\(\int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\) [241]

Optimal result

Integrand size = 28, antiderivative size = 99 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-a^2 x-\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4482, 2968, 3127, 3110, 3100, 2814, 3855} \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+a^2 (-x)-\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {a b \tan (c+d x) \sec (c+d x)}{3 d}+\frac {\tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+b)^2}{3 d} \]

[In]

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

[Out]

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

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2968

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 3100

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 3110

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] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3127

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*Si
n[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 3855

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

Rule 4482

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

Rubi steps \begin{align*} \text {integral}& = \int (b+a \cos (c+d x))^2 \sec ^2(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 ^4(c+d x) \, dx \\ & = \frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (b+a \cos (c+d x)) \left (2 a-b \cos (c+d x)-3 a \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (-2 \left (2 a^2-b^2\right )+6 a b \cos (c+d x)+6 a^2 \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {1}{6} \int \left (6 a b+6 a^2 \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = -a^2 x+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}-(a b) \int \sec (c+d x) \, dx \\ & = -a^2 x-\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x) \tan (c+d x)}{3 d}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {-3 a^2 \arctan (\tan (c+d x))-3 a b \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 a^2+3 a b \sec (c+d x)+b^2 \tan ^2(c+d x)\right )}{3 d} \]

[In]

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

[Out]

(-3*a^2*ArcTan[Tan[c + d*x]] - 3*a*b*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*a^2 + 3*a*b*Sec[c + d*x] + b^2*Ta
n[c + d*x]^2))/(3*d)

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {b^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(92\)
default	\(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {b^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(92\)
risch	\(-a^{2} x -\frac {2 i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-3 a^{2}+b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a b \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{d}\)	\(143\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.16 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {6 \, a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, a b \cos \left (d x + c\right ) + {\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**2*(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)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.83 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {2 \, b^{2} \tan \left (d x + c\right )^{3} - 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 3 \, a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

[In]

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

[Out]

1/6*(2*b^2*tan(d*x + c)^3 - 6*(d*x + c - tan(d*x + c))*a^2 - 3*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(
sin(d*x + c) + 1) - log(sin(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.99 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.60 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {3 \, {\left (d x + c\right )} a^{2} + 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.51 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.29 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{12}-\frac {b^2\,\sin \left (c+d\,x\right )}{4}-\frac {a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}-\frac {a^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,a^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,a\,b\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]

[In]

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

[Out]

-((b^2*sin(3*c + 3*d*x))/12 - (b^2*sin(c + d*x))/4 - (a^2*sin(3*c + 3*d*x))/4 - (a^2*sin(c + d*x))/4 + (3*a^2*
cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
*cos(3*c + 3*d*x))/2 - (a*b*sin(2*c + 2*d*x))/2 + (3*a*b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)
/2)))/2 + (a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2)/(d*((3*cos(c + d*x))/4 + cos(
3*c + 3*d*x)/4))