\(\int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\) [240]

Optimal result

Integrand size = 26, antiderivative size = 90 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-2 a b x+\frac {\left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {3 a^2 \sin (c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d}+\frac {(b+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4482, 2968, 3127, 3110, 3102, 2814, 3855} \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {3 a^2 \sin (c+d x)}{2 d}+\frac {a b \tan (c+d x)}{d}+\frac {\tan (c+d x) \sec (c+d x) (a \cos (c+d x)+b)^2}{2 d}-2 a b x \]

[In]

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

[Out]

-2*a*b*x + ((2*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(2*d) - (3*a^2*Sin[c + d*x])/(2*d) + (a*b*Tan[c + d*x])/d + (
(b + a*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*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 3102

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.83 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {-4 a b \arctan (\tan (c+d x))+\left (2 a^2-b^2\right ) \text {arctanh}(\sin (c+d x))-2 a^2 \sin (c+d x)+4 a b \tan (c+d x)+b^2 \sec (c+d x) \tan (c+d x)}{2 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {a^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+2 a b \left (\tan \left (d x +c \right )-d x -c \right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(98\)
default	\(\frac {a^{2} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+2 a b \left (\tan \left (d x +c \right )-d x -c \right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(98\)
risch	\(-2 x a b +\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 d}-\frac {i b \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )} a -b \,{\mathrm e}^{i \left (d x +c \right )}-4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right ) a^{2}}{d}-\frac {b^{2} \ln \left (i+{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(185\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.40 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {8 \, a b d x \cos \left (d x + c\right )^{2} - {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a b \cos \left (d x + c\right ) - b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

-1/4*(8*a*b*d*x*cos(d*x + c)^2 - (2*a^2 - b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*a^2 - b^2)*cos(d*x +
c)^2*log(-sin(d*x + c) + 1) + 2*(2*a^2*cos(d*x + c)^2 - 4*a*b*cos(d*x + c) - b^2)*sin(d*x + c))/(d*cos(d*x + c
)^2)

Sympy [F]

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

[In]

integrate(sec(d*x+c)*(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), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.13 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {8 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b + b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \]

[In]

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

[Out]

-1/4*(8*(d*x + c - tan(d*x + c))*a*b + b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(
sin(d*x + c) - 1)) - 2*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (84) = 168\).

Time = 0.94 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.90 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=-\frac {4 \, {\left (d x + c\right )} a b - {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + {\left (2 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {2 \, {\left (4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 22.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.63 \[ \int \sec (c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}-\frac {a^2\,\sin \left (c+d\,x\right )}{d}-\frac {b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {b^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}-\frac {4\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,a\,b\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

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