3.242 \(\int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=125 \[ -\frac {\left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}-\frac {2 a b \tan (c+d x)}{3 d}+\frac {a b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d} \]

[Out]

-1/8*(4*a^2+b^2)*arctanh(sin(d*x+c))/d-2/3*a*b*tan(d*x+c)/d+1/8*(2*a^2-b^2)*sec(d*x+c)*tan(d*x+c)/d+1/6*a*b*se
c(d*x+c)^2*tan(d*x+c)/d+1/4*(b+a*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.44, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {4397, 2889, 3048, 3031, 3021, 2748, 3767, 8, 3770} \[ -\frac {\left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}-\frac {2 a b \tan (c+d x)}{3 d}+\frac {a b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+b)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

-((4*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(8*d) - (2*a*b*Tan[c + d*x])/(3*d) + ((2*a^2 - b^2)*Sec[c + d*x]*Tan[c
+ d*x])/(8*d) + (a*b*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])
/(4*d)

Rule 8

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2889

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

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx &=\int (b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=\int (b+a \cos (c+d x))^2 \left (1-\cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (b+a \cos (c+d x)) \left (2 a-b \cos (c+d x)-3 a \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{12} \int \left (-3 \left (2 a^2-b^2\right )+8 a b \cos (c+d x)+9 a^2 \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (16 a b+3 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{3} (2 a b) \int \sec ^2(c+d x) \, dx-\frac {1}{8} \left (4 a^2+b^2\right ) \int \sec (c+d x) \, dx\\ &=-\frac {\left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {(2 a b) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=-\frac {\left (4 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {2 a b \tan (c+d x)}{3 d}+\frac {\left (2 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {(b+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.61, size = 336, normalized size = 2.69 \[ \frac {\sec ^4(c+d x) \left (12 \left (4 a^2+b^2\right ) \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 \left (4 a^2+b^2\right ) \cos (4 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+24 a^2 \sin (c+d x)+24 a^2 \sin (3 (c+d x))+36 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-36 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+32 a b \sin (2 (c+d x))-16 a b \sin (4 (c+d x))+42 b^2 \sin (c+d x)-6 b^2 \sin (3 (c+d x))+9 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-9 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{192 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^4*(36*a^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 9*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/
2]] + 12*(4*a^2 + b^2)*Cos[2*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin
[(c + d*x)/2]]) + 3*(4*a^2 + b^2)*Cos[4*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]]) - 36*a^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 9*b^2*Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]] + 24*a^2*Sin[c + d*x] + 42*b^2*Sin[c + d*x] + 32*a*b*Sin[2*(c + d*x)] + 24*a^2*Sin[3*(c + d*x)
] - 6*b^2*Sin[3*(c + d*x)] - 16*a*b*Sin[4*(c + d*x)]))/(192*d)

________________________________________________________________________________________

fricas [A]  time = 0.58, size = 129, normalized size = 1.03 \[ -\frac {3 \, {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, a b \cos \left (d x + c\right )^{3} - 16 \, a b \cos \left (d x + c\right ) - 3 \, {\left (4 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*(4*a^2 + b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*a^2 + b^2)*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 2*(16*a*b*cos(d*x + c)^3 - 16*a*b*cos(d*x + c) - 3*(4*a^2 - b^2)*cos(d*x + c)^2 - 6*b^2)*sin(d*x + c))
/(d*cos(d*x + c)^4)

________________________________________________________________________________________

giac [A]  time = 1.89, size = 226, normalized size = 1.81 \[ -\frac {3 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 64 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 64 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(4*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))
- 2*(12*a^2*tan(1/2*d*x + 1/2*c)^7 + 3*b^2*tan(1/2*d*x + 1/2*c)^7 - 12*a^2*tan(1/2*d*x + 1/2*c)^5 - 64*a*b*ta
n(1/2*d*x + 1/2*c)^5 + 21*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*a^2*tan(1/2*d*x + 1/2*c)^3 + 64*a*b*tan(1/2*d*x + 1/
2*c)^3 + 21*b^2*tan(1/2*d*x + 1/2*c)^3 + 12*a^2*tan(1/2*d*x + 1/2*c) + 3*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*
x + 1/2*c)^2 - 1)^4)/d

________________________________________________________________________________________

maple [A]  time = 0.11, size = 169, normalized size = 1.35 \[ \frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a^{2} \sin \left (d x +c \right )}{2 d}-\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a b \left (\sin ^{3}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {b^{2} \sin \left (d x +c \right )}{8 d}-\frac {b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*a^2*sin(d*x+c)^3/cos(d*x+c)^2+1/2*a^2*sin(d*x+c)/d-1/2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*a*b*sin(d*x
+c)^3/cos(d*x+c)^3+1/4/d*b^2*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*b^2*sin(d*x+c)^3/cos(d*x+c)^2+1/8*b^2*sin(d*x+c)/
d-1/8/d*b^2*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.39, size = 129, normalized size = 1.03 \[ \frac {32 \, a b \tan \left (d x + c\right )^{3} + 3 \, b^{2} {\left (\frac {2 \, {\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(32*a*b*tan(d*x + c)^3 + 3*b^2*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1)
- log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x
+ c) + 1) - log(sin(d*x + c) - 1)))/d

________________________________________________________________________________________

mupad [B]  time = 3.27, size = 177, normalized size = 1.42 \[ \frac {\left (a^2+\frac {b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-a^2-\frac {16\,a\,b}{3}+\frac {7\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-a^2+\frac {16\,a\,b}{3}+\frac {7\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (a^2+\frac {b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+\frac {b^2}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*((16*a*b)/3 - a^2 + (7*b^2)/4) + tan(c/2 + (d*x)/2)*(a^2 + b^2/4) + tan(c/2 + (d*x)/2)^7
*(a^2 + b^2/4) - tan(c/2 + (d*x)/2)^5*((16*a*b)/3 + a^2 - (7*b^2)/4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 +
(d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (atanh(tan(c/2 + (d*x)/2))*(a^2 + b^2/4))/
d

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________