3.1339 \(\int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=183 \[ \frac {a^2 \cos (c+d x)}{b d \left (a^2-b^2\right )}-\frac {b \cos (c+d x)}{d \left (a^2-b^2\right )}+\frac {a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac {a x}{a^2-b^2}-\frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{3/2}}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2902, 3473, 8, 2590, 14, 2746, 12, 2735, 2660, 618, 204} \[ \frac {a^2 \cos (c+d x)}{b d \left (a^2-b^2\right )}-\frac {b \cos (c+d x)}{d \left (a^2-b^2\right )}-\frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{3/2}}+\frac {a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}+\frac {a^3 x}{b^2 \left (a^2-b^2\right )}-\frac {a x}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

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 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2746

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b^2
*Cos[e + f*x])/(d*f), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]  time = 1.14, size = 186, normalized size = 1.02 \[ \frac {-\frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{3/2}}+\frac {-\left (a^3 (c+d x)\right )+a b^2 (c+d x)+b^3}{b^4-a^2 b^2}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\cos (c+d x)}{b}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 431, normalized size = 2.36 \[ \left [\frac {\sqrt {-a^{2} + b^{2}} a^{4} \cos \left (d x + c\right ) \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )}, \frac {\sqrt {a^{2} - b^{2}} a^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{2} b^{3} + b^{5} + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(-a^2 + b^2)*a^4*cos(d*x + c)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2
*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a
^2 - b^2)) - 2*a^2*b^3 + 2*b^5 + 2*(a^5 - 2*a^3*b^2 + a*b^4)*d*x*cos(d*x + c) + 2*(a^4*b - 2*a^2*b^3 + b^5)*co
s(d*x + c)^2 + 2*(a^3*b^2 - a*b^4)*sin(d*x + c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c)), (sqrt(a^2 - b^2
)*a^4*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*cos(d*x + c) - a^2*b^3 + b^5 + (a^5 - 2*a^3
*b^2 + a*b^4)*d*x*cos(d*x + c) + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c)^2 + (a^3*b^2 - a*b^4)*sin(d*x + c))/((
a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c))]

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giac [A]  time = 0.21, size = 173, normalized size = 0.95 \[ -\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{4}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}}} - \frac {{\left (d x + c\right )} a}{b^{2}} + \frac {2 \, {\left (a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2} - 2 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )} {\left (a^{2} b - b^{3}\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.37, size = 162, normalized size = 0.89 \[ -\frac {32}{d \left (32 a +32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {32}{d \left (32 a -32 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right ) \left (a +b \right ) b^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/d/(32*a+32*b)/(tan(1/2*d*x+1/2*c)-1)+2/d/b/(1+tan(1/2*d*x+1/2*c)^2)+2/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))-3
2/d/(32*a-32*b)/(tan(1/2*d*x+1/2*c)+1)-2/d/(a-b)/(a+b)*a^4/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2
*c)+2*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 14.28, size = 1656, normalized size = 9.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^5*sin(c + d*x) - 6*a^5*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(d*cos(c + d*x)*(a^2 - b^2
)*(a^4 + b^4 - 2*a^2*b^2)) + (2*a^7*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^2*d*(a^2 - b^2)*(a^4 + b^4
- 2*a^2*b^2)) + (a^6*cos(c + d*x) + a^6/2 + (a^6*cos(2*c + 2*d*x))/2)/(b*d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^
4 - 2*a^2*b^2)) + (b^3*(5*a^2*cos(c + d*x) + (7*a^2)/2 + (3*a^2*cos(2*c + 2*d*x))/2))/(d*cos(c + d*x)*(a^2 - b
^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^2*(2*a^3*sin(c + d*x) - 6*a^3*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 +
(d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^5*(2*cos(c + d*x) + cos(2*c + 2*d*x)/2
+ 3/2))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b^4*(a*sin(c + d*x) - 2*a*cos(c + d*x)*atan(si
n(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b*(4*a^4*cos(c
+ d*x) + (5*a^4)/2 + (3*a^4*cos(2*c + 2*d*x))/2))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (a^4*
atan(((2*b^14*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^14*sin(c/2 + (d*x)/2)*(b^6 -
a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^8*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 6*a^
3*b^11*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 15*a^5*b^9*cos(c/2 + (d*x)/2)*(b^6 - a^6
- 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 20*a^7*b^7*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 15
*a^9*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 6*a^11*b^3*cos(c/2 + (d*x)/2)*(b^6 - a
^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 3*a^6*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) - 1
3*a^2*b^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 36*a^4*b^10*sin(c/2 + (d*x)/2)*(b^6
- a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 56*a^6*b^8*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)
+ 54*a^8*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 33*a^10*b^4*sin(c/2 + (d*x)/2)*(b
^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 12*a^12*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(
1/2) + a^7*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + a*b^13*cos(c/2 + (d*x)/2)*(b^6 - a
^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + a^13*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))*1i)/(
(a^4 + b^4 - 2*a^2*b^2)*(2*b^13*sin(c/2 + (d*x)/2) + a*b^12*cos(c/2 + (d*x)/2) - 6*a^3*b^10*cos(c/2 + (d*x)/2)
+ 15*a^5*b^8*cos(c/2 + (d*x)/2) - 19*a^7*b^6*cos(c/2 + (d*x)/2) + 12*a^9*b^4*cos(c/2 + (d*x)/2) - 3*a^11*b^2*
cos(c/2 + (d*x)/2) - 12*a^2*b^11*sin(c/2 + (d*x)/2) + 30*a^4*b^9*sin(c/2 + (d*x)/2) - 38*a^6*b^7*sin(c/2 + (d*
x)/2) + 24*a^8*b^5*sin(c/2 + (d*x)/2) - 6*a^10*b^3*sin(c/2 + (d*x)/2))))*(-(a + b)^3*(a - b)^3)^(1/2)*2i)/(b^2
*d*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(c + d*x)**4*sec(c + d*x)**2/(a + b*sin(c + d*x)), x)

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