c+d x)\right )}{2 \sqrt {a} d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 54, normalized size = 0.76 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt [4]{b} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)] + ArcTanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(1/4)*d)

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fricas [B]  time = 110.36, size = 330, normalized size = 4.65 \[ \frac {1}{2} \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \arctan \left (a^{2} b d^{3} \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {3}{4}} \sin \left (d x + c\right ) + \sqrt {a^{2} d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} - \cos \left (d x + c\right )^{2} + 1} a^{2} b d^{3} \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {3}{4}}\right ) - \frac {1}{2} \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \arctan \left (-a^{2} b d^{3} \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {3}{4}} \sin \left (d x + c\right ) + \sqrt {a^{2} d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} - \cos \left (d x + c\right )^{2} + 1} a^{2} b d^{3} \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {3}{4}}\right ) + \frac {1}{8} \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {1}{4} \, a^{2} d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} + \frac {1}{2} \, a d \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) - \frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) - \frac {1}{8} \, \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \log \left (\frac {1}{4} \, a^{2} d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} - \frac {1}{2} \, a d \left (\frac {1}{a^{3} b d^{4}}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) - \frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1/(a^3*b*d^4))^(1/4)*arctan(a^2*b*d^3*(1/(a^3*b*d^4))^(3/4)*sin(d*x + c) + sqrt(a^2*d^2*sqrt(1/(a^3*b*d^4
)) - cos(d*x + c)^2 + 1)*a^2*b*d^3*(1/(a^3*b*d^4))^(3/4)) - 1/2*(1/(a^3*b*d^4))^(1/4)*arctan(-a^2*b*d^3*(1/(a^
3*b*d^4))^(3/4)*sin(d*x + c) + sqrt(a^2*d^2*sqrt(1/(a^3*b*d^4)) - cos(d*x + c)^2 + 1)*a^2*b*d^3*(1/(a^3*b*d^4)
)^(3/4)) + 1/8*(1/(a^3*b*d^4))^(1/4)*log(1/4*a^2*d^2*sqrt(1/(a^3*b*d^4)) + 1/2*a*d*(1/(a^3*b*d^4))^(1/4)*sin(d
*x + c) - 1/4*cos(d*x + c)^2 + 1/4) - 1/8*(1/(a^3*b*d^4))^(1/4)*log(1/4*a^2*d^2*sqrt(1/(a^3*b*d^4)) - 1/2*a*d*
(1/(a^3*b*d^4))^(1/4)*sin(d*x + c) - 1/4*cos(d*x + c)^2 + 1/4)

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giac [B]  time = 0.75, size = 224, normalized size = 3.15 \[ \frac {\frac {2 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{a b} + \frac {2 \, \sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{a b} + \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \log \left (\sin \left (d x + c\right )^{2} + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{a b} - \frac {\sqrt {2} \left (-a b^{3}\right )^{\frac {1}{4}} \log \left (\sin \left (d x + c\right )^{2} - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{a b}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(2)*(-a*b^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) + 2*sin(d*x + c))/(-a/b)^(1/4))/(a*b) +
2*sqrt(2)*(-a*b^3)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) - 2*sin(d*x + c))/(-a/b)^(1/4))/(a*b) + sq
rt(2)*(-a*b^3)^(1/4)*log(sin(d*x + c)^2 + sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/(a*b) - sqrt(2)*(-a*
b^3)^(1/4)*log(sin(d*x + c)^2 - sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/(a*b))/d

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maple [A]  time = 0.23, size = 81, normalized size = 1.14 \[ \frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 d a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 d a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(a-b*sin(d*x+c)^4),x)

[Out]

1/4/d*(a/b)^(1/4)/a*ln((sin(d*x+c)+(a/b)^(1/4))/(sin(d*x+c)-(a/b)^(1/4)))+1/2/d*(a/b)^(1/4)/a*arctan(sin(d*x+c
)/(a/b)^(1/4))

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maxima [A]  time = 0.65, size = 100, normalized size = 1.41 \[ \frac {\frac {2 \, \arctan \left (\frac {\sqrt {b} \sin \left (d x + c\right )}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} - \frac {\log \left (\frac {\sqrt {b} \sin \left (d x + c\right ) - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sin \left (d x + c\right ) + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*arctan(sqrt(b)*sin(d*x + c)/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) - log((sqrt(b)*sin(d
*x + c) - sqrt(sqrt(a)*sqrt(b)))/(sqrt(b)*sin(d*x + c) + sqrt(sqrt(a)*sqrt(b))))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b)
)))/d

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mupad [B]  time = 0.11, size = 40, normalized size = 0.56 \[ \frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sin \left (c+d\,x\right )}{a^{1/4}}\right )+\mathrm {atanh}\left (\frac {b^{1/4}\,\sin \left (c+d\,x\right )}{a^{1/4}}\right )}{2\,a^{3/4}\,b^{1/4}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)/(a - b*sin(c + d*x)^4),x)

[Out]

(atan((b^(1/4)*sin(c + d*x))/a^(1/4)) + atanh((b^(1/4)*sin(c + d*x))/a^(1/4)))/(2*a^(3/4)*b^(1/4)*d)

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sympy [A]  time = 8.90, size = 155, normalized size = 2.18 \[ \begin {cases} \frac {\tilde {\infty } x \cos {\relax (c )}}{\sin ^{4}{\relax (c )}} & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {1}{3 b d \sin ^{3}{\left (c + d x \right )}} & \text {for}\: a = 0 \\\frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: b = 0 \\\frac {x \cos {\relax (c )}}{a - b \sin ^{4}{\relax (c )}} & \text {for}\: d = 0 \\- \frac {\sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sin {\left (c + d x \right )} \right )}}{4 a^{\frac {3}{4}} d} + \frac {\sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sin {\left (c + d x \right )} \right )}}{4 a^{\frac {3}{4}} d} + \frac {\sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\sin {\left (c + d x \right )}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{2 a^{\frac {3}{4}} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a-b*sin(d*x+c)**4),x)

[Out]

Piecewise((zoo*x*cos(c)/sin(c)**4, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (1/(3*b*d*sin(c + d*x)**3), Eq(a, 0)), (si
n(c + d*x)/(a*d), Eq(b, 0)), (x*cos(c)/(a - b*sin(c)**4), Eq(d, 0)), (-(1/b)**(1/4)*log(-a**(1/4)*(1/b)**(1/4)
+ sin(c + d*x))/(4*a**(3/4)*d) + (1/b)**(1/4)*log(a**(1/4)*(1/b)**(1/4) + sin(c + d*x))/(4*a**(3/4)*d) + (1/b
)**(1/4)*atan(sin(c + d*x)/(a**(1/4)*(1/b)**(1/4)))/(2*a**(3/4)*d), True))

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