3.1.21 \(\int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx\) [21]
Optimal. Leaf size=71 \[ -\frac {5 a^2 x}{2}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d} \]
[Out]
-5/2*a^2*x+2*a^2*cos(d*x+c)/d+2*a^2*cos(d*x+c)/d/(1-sin(d*x+c))+1/2*a^2*cos(d*x+c)*sin(d*x+c)/d
________________________________________________________________________________________
Rubi [A]
time = 0.07, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2788, 2727,
2718, 2715, 8} \begin {gather*} \frac {2 a^2 \cos (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {5 a^2 x}{2} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]
[Out]
(-5*a^2*x)/2 + (2*a^2*Cos[c + d*x])/d + (2*a^2*Cos[c + d*x])/(d*(1 - Sin[c + d*x])) + (a^2*Cos[c + d*x]*Sin[c
+ d*x])/(2*d)
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 2715
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]
Rule 2718
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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 2788
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])
Rubi steps
\begin {align*} \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx &=a^2 \int \left (-2-\frac {2}{-1+\sin (c+d x)}-2 \sin (c+d x)-\sin ^2(c+d x)\right ) \, dx\\ &=-2 a^2 x-a^2 \int \sin ^2(c+d x) \, dx-\left (2 a^2\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx-\left (2 a^2\right ) \int \sin (c+d x) \, dx\\ &=-2 a^2 x+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {1}{2} a^2 \int 1 \, dx\\ &=-\frac {5 a^2 x}{2}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(145\) vs. \(2(71)=142\).
time = 0.29, size = 145, normalized size = 2.04 \begin {gather*} -\frac {a^2 (1+\sin (c+d x))^2 \left (\cos \left (\frac {1}{2} (c+d x)\right ) (10 (c+d x)-8 \cos (c+d x)-\sin (2 (c+d x)))+\sin \left (\frac {1}{2} (c+d x)\right ) (-2 (8+5 c+5 d x)+8 \cos (c+d x)+\sin (2 (c+d x)))\right )}{4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[c + d*x])^2*Tan[c + d*x]^2,x]
[Out]
-1/4*(a^2*(1 + Sin[c + d*x])^2*(Cos[(c + d*x)/2]*(10*(c + d*x) - 8*Cos[c + d*x] - Sin[2*(c + d*x)]) + Sin[(c +
d*x)/2]*(-2*(8 + 5*c + 5*d*x) + 8*Cos[c + d*x] + Sin[2*(c + d*x)])))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)
________________________________________________________________________________________
Maple [A]
time = 0.20, size = 117, normalized size = 1.65
| | |
| method |
result |
size |
| | |
| risch |
\(-\frac {5 a^{2} x}{2}+\frac {a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {4 a^{2}}{d \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}+\frac {a^{2} \sin \left (2 d x +2 c \right )}{4 d}\) |
\(79\) |
| derivativedivides |
\(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) |
\(117\) |
| default |
\(\frac {a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+2 a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+a^{2} \left (\tan \left (d x +c \right )-d x -c \right )}{d}\) |
\(117\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(d*x+c))^2*tan(d*x+c)^2,x,method=_RETURNVERBOSE)
[Out]
1/d*(a^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)+2*a^2*(sin(d*x+c)^4/
cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+a^2*(tan(d*x+c)-d*x-c))
________________________________________________________________________________________
Maxima [A]
time = 0.49, size = 84, normalized size = 1.18 \begin {gather*} -\frac {{\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 )} a^{2} + 2 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 4 \, a^{2} {\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+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="maxima")
[Out]
-1/2*((3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*a^2 + 2*(d*x + c - tan(d*x + c))*a^2
- 4*a^2*(1/cos(d*x + c) + cos(d*x + c)))/d
________________________________________________________________________________________
Fricas [A]
time = 0.36, size = 125, normalized size = 1.76 \begin {gather*} \frac {a^{2} \cos \left (d x + c\right )^{3} - 5 \, a^{2} d x + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{2} - {\left (5 \, a^{2} d x - 7 \, a^{2}\right )} \cos \left (d x + c\right ) + {\left (5 \, a^{2} d x + a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="fricas")
[Out]
1/2*(a^2*cos(d*x + c)^3 - 5*a^2*d*x + 4*a^2*cos(d*x + c)^2 + 4*a^2 - (5*a^2*d*x - 7*a^2)*cos(d*x + c) + (5*a^2
*d*x + a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c) + 4*a^2)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int 2 \sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))**2*tan(d*x+c)**2,x)
[Out]
a**2*(Integral(2*sin(c + d*x)*tan(c + d*x)**2, x) + Integral(sin(c + d*x)**2*tan(c + d*x)**2, x) + Integral(ta
n(c + d*x)**2, x))
________________________________________________________________________________________
Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 5370 vs.
\(2 (65) = 130\).
time = 12.03, size = 5370, normalized size = 75.63 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(d*x+c))^2*tan(d*x+c)^2,x, algorithm="giac")
[Out]
-1/2*(5*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 + 5*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*
c)^4*tan(c) - 5*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 20*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3
*tan(1/2*c)^3*tan(c)^3 + 5*a^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^3 - 8*a^2*tan(d*x)^3*tan(1/2*d*
x)^4*tan(1/2*c)^4*tan(c)^3 + 5*a^2*tan(d*x)^3*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 + 5*a^2*tan(d*x)^2*tan(1/2*
d*x)^4*tan(1/2*c)^4*tan(c)^3 - 5*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4 - 20*a^2*d*x*tan(d*x)^3*tan(1/
2*d*x)^3*tan(1/2*c)^3*tan(c) + 5*a^2*d*x*tan(d*x)*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)^4*tan(1/2*c)^4*tan(c) + 20*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 - 5*a^2*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)^4*tan(1/2*c)^4*tan(c)^2 - 5*a^2*d*x*tan(d*x)^3*ta
n(1/2*d*x)^4*tan(c)^3 - 20*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 20*a^2*d*x*tan(d*x)^3*tan(1
/2*d*x)*tan(1/2*c)^3*tan(c)^3 - 20*a^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 + 32*a^2*tan(d*x)^3*t
an(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 - 5*a^2*d*x*tan(d*x)^3*tan(1/2*c)^4*tan(c)^3 - 8*a^2*tan(d*x)*tan(1/2*d*x)
^4*tan(1/2*c)^4*tan(c)^3 + 4*a^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) - 20*a^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 - 20*a^2*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 + 4*a^2*tan(1/2*d*x)^4*tan(1/2*c)^4*
tan(c)^3 + 20*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3 - 5*a^2*d*x*tan(1/2*d*x)^4*tan(1/2*c)^4 + 8*a^2*t
an(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^4 - 5*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)^4*tan(c) - 20*a^2*d*x*tan(d*x)^3*tan
(1/2*d*x)^3*tan(1/2*c)*tan(c) - 20*a^2*d*x*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) - 20*a^2*d*x*tan(d*x)*t
an(1/2*d*x)^3*tan(1/2*c)^3*tan(c) + 32*a^2*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) - 5*a^2*d*x*tan(d*x)^
3*tan(1/2*c)^4*tan(c) - 8*a^2*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) + 5*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^
4*tan(c)^2 + 20*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^2 + 20*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)*tan
(1/2*c)^3*tan(c)^2 + 20*a^2*d*x*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 - 32*a^2*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/
2*c)^3*tan(c)^2 + 5*a^2*d*x*tan(d*x)^2*tan(1/2*c)^4*tan(c)^2 + 8*a^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c)^2 - 5*
a^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(c)^3 - 8*a^2*tan(d*x)^3*tan(1/2*d*x)^4*tan(c)^3 - 20*a^2*d*x*tan(d*x)^3*ta
n(1/2*d*x)*tan(1/2*c)*tan(c)^3 - 20*a^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 32*a^2*tan(d*x)^3*ta
n(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 96*a^2*tan(d*x)^3*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(c)^3 - 20*a^2*d*x*tan(d*x
)*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^3 - 32*a^2*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^3 + 32*a^2*tan(d*x)*
tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 - 5*a^2*d*x*tan(d*x)*tan(1/2*c)^4*tan(c)^3 - 8*a^2*tan(d*x)^3*tan(1/2*c)^
4*tan(c)^3 - 16*a^2*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)^3 + 5*a^2*tan(d*x)*tan(1/2*d*x)^4*tan(1/2*c)^4 - 8*a^
2*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c) + 5*a^2*tan(1/2*d*x)^4*tan(1/2*c)^4*tan(c) - 5*a^2*tan(d*x)^3*
tan(1/2*d*x)^4*tan(c)^2 - 20*a^2*tan(d*x)^3*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^2 - 20*a^2*tan(d*x)^3*tan(1/2*d*x
)*tan(1/2*c)^3*tan(c)^2 - 8*a^2*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 - 5*a^2*tan(d*x)^3*tan(1/2*c)^4*
tan(c)^2 - 5*a^2*tan(d*x)^2*tan(1/2*d*x)^4*tan(c)^3 - 20*a^2*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^3 - 2
0*a^2*tan(d*x)^2*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^3 - 16*a^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^3 - 5*a^2*tan(
d*x)^2*tan(1/2*c)^4*tan(c)^3 + 5*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^4 + 20*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)^3*tan(
1/2*c) + 20*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)*tan(1/2*c)^3 + 20*a^2*d*x*tan(1/2*d*x)^3*tan(1/2*c)^3 - 32*a^2*tan
(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)^3 + 5*a^2*d*x*tan(d*x)^2*tan(1/2*c)^4 + 8*a^2*tan(1/2*d*x)^4*tan(1/2*c)^4 -
5*a^2*d*x*tan(d*x)*tan(1/2*d*x)^4*tan(c) - 8*a^2*tan(d*x)^3*tan(1/2*d*x)^4*tan(c) - 20*a^2*d*x*tan(d*x)^3*tan(
1/2*d*x)*tan(1/2*c)*tan(c) - 20*a^2*d*x*tan(d*x)*tan(1/2*d*x)^3*tan(1/2*c)*tan(c) - 32*a^2*tan(d*x)^3*tan(1/2*
d*x)^3*tan(1/2*c)*tan(c) - 96*a^2*tan(d*x)^3*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(c) - 20*a^2*d*x*tan(d*x)*tan(1/2*
d*x)*tan(1/2*c)^3*tan(c) - 32*a^2*tan(d*x)^3*tan(1/2*d*x)*tan(1/2*c)^3*tan(c) + 32*a^2*tan(d*x)*tan(1/2*d*x)^3
*tan(1/2*c)^3*tan(c) - 5*a^2*d*x*tan(d*x)*tan(1/2*c)^4*tan(c) - 8*a^2*tan(d*x)^3*tan(1/2*c)^4*tan(c) + 5*a^2*d
*x*tan(1/2*d*x)^4*tan(c)^2 + 8*a^2*tan(d*x)^2*tan(1/2*d*x)^4*tan(c)^2 + 20*a^2*d*x*tan(d*x)^2*tan(1/2*d*x)*tan
(1/2*c)*tan(c)^2 + 20*a^2*d*x*tan(1/2*d*x)^3*tan(1/2*c)*tan(c)^2 + 32*a^2*tan(d*x)^2*tan(1/2*d*x)^3*tan(1/2*c)
*tan(c)^2 + 96*a^2*tan(d*x)^2*tan(1/2*d*x)^2*tan(1/2*c)^2*tan(c)^2 + 20*a^2*d*x*tan(1/2*d*x)*tan(1/2*c)^3*tan(
c)^2 + 32*a^2*tan(d*x)^2*tan(1/2*d*x)*tan(1/2*c)^3*tan(c)^2 - 32*a^2*tan(1/2*d*x)^3*tan(1/2*c)^3*tan(c)^2 + 5*
a^2*d*x*tan(1/2*c)^4*tan(c)^2 + 8*a^2*tan(d*x)^...
________________________________________________________________________________________
Mupad [B]
time = 8.69, size = 213, normalized size = 3.00 \begin {gather*} -\frac {5\,a^2\,x}{2}-\frac {\frac {5\,a^2\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (5\,c+5\,d\,x-6\right )}{2}\right )-\frac {a^2\,\left (5\,c+5\,d\,x-16\right )}{2}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (5\,c+5\,d\,x-10\right )}{2}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (10\,c+10\,d\,x-10\right )}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (10\,c+10\,d\,x-22\right )}{2}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(c + d*x)^2*(a + a*sin(c + d*x))^2,x)
[Out]
- (5*a^2*x)/2 - ((5*a^2*(c + d*x))/2 - tan(c/2 + (d*x)/2)*((5*a^2*(c + d*x))/2 - (a^2*(5*c + 5*d*x - 6))/2) -
(a^2*(5*c + 5*d*x - 16))/2 + tan(c/2 + (d*x)/2)^4*((5*a^2*(c + d*x))/2 - (a^2*(5*c + 5*d*x - 10))/2) - tan(c/2
+ (d*x)/2)^3*(5*a^2*(c + d*x) - (a^2*(10*c + 10*d*x - 10))/2) + tan(c/2 + (d*x)/2)^2*(5*a^2*(c + d*x) - (a^2*
(10*c + 10*d*x - 22))/2))/(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)
________________________________________________________________________________________