3.5.15 \(\int \frac {\cos ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [415]

Optimal. Leaf size=125 \[ -\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan ^{-1}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} d}+\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan ^{-1}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} d} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3303, 1107, 211} \begin {gather*} \frac {\sqrt {\sqrt {a}+\sqrt {b}} \text {ArcTan}\left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} d}-\frac {\sqrt {\sqrt {a}-\sqrt {b}} \text {ArcTan}\left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \sqrt {b} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a - b*Sin[c + d*x]^4),x]

[Out]

-1/2*(Sqrt[Sqrt[a] - Sqrt[b]]*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(a^(3/4)*Sqrt[b]*d) + (S
qrt[Sqrt[a] + Sqrt[b]]*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*a^(3/4)*Sqrt[b]*d)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1107

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

Rule 3303

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(m/2 + 2
*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.20, size = 158, normalized size = 1.26 \begin {gather*} \frac {\frac {\left (\sqrt {a} \sqrt {b}+b\right ) \tan ^{-1}\left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {\left (\sqrt {a} \sqrt {b}-b\right ) \tanh ^{-1}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {-a+\sqrt {a} \sqrt {b}}}}{2 \sqrt {a} b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a - b*Sin[c + d*x]^4),x]

[Out]

(((Sqrt[a]*Sqrt[b] + b)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqrt[b]]])/Sqrt[a + Sqrt[a]
*Sqrt[b]] + ((Sqrt[a]*Sqrt[b] - b)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/Sqr
t[-a + Sqrt[a]*Sqrt[b]])/(2*Sqrt[a]*b*d)

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Maple [A]
time = 0.56, size = 115, normalized size = 0.92

method	result	size
risch	\(\munderset {\textit {\_R} =\RootOf \left (256 a^{3} b^{2} d^{4} \textit {\_Z}^{4}+32 a^{2} b \,d^{2} \textit {\_Z}^{2}+a -b \right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+32 a^{2} d^{2} \textit {\_R}^{2}+8 i a d \textit {\_R} +\frac {2 a}{b}-1\right )\)	\(73\)
derivativedivides	\(\frac {\left (a -b \right ) \left (-\frac {\arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\arctanh \left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\)	\(115\)
default	\(\frac {\left (a -b \right ) \left (-\frac {\arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\arctanh \left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a-b*sin(d*x+c)^4),x, algorithm="maxima")

[Out]

-integrate(cos(d*x + c)^2/(b*sin(d*x + c)^4 - a), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (85) = 170\).
time = 0.50, size = 541, normalized size = 4.33 \begin {gather*} -\frac {1}{8} \, \sqrt {-\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} + 1}{a b d^{2}}} \log \left (\frac {1}{2} \, a d \sqrt {-\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} + 1}{a b d^{2}}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} d^{2} \cos \left (d x + c\right )^{2} - a^{2} d^{2}\right )} \sqrt {\frac {1}{a^{3} b d^{4}}} - \frac {1}{4}\right ) + \frac {1}{8} \, \sqrt {-\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} + 1}{a b d^{2}}} \log \left (-\frac {1}{2} \, a d \sqrt {-\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} + 1}{a b d^{2}}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} d^{2} \cos \left (d x + c\right )^{2} - a^{2} d^{2}\right )} \sqrt {\frac {1}{a^{3} b d^{4}}} - \frac {1}{4}\right ) + \frac {1}{8} \, \sqrt {\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} - 1}{a b d^{2}}} \log \left (\frac {1}{2} \, a d \sqrt {\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} - 1}{a b d^{2}}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} d^{2} \cos \left (d x + c\right )^{2} - a^{2} d^{2}\right )} \sqrt {\frac {1}{a^{3} b d^{4}}} + \frac {1}{4}\right ) - \frac {1}{8} \, \sqrt {\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} - 1}{a b d^{2}}} \log \left (-\frac {1}{2} \, a d \sqrt {\frac {a b d^{2} \sqrt {\frac {1}{a^{3} b d^{4}}} - 1}{a b d^{2}}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4} \, {\left (2 \, a^{2} d^{2} \cos \left (d x + c\right )^{2} - a^{2} d^{2}\right )} \sqrt {\frac {1}{a^{3} b d^{4}}} + \frac {1}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a-b*sin(d*x+c)^4),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a-b*sin(d*x+c)**4),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (85) = 170\).
time = 0.83, size = 558, normalized size = 4.46 \begin {gather*} \frac {\frac {{\left (3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a^{2} b - 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} a b^{2} - \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} b^{3} - 3 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} + 6 \, \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b + \sqrt {a^{2} - a b + \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a + \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} b - 12 \, a^{4} b^{2} + 14 \, a^{3} b^{3} - 4 \, a^{2} b^{4} - a b^{5}} + \frac {{\left (3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a^{2} b - 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} a b^{2} - \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} b^{3} - 3 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a^{2} + 6 \, \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} a b + \sqrt {a^{2} - a b - \sqrt {a b} {\left (a - b\right )}} \sqrt {a b} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \tan \left (d x + c\right )}{\sqrt {\frac {4 \, a - \sqrt {-16 \, {\left (a - b\right )} a + 16 \, a^{2}}}{a - b}}}\right )\right )} {\left | a - b \right |}}{3 \, a^{5} b - 12 \, a^{4} b^{2} + 14 \, a^{3} b^{3} - 4 \, a^{2} b^{4} - a b^{5}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a-b*sin(d*x+c)^4),x, algorithm="giac")

[Out]

1/2*((3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a^2*b - 6*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*a*b^2 - sqrt(a^2 - a
*b + sqrt(a*b)*(a - b))*b^3 - 3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2 + 6*sqrt(a^2 - a*b + sqrt(a*
b)*(a - b))*sqrt(a*b)*a*b + sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*b^2)*(pi*floor((d*x + c)/pi + 1/2) +
arctan(2*tan(d*x + c)/sqrt((4*a + sqrt(-16*(a - b)*a + 16*a^2))/(a - b))))*abs(a - b)/(3*a^5*b - 12*a^4*b^2 +
14*a^3*b^3 - 4*a^2*b^4 - a*b^5) + (3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b - 6*sqrt(a^2 - a*b - sqrt(a*b)
*(a - b))*a*b^2 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*b^3 - 3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^
2 + 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b + sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^2)*(
pi*floor((d*x + c)/pi + 1/2) + arctan(2*tan(d*x + c)/sqrt((4*a - sqrt(-16*(a - b)*a + 16*a^2))/(a - b))))*abs(
a - b)/(3*a^5*b - 12*a^4*b^2 + 14*a^3*b^3 - 4*a^2*b^4 - a*b^5))/d

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Mupad [B]
time = 15.66, size = 1409, normalized size = 11.27 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )+\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2+\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\,1{}\mathrm {i}+\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )-\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2-\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\,1{}\mathrm {i}}{\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )+\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2+\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}-\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )-\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2-\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}}\right )\,\sqrt {-\frac {\sqrt {a^3\,b^3}+a^2\,b}{16\,a^3\,b^2}}\,2{}\mathrm {i}}{d}+\frac {\mathrm {atan}\left (\frac {\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )+\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2+\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\,1{}\mathrm {i}+\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )-\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2-\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\,1{}\mathrm {i}}{\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )+\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2+\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}-\left (\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a^3-12\,a^2\,b+12\,a\,b^2-4\,b^3\right )-\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\,\left (16\,a\,b^3+16\,a^3\,b-32\,a^2\,b^2-\mathrm {tan}\left (c+d\,x\right )\,\left (64\,a^4\,b-128\,a^3\,b^2+64\,a^2\,b^3\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\right )\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}}\right )\,\sqrt {\frac {\sqrt {a^3\,b^3}-a^2\,b}{16\,a^3\,b^2}}\,2{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(a - b*sin(c + d*x)^4),x)

[Out]

(atan(((tan(c + d*x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3) + (-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)*(
16*a*b^3 + 16*a^3*b - 32*a^2*b^2 + tan(c + d*x)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2)*(-((a^3*b^3)^(1/2) + a^2
*b)/(16*a^3*b^2))^(1/2)))*(-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)*1i + (tan(c + d*x)*(12*a*b^2 - 12*a^
2*b + 4*a^3 - 4*b^3) - (-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)*(16*a*b^3 + 16*a^3*b - 32*a^2*b^2 - tan
(c + d*x)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2)*(-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)))*(-((a^3*b^3)
^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)*1i)/((tan(c + d*x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3) + (-((a^3*b^3)^(1
/2) + a^2*b)/(16*a^3*b^2))^(1/2)*(16*a*b^3 + 16*a^3*b - 32*a^2*b^2 + tan(c + d*x)*(64*a^4*b + 64*a^2*b^3 - 128
*a^3*b^2)*(-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)))*(-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2) -
(tan(c + d*x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3) - (-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)*(16*a*b^
3 + 16*a^3*b - 32*a^2*b^2 - tan(c + d*x)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2)*(-((a^3*b^3)^(1/2) + a^2*b)/(16
*a^3*b^2))^(1/2)))*(-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))^(1/2)))*(-((a^3*b^3)^(1/2) + a^2*b)/(16*a^3*b^2))
^(1/2)*2i)/d + (atan(((tan(c + d*x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3) + (((a^3*b^3)^(1/2) - a^2*b)/(16*a^3
*b^2))^(1/2)*(16*a*b^3 + 16*a^3*b - 32*a^2*b^2 + tan(c + d*x)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2)*(((a^3*b^3
)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)))*(((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)*1i + (tan(c + d*x)*(12*a
*b^2 - 12*a^2*b + 4*a^3 - 4*b^3) - (((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)*(16*a*b^3 + 16*a^3*b - 32*a^
2*b^2 - tan(c + d*x)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2)*(((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)))*((
(a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)*1i)/((tan(c + d*x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3) + (((a^3
*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)*(16*a*b^3 + 16*a^3*b - 32*a^2*b^2 + tan(c + d*x)*(64*a^4*b + 64*a^2*b
^3 - 128*a^3*b^2)*(((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)))*(((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1
/2) - (tan(c + d*x)*(12*a*b^2 - 12*a^2*b + 4*a^3 - 4*b^3) - (((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)*(16
*a*b^3 + 16*a^3*b - 32*a^2*b^2 - tan(c + d*x)*(64*a^4*b + 64*a^2*b^3 - 128*a^3*b^2)*(((a^3*b^3)^(1/2) - a^2*b)
/(16*a^3*b^2))^(1/2)))*(((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2))^(1/2)))*(((a^3*b^3)^(1/2) - a^2*b)/(16*a^3*b^2
))^(1/2)*2i)/d

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