3.406 \(\int \frac{\cos ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=95 \[ \frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4} d}-\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4} d} \]

[Out]

((Sqrt[a] + Sqrt[b])*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(3/4)*d) - ((Sqrt[a] - Sqrt[b])*ArcT
anh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(3/4)*d)

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Rubi [A]  time = 0.101772, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3223, 1167, 205, 208} \[ \frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4} d}-\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{3/4} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]

[Out]

((Sqrt[a] + Sqrt[b])*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(3/4)*d) - ((Sqrt[a] - Sqrt[b])*ArcT
anh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(3/4)*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 1167

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

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]

Rubi steps

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

Mathematica [C]  time = 0.081646, size = 160, normalized size = 1.68 \[ \frac{\left (\sqrt{a}-\sqrt{b}\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )+i \left (\sqrt{a}+\sqrt{b}\right ) \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )-i \left (\sqrt{a}+\sqrt{b}\right ) \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )-\left (\sqrt{a}-\sqrt{b}\right ) \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )}{4 a^{3/4} b^{3/4} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a] - Sqrt[b])*Log[a^(1/4) - b^(1/4)*Sin[c + d*x]] + I*(Sqrt[a] + Sqrt[b])*Log[a^(1/4) - I*b^(1/4)*Sin[c
+ d*x]] - I*(Sqrt[a] + Sqrt[b])*Log[a^(1/4) + I*b^(1/4)*Sin[c + d*x]] - (Sqrt[a] - Sqrt[b])*Log[a^(1/4) + b^(
1/4)*Sin[c + d*x]])/(4*a^(3/4)*b^(3/4)*d)

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Maple [B]  time = 0.071, size = 160, normalized size = 1.7 \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 ) }+{\frac{1}{2\,bd}\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}}-{\frac{1}{4\,bd}\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}{\sqrt [4]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(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))+1/2/d/b/(a/b)^(1/4)*arctan(sin(d*x+c)/(a/b)^(1/4))-1/4/d/b/(a/b)^(1/4)*ln((sin(d*x+c)+(a/b)^(1/
4))/(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)^3/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.87559, size = 1411, normalized size = 14.85 \begin{align*} \frac{1}{4} \, \sqrt{-\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}} \log \left (\frac{1}{2} \,{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac{1}{2} \,{\left (a^{3} b^{2} d^{3} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} -{\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt{-\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}}\right ) - \frac{1}{4} \, \sqrt{\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}} \log \left (\frac{1}{2} \,{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac{1}{2} \,{\left (a^{3} b^{2} d^{3} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} +{\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt{\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}}\right ) - \frac{1}{4} \, \sqrt{-\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}} \log \left (-\frac{1}{2} \,{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac{1}{2} \,{\left (a^{3} b^{2} d^{3} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} -{\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt{-\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}}\right ) + \frac{1}{4} \, \sqrt{\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}} \log \left (-\frac{1}{2} \,{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac{1}{2} \,{\left (a^{3} b^{2} d^{3} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} +{\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt{\frac{a b d^{2} \sqrt{\frac{a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(-(a*b*d^2*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) + 2)/(a*b*d^2))*log(1/2*(a^2 - b^2)*sin(d*x + c) +
1/2*(a^3*b^2*d^3*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) - (a^2*b + a*b^2)*d)*sqrt(-(a*b*d^2*sqrt((a^2 + 2*a*b
+ b^2)/(a^3*b^3*d^4)) + 2)/(a*b*d^2))) - 1/4*sqrt((a*b*d^2*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) - 2)/(a*b*
d^2))*log(1/2*(a^2 - b^2)*sin(d*x + c) + 1/2*(a^3*b^2*d^3*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) + (a^2*b + a
*b^2)*d)*sqrt((a*b*d^2*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) - 2)/(a*b*d^2))) - 1/4*sqrt(-(a*b*d^2*sqrt((a^2
+ 2*a*b + b^2)/(a^3*b^3*d^4)) + 2)/(a*b*d^2))*log(-1/2*(a^2 - b^2)*sin(d*x + c) + 1/2*(a^3*b^2*d^3*sqrt((a^2
+ 2*a*b + b^2)/(a^3*b^3*d^4)) - (a^2*b + a*b^2)*d)*sqrt(-(a*b*d^2*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) + 2)
/(a*b*d^2))) + 1/4*sqrt((a*b*d^2*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) - 2)/(a*b*d^2))*log(-1/2*(a^2 - b^2)*
sin(d*x + c) + 1/2*(a^3*b^2*d^3*sqrt((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) + (a^2*b + a*b^2)*d)*sqrt((a*b*d^2*sqr
t((a^2 + 2*a*b + b^2)/(a^3*b^3*d^4)) - 2)/(a*b*d^2)))

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Sympy [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)**3/(a-b*sin(d*x+c)**4),x)

[Out]

Timed out

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Giac [B]  time = 5.90468, size = 378, normalized size = 3.98 \begin{align*} \frac{\frac{2 \, \sqrt{2}{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \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^{3}} + \frac{2 \, \sqrt{2}{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \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^{3}} + \frac{\sqrt{2}{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} + \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \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^{3}} - \frac{\sqrt{2}{\left (\left (-a b^{3}\right )^{\frac{1}{4}} b^{2} + \left (-a b^{3}\right )^{\frac{3}{4}}\right )} \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^{3}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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