3.389 \(\int \frac {\cos ^2(c+d x)}{a+b \sin ^3(c+d x)} \, dx\)

Optimal. Leaf size=484 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{2/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {2 \tanh ^{-1}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}} \]

[Out]

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

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Rubi [A]  time = 0.63, antiderivative size = 484, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3226, 3213, 2660, 618, 204, 3220, 206} \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {a^{2/3}-b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {a^{2/3}-b^{2/3}}}+\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+(-1)^{2/3} \sqrt [3]{b}}{\sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+\sqrt [3]{-1} b^{2/3}}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{-1} \left ((-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}\right )}{\sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}\right )}{3 b^{2/3} d \sqrt {b^{2/3}-(-1)^{2/3} a^{2/3}}}+\frac {2 \tanh ^{-1}\left (\frac {(-1)^{2/3} \sqrt [3]{a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt [3]{b}}{\sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}}\right )}{3 b^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}+b^{2/3}}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + b*Sin[c + d*x]^3),x]

[Out]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 3226

Int[cos[(e_.) + (f_.)*(x_)]^(m_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_)), x_Symbol] :> Int[Expand[(1 - Sin
[e + f*x]^2)^(m/2)/(a + b*Sin[e + f*x]^n), x], x] /; FreeQ[{a, b, e, f}, x] && IGtQ[m/2, 0] && IntegerQ[(n - 1
)/2]

Rubi steps

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

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Mathematica [C]  time = 0.10, size = 231, normalized size = 0.48 \[ -\frac {i \text {RootSum}\left [i \text {$\#$1}^6 b-3 i \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 i \text {$\#$1}^2 b-i b\& ,\frac {2 \text {$\#$1}^4 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-2 i \text {$\#$1}^2 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )-i \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )+4 \text {$\#$1}^2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i \text {$\#$1}^4 \log \left (\text {$\#$1}^2-2 \text {$\#$1} \cos (c+d x)+1\right )+2 \tan ^{-1}\left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )}{\text {$\#$1}^5 b-2 \text {$\#$1}^3 b-4 i \text {$\#$1}^2 a+\text {$\#$1} b}\& \right ]}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + b*Sin[c + d*x]^3),x]

[Out]

((-1/6*I)*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^6 & , (2*ArcTan[Sin[c + d*x]/(Cos[c
+ d*x] - #1)] - I*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + 4*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - (2*I)
*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + 2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*Log[1 - 2*Cos[c
+ d*x]*#1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1^5) & ])/d

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fricas [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)^2/(a+b*sin(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^2/(b*sin(d*x + c)^3 + a), x)

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maple [C]  time = 0.89, size = 83, normalized size = 0.17 \[ \frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 \textit {\_R}^{2}+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+b*sin(d*x+c)^3),x)

[Out]

1/3/d*sum((_R^4-2*_R^2+1)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a+3*_Z^4*a+
8*_Z^3*b+3*_Z^2*a+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a+b*sin(d*x+c)^3),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^2/(b*sin(d*x + c)^3 + a), x)

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mupad [B]  time = 17.40, size = 951, normalized size = 1.96 \[ \frac {\sum _{k=1}^6\ln \left (24576\,a^4-24576\,a^2\,b^2-\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )\,a^2\,b^3\,122880-\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,24576-32768\,a\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+32768\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^2\,a^2\,b^4\,294912+{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^2\,a^4\,b^2\,294912+{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^4\,a^4\,b^4\,663552-{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^4\,a^6\,b^2\,663552-{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^5\,a^4\,b^5\,7962624+{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^5\,a^6\,b^3\,5971968+\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )\,a^4\,b\,49152+\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,147456+{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^2\,a^5\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,294912-{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^2\,a^3\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,294912+{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^3\,a^3\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1769472-{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^3\,a^5\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1769472-{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^4\,a^3\,b^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5308416+{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^4\,a^5\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5308416-{\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}^5\,a^5\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1990656-\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )\,a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,196608\right )\,\mathrm {root}\left (729\,a^4\,b^4\,d^6+27\,a^2\,b^2\,d^2+a^2-b^2,d,k\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(a + b*sin(c + d*x)^3),x)

[Out]

symsum(log(24576*a^4 - 24576*a^2*b^2 - 122880*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)*a^2*b^3
- 24576*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)*a^5*tan(c/2 + (d*x)/2) - 32768*a*b^3*tan(c/2
+ (d*x)/2) + 32768*a^3*b*tan(c/2 + (d*x)/2) - 294912*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)
^2*a^2*b^4 + 294912*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^2*a^4*b^2 + 663552*root(729*a^4*b
^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^4*a^4*b^4 - 663552*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^
2, d, k)^4*a^6*b^2 - 7962624*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^5*a^4*b^5 + 5971968*root
(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^5*a^6*b^3 + 49152*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 +
a^2 - b^2, d, k)*a^4*b + 147456*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)*a^3*b^2*tan(c/2 + (d
*x)/2) + 294912*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^2*a^5*b*tan(c/2 + (d*x)/2) - 294912*r
oot(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^2*a^3*b^3*tan(c/2 + (d*x)/2) + 1769472*root(729*a^4*b^
4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^3*a^3*b^4*tan(c/2 + (d*x)/2) - 1769472*root(729*a^4*b^4*d^6 + 27*a^2
*b^2*d^2 + a^2 - b^2, d, k)^3*a^5*b^2*tan(c/2 + (d*x)/2) - 5308416*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2
- b^2, d, k)^4*a^3*b^5*tan(c/2 + (d*x)/2) + 5308416*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^
4*a^5*b^3*tan(c/2 + (d*x)/2) - 1990656*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)^5*a^5*b^4*tan(
c/2 + (d*x)/2) - 196608*root(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k)*a*b^4*tan(c/2 + (d*x)/2))*roo
t(729*a^4*b^4*d^6 + 27*a^2*b^2*d^2 + a^2 - b^2, d, k), k, 1, 6)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+b*sin(d*x+c)**3),x)

[Out]

Integral(cos(c + d*x)**2/(a + b*sin(c + d*x)**3), x)

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