3.2.88 \(\int \frac {\tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [188]

Optimal. Leaf size=200 \[ -\frac {2 a^3 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac {4 a b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}+\frac {\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac {\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac {a^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2810, 2727, 2743, 12, 2739, 632, 210} \begin {gather*} -\frac {4 a b^2 \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}-\frac {a^2 b \cos (c+d x)}{d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}-\frac {2 a^3 \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{5/2}}+\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^3*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*d) - (4*a*b^2*ArcTan[(b + a*Tan[(
c + d*x)/2])/Sqrt[a^2 - b^2]])/((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^2*b*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 210

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

Rule 632

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 2727

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 2739

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 2743

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] && Integ
erQ[2*n]

Rule 2810

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Int[ExpandIntegrand
[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^m/(1 - Sin[e + f*x]^2)^(p/2)), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a
^2 - b^2, 0] && IntegersQ[m, p/2]

Rubi steps

\begin {align*} \int \frac {\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^2}{\left (a^2-b^2\right ) (a+b \sin (c+d x))^2}-\frac {2 a b^2}{\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 (2 a b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {a^2 \int \frac {1}{(a+b \sin (c+d x))^2} \, dx}{a^2-b^2}\\ &=\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^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {a^2 \int \frac {a}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}-\frac {\left (4 a b^2\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\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^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {a^3 \int \frac {1}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (8 a b^2\right ) \text {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 )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {4 a b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\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^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {4 a b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\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^2 b \cos (c+d x)}{\left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac {\left (4 a^3\right ) \text {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 )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {2 a^3 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} d}-\frac {4 a b^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\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^2 b \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 = 0.72, size = 169, normalized size = 0.84 \begin {gather*} \frac {-\frac {2 a \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )-\frac {a^2 b \cos (c+d x)}{(a-b)^2 (a+b)^2 (a+b \sin (c+d x))}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.38, size = 162, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(162\)
default	\(\frac {-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \left (\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a b}{a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a}+\frac {\left (a^{2}+2 b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d}\)	\(162\)
risch	\(\frac {2 i \left (3 a^{3} {\mathrm e}^{i \left (d x +c \right )}+4 i a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 i a^{2} b +i b^{3}+a^{3} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right ) \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right ) \left (a^{2}-b^{2}\right )^{2} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {2 b^{2} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {2 b^{2} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\)	\(476\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(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 de

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Fricas [A]
time = 0.40, size = 569, normalized size = 2.84 \begin {gather*} \left [-\frac {2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} + 2 \, {\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{4} + 2 \, a^{2} b^{2}\right )} \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^{5} - 2 \, a^{3} b^{2} + a b^{4}\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^{4} b - 2 \, a^{2} b^{3} + b^{5} + {\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} - {\left ({\left (a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{4} + 2 \, a^{2} b^{2}\right )} \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^{5} - 2 \, a^{3} b^{2} + a b^{4}\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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*a^4*b - 4*a^2*b^3 + 2*b^5 + 2*(2*a^4*b - a^2*b^3 - b^5)*cos(d*x + c)^2 + ((a^3*b + 2*a*b^3)*cos(d*x +
c)*sin(d*x + c) + (a^4 + 2*a^2*b^2)*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^5 - 2*a^3*b^2 + a*b^4)*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^4*b - 2*
a^2*b^3 + b^5 + (2*a^4*b - a^2*b^3 - b^5)*cos(d*x + c)^2 - ((a^3*b + 2*a*b^3)*cos(d*x + c)*sin(d*x + c) + (a^4
+ 2*a^2*b^2)*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^
5 - 2*a^3*b^2 + a*b^4)*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.17, size = 251, normalized size = 1.26 \begin {gather*} -\frac {2 \, {\left (\frac {{\left (a^{3} + 2 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a^{2} b}{{\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((a^3 + 2*a*b^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 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (a^3*tan(1/2*d*x + 1/2*c)^3 + 2*a*b^2*tan(1/2*d*x + 1/2*c)^
3 + a^2*b*tan(1/2*d*x + 1/2*c)^2 + 2*b^3*tan(1/2*d*x + 1/2*c)^2 + a^3*tan(1/2*d*x + 1/2*c) - 4*a*b^2*tan(1/2*d
*x + 1/2*c) - 3*a^2*b)/((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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Mupad [B]
time = 8.68, size = 313, normalized size = 1.56 \begin {gather*} -\frac {\frac {6\,a^2\,b}{{\left (a^2-b^2\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a\,b^2-a^3\right )}{{\left (a^2-b^2\right )}^2}-\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2+2\,b^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (a^2+2\,b^2\right )}{{\left (a^2-b^2\right )}^2}}{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 {2\,a\,\mathrm {atan}\left (\frac {\frac {a\,\left (a^2+2\,b^2\right )\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}+\frac {2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+2\,b^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}}}{2\,a^3+4\,a\,b^2}\right )\,\left (a^2+2\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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