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]

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)

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Rubi [A]  time = 0.0673509, antiderivative size = 71, normalized size of antiderivative = 1., 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 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])

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 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 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]

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*}

Mathematica [A]  time = 0.0235646, 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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Maple [A]  time = 0.033, size = 81, normalized size = 1.1 \begin{align*}{\frac{1}{4\,da}\sqrt [4]{{\frac{a}{b}}}\ln \left ({ \left ( \sin \left ( dx+c \right ) +\sqrt [4]{{\frac{a}{b}}} \right ) \left ( \sin \left ( dx+c \right ) -\sqrt [4]{{\frac{a}{b}}} \right ) ^{-1}} \right ) }+{\frac{1}{2\,da}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ) } \end{align*}

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

[Out]

Exception raised: ValueError

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

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]

Timed out

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Sympy [A]  time = 16.0517, size = 165, normalized size = 2.32 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \cos{\left (c \right )}}{\sin ^{4}{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{\sin{\left (c + d x \right )}}{a d} & \text{for}\: b = 0 \\\frac{x \cos{\left (c \right )}}{a - b \sin ^{4}{\left (c \right )}} & \text{for}\: d = 0 \\\frac{1}{3 b d \sin ^{3}{\left (c + d x \right )}} & \text{for}\: a = 0 \\- \frac{\log{\left (- \sqrt [4]{a} \sqrt [4]{\frac{1}{b}} + \sin{\left (c + d x \right )} \right )}}{4 a^{\frac{3}{4}} b^{2} d \left (\frac{1}{b}\right )^{\frac{7}{4}}} + \frac{\log{\left (\sqrt [4]{a} \sqrt [4]{\frac{1}{b}} + \sin{\left (c + d x \right )} \right )}}{4 a^{\frac{3}{4}} b^{2} d \left (\frac{1}{b}\right )^{\frac{7}{4}}} + \frac{\operatorname{atan}{\left (\frac{\sin{\left (c + d x \right )}}{\sqrt [4]{a} \sqrt [4]{\frac{1}{b}}} \right )}}{2 a^{\frac{3}{4}} b^{2} d \left (\frac{1}{b}\right )^{\frac{7}{4}}} & \text{otherwise} \end{cases} \end{align*}

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)), (sin(c + d*x)/(a*d), Eq(b, 0)), (x*cos(c)/
(a - b*sin(c)**4), Eq(d, 0)), (1/(3*b*d*sin(c + d*x)**3), Eq(a, 0)), (-log(-a**(1/4)*(1/b)**(1/4) + sin(c + d*
x))/(4*a**(3/4)*b**2*d*(1/b)**(7/4)) + log(a**(1/4)*(1/b)**(1/4) + sin(c + d*x))/(4*a**(3/4)*b**2*d*(1/b)**(7/
4)) + atan(sin(c + d*x)/(a**(1/4)*(1/b)**(1/4)))/(2*a**(3/4)*b**2*d*(1/b)**(7/4)), True))

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Giac [B]  time = 6.58507, size = 302, normalized size = 4.25 \begin{align*} \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} \end{align*}

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