3.405 \(\int \frac{\cos ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

Optimal. Leaf size=113 \[ \frac{\left (\sqrt{a}+\sqrt{b}\right )^2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}+\frac{\left (-2 \sqrt{a} \sqrt{b}+a+b\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}-\frac{\sin (c+d x)}{b d} \]

[Out]

((Sqrt[a] + Sqrt[b])^2*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(5/4)*d) + ((a - 2*Sqrt[a]*Sqrt[b]
+ b)*ArcTanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(5/4)*d) - Sin[c + d*x]/(b*d)

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Rubi [A]  time = 0.161679, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3223, 1171, 1167, 205, 208} \[ \frac{\left (\sqrt{a}+\sqrt{b}\right )^2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}+\frac{\left (-2 \sqrt{a} \sqrt{b}+a+b\right ) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}-\frac{\sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]

[Out]

((Sqrt[a] + Sqrt[b])^2*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(5/4)*d) + ((a - 2*Sqrt[a]*Sqrt[b]
+ b)*ArcTanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*b^(5/4)*d) - Sin[c + d*x]/(b*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 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + c*x^
4), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[q]

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 ^5(c+d x)}{a-b \sin ^4(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{a-b x^4} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{b}+\frac{a+b-2 b x^2}{b \left (a-b x^4\right )}\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{\sin (c+d x)}{b d}+\frac{\operatorname{Subst}\left (\int \frac{a+b-2 b x^2}{a-b x^4} \, dx,x,\sin (c+d x)\right )}{b d}\\ &=-\frac{\sin (c+d x)}{b d}-\frac{\left (2 \sqrt{b}-\frac{a+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 \sqrt{b} d}-\frac{\left (2 \sqrt{b}+\frac{a+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 \sqrt{b} d}\\ &=\frac{\left (\sqrt{a}+\sqrt{b}\right )^2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}+\frac{\left (\sqrt{a}-\sqrt{b}\right )^2 \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} b^{5/4} d}-\frac{\sin (c+d x)}{b d}\\ \end{align*}

Mathematica [C]  time = 0.210647, size = 189, normalized size = 1.67 \[ \frac{-4 a^{3/4} \sqrt [4]{b} \sin (c+d x)+\left (\sqrt{a}-\sqrt{b}\right )^2 \left (-\log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )\right )+i \left (-i \left (\sqrt{a}-\sqrt{b}\right )^2 \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )+\left (\sqrt{a}+\sqrt{b}\right )^2 \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )-\left (\sqrt{a}+\sqrt{b}\right )^2 \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )\right )}{4 a^{3/4} b^{5/4} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a - b*Sin[c + d*x]^4),x]

[Out]

(-((Sqrt[a] - Sqrt[b])^2*Log[a^(1/4) - b^(1/4)*Sin[c + d*x]]) + I*((Sqrt[a] + Sqrt[b])^2*Log[a^(1/4) - I*b^(1/
4)*Sin[c + d*x]] - (Sqrt[a] + Sqrt[b])^2*Log[a^(1/4) + I*b^(1/4)*Sin[c + d*x]] - I*(Sqrt[a] - Sqrt[b])^2*Log[a
^(1/4) + b^(1/4)*Sin[c + d*x]]) - 4*a^(3/4)*b^(1/4)*Sin[c + d*x])/(4*a^(3/4)*b^(5/4)*d)

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Maple [B]  time = 0.069, size = 252, normalized size = 2.2 \begin{align*} -{\frac{\sin \left ( dx+c \right ) }{bd}}+{\frac{1}{2\,bd}\sqrt [4]{{\frac{a}{b}}}\arctan \left ({\sin \left ( dx+c \right ){\frac{1}{\sqrt [4]{{\frac{a}{b}}}}}} \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}{4\,bd}\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}{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}{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}{2\,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)^5/(a-b*sin(d*x+c)^4),x)

[Out]

-sin(d*x+c)/b/d+1/2/d/b*(a/b)^(1/4)*arctan(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))+1/4/d/b*(a/b)^(1/4)*ln((sin(d*x+c)+(a/b)^(1/4))/(sin(d*x+c)-(a/b)^(1/4)))+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/d/b/(a/b)^(1/4)*arctan(sin(d*x+c)/(a/b)^(1/4))-1/2/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)^5/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.8742, size = 2337, normalized size = 20.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

-1/4*(b*d*sqrt(-(a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4*a + 4*b)/(a*
b^2*d^2))*log(1/2*(a^4 + 4*a^3*b - 10*a^2*b^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b^4*d^3*sqrt((a^4 + 1
2*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) - (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt(-(a*b^
2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4*a + 4*b)/(a*b^2*d^2))) - b*d*sqrt
((a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) - 4*a - 4*b)/(a*b^2*d^2))*log(1
/2*(a^4 + 4*a^3*b - 10*a^2*b^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b^4*d^3*sqrt((a^4 + 12*a^3*b + 38*a^
2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt((a*b^2*d^2*sqrt((a^4
+ 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) - 4*a - 4*b)/(a*b^2*d^2))) - b*d*sqrt(-(a*b^2*d^2*sqr
t((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4*a + 4*b)/(a*b^2*d^2))*log(-1/2*(a^4 + 4*a^
3*b - 10*a^2*b^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b^4*d^3*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b
^3 + b^4)/(a^3*b^5*d^4)) - (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt(-(a*b^2*d^2*sqrt((a^4 + 12*a^3*b +
38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) + 4*a + 4*b)/(a*b^2*d^2))) + b*d*sqrt((a*b^2*d^2*sqrt((a^4 + 12*a^
3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3*b^5*d^4)) - 4*a - 4*b)/(a*b^2*d^2))*log(-1/2*(a^4 + 4*a^3*b - 10*a^2*b
^2 + 4*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*a^3*b^4*d^3*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12*a*b^3 + b^4)/(a^3
*b^5*d^4)) + (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4)*d)*sqrt((a*b^2*d^2*sqrt((a^4 + 12*a^3*b + 38*a^2*b^2 + 12
*a*b^3 + b^4)/(a^3*b^5*d^4)) - 4*a - 4*b)/(a*b^2*d^2))) + 4*sin(d*x + c))/(b*d)

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

[Out]

Timed out

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Giac [B]  time = 6.39496, size = 420, normalized size = 3.72 \begin{align*} -\frac{\frac{8 \, \sin \left (d x + c\right )}{b} - \frac{2 \, \sqrt{2}{\left (\left (-a b^{3}\right )^{\frac{1}{4}}{\left (a b + b^{2}\right )} - 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}}{\left (a b + b^{2}\right )} - 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}}{\left (a b + b^{2}\right )} + 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}}{\left (a b + b^{2}\right )} + 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)^5/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

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