3.1465 \(\int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=212 \[ -\frac {2 a^2 \left (a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{5/2}}+\frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{5/2}}+\frac {a^3 \cos (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

[Out]

2*a^4*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b/(a^2-b^2)^(5/2)/d-2*a^2*(a^2-3*b^2)*arctan((b+a*tan(1
/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b/(a^2-b^2)^(5/2)/d+1/2*cos(d*x+c)/(a+b)^2/d/(1-sin(d*x+c))+1/2*cos(d*x+c)/(a-
b)^2/d/(1+sin(d*x+c))+a^3*cos(d*x+c)/(a^2-b^2)^2/d/(a+b*sin(d*x+c))

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Rubi [A]  time = 0.30, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2897, 2648, 2664, 12, 2660, 618, 204} \[ \frac {2 a^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{5/2}}-\frac {2 a^2 \left (a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{5/2}}+\frac {a^3 \cos (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}+\frac {\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b*(a^2 - b^2)^(5/2)*d) - (2*a^2*(a^2 - 3*b^2)*ArcTan
[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b*(a^2 - b^2)^(5/2)*d) + Cos[c + d*x]/(2*(a + b)^2*d*(1 - Sin[c +
d*x])) + Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin[c + d*x])) + (a^3*Cos[c + d*x])/((a^2 - b^2)^2*d*(a + b*Sin[c +
d*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 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 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 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 2664

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

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rubi steps

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

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Mathematica [A]  time = 0.98, size = 162, normalized size = 0.76 \[ \frac {\frac {a^3 \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac {6 a^2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\sin \left (\frac {1}{2} (c+d x)\right ) \left (\frac {1}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{(a-b)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 534, normalized size = 2.52 \[ \left [\frac {2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} + 2 \, {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{3} b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )\right )}}, \frac {a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{5} + a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (a^{2} b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a^{3} b \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )}{{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a^5 - 4*a^3*b^2 + 2*a*b^4 + 2*(a^5 + a^3*b^2 - 2*a*b^4)*cos(d*x + c)^2 - 3*(a^2*b^2*cos(d*x + c)*sin(d
*x + c) + a^3*b*cos(d*x + c))*sqrt(-a^2 + b^2)*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^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c))/((a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x
+ c)*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c)), (a^5 - 2*a^3*b^2 + a*b^4 + (a^5 +
a^3*b^2 - 2*a*b^4)*cos(d*x + c)^2 - 3*(a^2*b^2*cos(d*x + c)*sin(d*x + c) + a^3*b*cos(d*x + c))*sqrt(a^2 - b^2)
*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (a^4*b - 2*a^2*b^3 + b^5)*sin(d*x + c))/((a^6*
b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*cos(d*x + c)*sin(d*x + c) + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x
+ c))]

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giac [A]  time = 0.24, size = 222, normalized size = 1.05 \[ \frac {2 \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{2} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a^{3} - a b^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )} {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.53, size = 219, normalized size = 1.03 \[ -\frac {1}{d \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{d \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 a^{3}}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {6 a^{2} b \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)+1/d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)+2/d*a^2/(a-b)^2/(a+b)^2/(tan(1/2*d*x+1/
2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)*b+2/d*a^3/(a-b)^2/(a+b)^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(
1/2*d*x+1/2*c)*b+a)+6/d*a^2/(a-b)^2/(a+b)^2*b/(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)^3/(a+b*sin(d*x+c))^2,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 = 15.28, size = 276, normalized size = 1.30 \[ \frac {\frac {2\,a\,\left (2\,a^2+b^2\right )}{{\left (a^2-b^2\right )}^2}-\frac {6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{{\left (a^2-b^2\right )}^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^2\right )}{{\left (a^2-b^2\right )}^2}-\frac {6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {6\,a^2\,b\,\mathrm {atan}\left (\frac {\frac {a^2\,b\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}}{a^2\,b}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*a*(2*a^2 + b^2))/(a^2 - b^2)^2 - (6*a^2*b*tan(c/2 + (d*x)/2)^3)/(a^2 - b^2)^2 + (2*b*tan(c/2 + (d*x)/2)*(a
^2 + 2*b^2))/(a^2 - b^2)^2 - (6*a*b^2*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2))/(d*(a + 2*b*tan(c/2 + (d*
x)/2) - a*tan(c/2 + (d*x)/2)^4 - 2*b*tan(c/2 + (d*x)/2)^3)) + (6*a^2*b*atan(((a^2*b*(2*a^4*b + 2*b^5 - 4*a^2*b
^3))/(2*(a + b)^(5/2)*(a - b)^(5/2)) + (a^3*b*tan(c/2 + (d*x)/2)*(a^4 + b^4 - 2*a^2*b^2))/((a + b)^(5/2)*(a -
b)^(5/2)))/(a^2*b)))/(d*(a + b)^(5/2)*(a - b)^(5/2))

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

[Out]

Timed out

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