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]

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

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Rubi [A]  time = 0.10, antiderivative size = 95, normalized size of antiderivative = 1.00, 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 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 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 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 ^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*}

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Mathematica [C]  time = 0.08, 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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fricas [B]  time = 0.56, size = 631, normalized size = 6.64 \[ \frac {1}{4} \, \sqrt {-\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}} \log \left (\frac {1}{2} \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (a^{3} b^{2} d^{3} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - {\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt {-\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}}\right ) - \frac {1}{4} \, \sqrt {\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}} \log \left (\frac {1}{2} \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (a^{3} b^{2} d^{3} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + {\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt {\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}}\right ) - \frac {1}{4} \, \sqrt {-\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}} \log \left (-\frac {1}{2} \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (a^{3} b^{2} d^{3} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - {\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt {-\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + 2}{a b d^{2}}}\right ) + \frac {1}{4} \, \sqrt {\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}} \log \left (-\frac {1}{2} \, {\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right ) + \frac {1}{2} \, {\left (a^{3} b^{2} d^{3} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} + {\left (a^{2} b + a b^{2}\right )} d\right )} \sqrt {\frac {a b d^{2} \sqrt {\frac {a^{2} + 2 \, a b + b^{2}}{a^{3} b^{3} d^{4}}} - 2}{a b d^{2}}}\right ) \]

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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giac [B]  time = 0.77, size = 280, normalized size = 2.95 \[ \frac {\frac {2 \, \sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{a b^{3}} + \frac {2 \, \sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} - \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{a b^{3}} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} + \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{a b^{3}} - \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} + \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{a b^{3}}}{8 \, d} \]

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

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maple [B]  time = 0.61, size = 160, normalized size = 1.68 \[ \frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 d a}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 d a}+\frac {\arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 d b \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 d b \left (\frac {a}{b}\right )^{\frac {1}{4}}} \]

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 [A]  time = 0.54, size = 121, normalized size = 1.27 \[ \frac {\frac {2 \, {\left (\sqrt {a} + \sqrt {b}\right )} \arctan \left (\frac {\sqrt {b} \sin \left (d x + c\right )}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {{\left (\sqrt {a} - \sqrt {b}\right )} \log \left (\frac {\sqrt {b} \sin \left (d x + c\right ) - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sin \left (d x + c\right ) + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}}{4 \, d} \]

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]

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

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mupad [B]  time = 15.81, size = 489, normalized size = 5.15 \[ -\frac {2\,\mathrm {atanh}\left (\frac {8\,b^3\,\sin \left (c+d\,x\right )\,\sqrt {-\frac {1}{8\,a\,b}-\frac {\sqrt {a^3\,b^3}}{16\,a^2\,b^3}-\frac {\sqrt {a^3\,b^3}}{16\,a^3\,b^2}}}{2\,a\,b+\frac {2\,\sqrt {a^3\,b^3}}{a}+2\,b^2+\frac {2\,b\,\sqrt {a^3\,b^3}}{a^2}}+\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )\,\sqrt {-\frac {1}{8\,a\,b}-\frac {\sqrt {a^3\,b^3}}{16\,a^2\,b^3}-\frac {\sqrt {a^3\,b^3}}{16\,a^3\,b^2}}}{2\,a\,b+\frac {2\,\sqrt {a^3\,b^3}}{a}+2\,b^2+\frac {2\,b\,\sqrt {a^3\,b^3}}{a^2}}\right )\,\sqrt {-\frac {a\,\sqrt {a^3\,b^3}+b\,\sqrt {a^3\,b^3}+2\,a^2\,b^2}{16\,a^3\,b^3}}}{d}-\frac {2\,\mathrm {atanh}\left (\frac {8\,b^3\,\sin \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}}{16\,a^2\,b^3}-\frac {1}{8\,a\,b}+\frac {\sqrt {a^3\,b^3}}{16\,a^3\,b^2}}}{2\,a\,b-\frac {2\,\sqrt {a^3\,b^3}}{a}+2\,b^2-\frac {2\,b\,\sqrt {a^3\,b^3}}{a^2}}+\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}}{16\,a^2\,b^3}-\frac {1}{8\,a\,b}+\frac {\sqrt {a^3\,b^3}}{16\,a^3\,b^2}}}{2\,a\,b-\frac {2\,\sqrt {a^3\,b^3}}{a}+2\,b^2-\frac {2\,b\,\sqrt {a^3\,b^3}}{a^2}}\right )\,\sqrt {\frac {a\,\sqrt {a^3\,b^3}+b\,\sqrt {a^3\,b^3}-2\,a^2\,b^2}{16\,a^3\,b^3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a - b*sin(c + d*x)^4),x)

[Out]

- (2*atanh((8*b^3*sin(c + d*x)*(- 1/(8*a*b) - (a^3*b^3)^(1/2)/(16*a^2*b^3) - (a^3*b^3)^(1/2)/(16*a^3*b^2))^(1/
2))/(2*a*b + (2*(a^3*b^3)^(1/2))/a + 2*b^2 + (2*b*(a^3*b^3)^(1/2))/a^2) + (8*a*b^2*sin(c + d*x)*(- 1/(8*a*b) -
(a^3*b^3)^(1/2)/(16*a^2*b^3) - (a^3*b^3)^(1/2)/(16*a^3*b^2))^(1/2))/(2*a*b + (2*(a^3*b^3)^(1/2))/a + 2*b^2 +
(2*b*(a^3*b^3)^(1/2))/a^2))*(-(a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + 2*a^2*b^2)/(16*a^3*b^3))^(1/2))/d - (2*
atanh((8*b^3*sin(c + d*x)*((a^3*b^3)^(1/2)/(16*a^2*b^3) - 1/(8*a*b) + (a^3*b^3)^(1/2)/(16*a^3*b^2))^(1/2))/(2*
a*b - (2*(a^3*b^3)^(1/2))/a + 2*b^2 - (2*b*(a^3*b^3)^(1/2))/a^2) + (8*a*b^2*sin(c + d*x)*((a^3*b^3)^(1/2)/(16*
a^2*b^3) - 1/(8*a*b) + (a^3*b^3)^(1/2)/(16*a^3*b^2))^(1/2))/(2*a*b - (2*(a^3*b^3)^(1/2))/a + 2*b^2 - (2*b*(a^3
*b^3)^(1/2))/a^2))*((a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) - 2*a^2*b^2)/(16*a^3*b^3))^(1/2))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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