3.2.56 \(\int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx\) [156]

Optimal. Leaf size=94 \[ -a^2 x-\frac {3 b^2 x}{2}+\frac {2 a b \cos (c+d x)}{d}+\frac {2 a b \sec (c+d x)}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {3 b^2 \tan (c+d x)}{2 d}-\frac {b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2801, 3554, 8, 2670, 14, 2671, 294, 327, 209} \begin {gather*} \frac {a^2 \tan (c+d x)}{d}+a^2 (-x)+\frac {2 a b \cos (c+d x)}{d}+\frac {2 a b \sec (c+d x)}{d}+\frac {3 b^2 \tan (c+d x)}{2 d}-\frac {b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {3 b^2 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 209

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

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2670

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 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 2801

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

Rule 3554

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

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Mathematica [A]
time = 0.33, size = 77, normalized size = 0.82 \begin {gather*} \frac {-4 \left (2 a^2+3 b^2\right ) (c+d x)+b \sec (c+d x) (24 a+8 a \cos (2 (c+d x))+b \sin (3 (c+d x)))+\left (8 a^2+9 b^2\right ) \tan (c+d x)}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(2*a^2 + 3*b^2)*(c + d*x) + b*Sec[c + d*x]*(24*a + 8*a*Cos[2*(c + d*x)] + b*Sin[3*(c + d*x)]) + (8*a^2 + 9
*b^2)*Tan[c + d*x])/(8*d)

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Maple [A]
time = 0.23, size = 116, normalized size = 1.23

method	result	size
derivativedivides	\(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\)	\(116\)
default	\(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\)	\(116\)
risch	\(-a^{2} x -\frac {3 b^{2} x}{2}-\frac {i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{2}+2 i b^{2}+4 a \,{\mathrm e}^{i \left (d x +c \right )} b}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.72, size = 83, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} + {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b^{2} - 4 \, a b {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.40, size = 81, normalized size = 0.86 \begin {gather*} -\frac {{\left (2 \, a^{2} + 3 \, b^{2}\right )} d x \cos \left (d x + c\right ) - 4 \, a b \cos \left (d x + c\right )^{2} - 4 \, a b - {\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((2*a^2 + 3*b^2)*d*x*cos(d*x + c) - 4*a*b*cos(d*x + c)^2 - 4*a*b - (b^2*cos(d*x + c)^2 + 2*a^2 + 2*b^2)*s
in(d*x + c))/(d*cos(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 7670 vs. \(2 (88) = 176\).
time = 29.43, size = 7670, normalized size = 81.60 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 3*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*
c)^4*tan(c)^3 + 2*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 3*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*
tan(1/2*c)^4*tan(c) - 2*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 3*b^2*d*x*tan(d*x)^2*tan(1/2
*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 8*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 - 12*b^2*d*x*tan(d*x
)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 + 2*a^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 3*b^2*d*x
*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 - 8*a*b*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 2*a^2
*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 + 3*b^2*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 + 2*a
^2*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 3*b^2*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 - 2
*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4 - 3*b^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4 - 8*a^2*d*x
*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) - 12*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) + 2*
a^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 3*b^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) -
8*a*b*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 8*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^
2 + 12*b^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 - 2*a^2*d*x*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^
2 - 3*b^2*d*x*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 + 8*a*b*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 2
*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(c)^3 - 3*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(c)^3 - 8*a^2*d*x*tan(d*x
)^3*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 12*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 8*a^2*d*x*
tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^3 - 12*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^3 - 8*a
^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 - 12*b^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^
3 + 32*a*b*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 - 2*a^2*d*x*tan(d*x)^3*tan(1/2*c)^4*tan(c)^3 - 3*b^
2*d*x*tan(d*x)^3*tan(1/2*c)^4*tan(c)^3 - 8*a*b*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 2*a^2*tan(d*x)^
3*tan(1/2*d*x)^4*tan(1/2*c)^4 + 2*b^2*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4 + 2*a^2*tan(d*x)^2*tan(1/2*d*x)^4
*tan(1/2*c)^4*tan(c) - 8*a^2*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 - 12*b^2*tan(d*x)^3*tan(1/2*d*x)^
3*tan(1/2*c)^3*tan(c)^2 + 2*a^2*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 8*a^2*tan(d*x)^2*tan(1/2*d*x)^
3*tan(1/2*c)^3*tan(c)^3 - 12*b^2*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 + 2*a^2*tan(1/2*d*x)^4*tan(1/
2*c)^4*tan(c)^3 + 2*b^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 8*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^
3 + 12*b^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3 - 2*a^2*d*x*tan(1/2*d*x)^4*tan(1/2*c)^4 - 3*b^2*d*x*tan(
1/2*d*x)^4*tan(1/2*c)^4 + 8*a*b*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4 - 2*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*t
an(c) - 3*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(c) - 8*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 1
2*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 8*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)
- 12*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) - 8*a^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(
c) - 12*b^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) + 32*a*b*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*ta
n(c) - 2*a^2*d*x*tan(d*x)^3*tan(1/2*c)^4*tan(c) - 3*b^2*d*x*tan(d*x)^3*tan(1/2*c)^4*tan(c) - 8*a*b*tan(d*x)*ta
n(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 2*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(c)^2 + 3*b^2*d*x*tan(d*x)^2*tan(1/2
*d*x)^4*tan(c)^2 + 8*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^2 + 12*b^2*d*x*tan(d*x)^2*tan(1/2*d*x
)^3*tan(1/2*c)*tan(c)^2 + 8*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^2 + 12*b^2*d*x*tan(d*x)^2*tan(
1/2*d*x)*tan(1/2*c)^3*tan(c)^2 + 8*a^2*d*x*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 + 12*b^2*d*x*tan(1/2*d*x)^3*ta
n(1/2*c)^3*tan(c)^2 - 32*a*b*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 + 2*a^2*d*x*tan(d*x)^2*tan(1/2*c)
^4*tan(c)^2 + 3*b^2*d*x*tan(d*x)^2*tan(1/2*c)^4*tan(c)^2 + 8*a*b*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 2*a^2*
d*x*tan(d*x)*tan(1/2*d*x)^4*tan(c)^3 - 3*b^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(c)^3 - 8*a*b*tan(d*x)^3*tan(1/2*d
*x)^4*tan(c)^3 - 8*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)*tan(c)^3 - 12*b^2*d*x*tan(d*x)^3*tan(1/2*d*x)*ta
n(1/2*c)*tan(c)^3 - 8*a^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 12*b^2*d*x*tan(d*x)*tan(1/2*d*x)^3
*tan(1/2*c)*tan(c)^3 - 32*a*b*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 96*a*b*tan(d*x)^3*tan(1/2*d*x)^2
*tan(1/2*c)^2*tan(c)^3 - 8*a^2*d*x*tan(d*x)*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^3 - 12*b^2*d*x*tan(d*x)*tan(1/2*d
*x)*tan(1/2*c)^3*tan(c)^3 - 32*a*b*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^3 + 32*a*b*tan(d*x)*tan(1/2*d*x
)^3*tan(1/2*c)^3*tan(c)^3 - 2*a^2*d*x*tan(d*x)*tan(1/2*c)^4*tan(c)^3 - 3*b^2*d*x*tan(d*x)*tan(1/2*c)^4*tan(c)^
3 - 8*a*b*tan(d*x)^3*tan(1/2*c)^4*tan(c)^3 - 8*...

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Mupad [B]
time = 9.29, size = 147, normalized size = 1.56 \begin {gather*} \frac {\left (2\,a^2+3\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,a^2+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (2\,a^2+3\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8\,a\,b}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {x\,\left (2\,a^2+3\,b^2\right )}{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]

(8*a*b + tan(c/2 + (d*x)/2)^3*(4*a^2 + 2*b^2) + tan(c/2 + (d*x)/2)^5*(2*a^2 + 3*b^2) + tan(c/2 + (d*x)/2)*(2*a
^2 + 3*b^2) + 8*a*b*tan(c/2 + (d*x)/2)^2)/(d*(tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)
^6 + 1)) - (x*(2*a^2 + 3*b^2))/2

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