\(\int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\) [248]

Optimal result

Integrand size = 28, antiderivative size = 111 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {a^3 \cos (c+d x)}{d}+\frac {3 a^2 b \log (\cos (c+d x))}{d}+\frac {a \left (a^2-3 b^2\right ) \sec (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {a b^2 \sec ^3(c+d x)}{d}+\frac {b^3 \sec ^4(c+d x)}{4 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4482, 2916, 12, 908} \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {a^3 \cos (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {a \left (a^2-3 b^2\right ) \sec (c+d x)}{d}+\frac {3 a^2 b \log (\cos (c+d x))}{d}+\frac {a b^2 \sec ^3(c+d x)}{d}+\frac {b^3 \sec ^4(c+d x)}{4 d} \]

[In]

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

[Out]

(a^3*Cos[c + d*x])/d + (3*a^2*b*Log[Cos[c + d*x]])/d + (a*(a^2 - 3*b^2)*Sec[c + d*x])/d + (b*(3*a^2 - b^2)*Sec
[c + d*x]^2)/(2*d) + (a*b^2*Sec[c + d*x]^3)/d + (b^3*Sec[c + d*x]^4)/(4*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2916

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

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))^3 \sec ^2(c+d x) \tan ^3(c+d x) \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {a^5 (b+x)^3 \left (a^2-x^2\right )}{x^5} \, dx,x,a \cos (c+d x)\right )}{a^3 d} \\ & = -\frac {a^2 \text {Subst}\left (\int \frac {(b+x)^3 \left (a^2-x^2\right )}{x^5} \, dx,x,a \cos (c+d x)\right )}{d} \\ & = -\frac {a^2 \text {Subst}\left (\int \left (-1+\frac {a^2 b^3}{x^5}+\frac {3 a^2 b^2}{x^4}+\frac {3 a^2 b-b^3}{x^3}+\frac {a^2-3 b^2}{x^2}-\frac {3 b}{x}\right ) \, dx,x,a \cos (c+d x)\right )}{d} \\ & = \frac {a^3 \cos (c+d x)}{d}+\frac {3 a^2 b \log (\cos (c+d x))}{d}+\frac {a \left (a^2-3 b^2\right ) \sec (c+d x)}{d}+\frac {b \left (3 a^2-b^2\right ) \sec ^2(c+d x)}{2 d}+\frac {a b^2 \sec ^3(c+d x)}{d}+\frac {b^3 \sec ^4(c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.06 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.87 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {4 a^3 \cos (c+d x)+12 a^2 b \log (\cos (c+d x))+4 a \left (a^2-3 b^2\right ) \sec (c+d x)+\left (6 a^2 b-2 b^3\right ) \sec ^2(c+d x)+4 a b^2 \sec ^3(c+d x)+b^3 \sec ^4(c+d x)}{4 d} \]

[In]

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

[Out]

(4*a^3*Cos[c + d*x] + 12*a^2*b*Log[Cos[c + d*x]] + 4*a*(a^2 - 3*b^2)*Sec[c + d*x] + (6*a^2*b - 2*b^3)*Sec[c +
d*x]^2 + 4*a*b^2*Sec[c + d*x]^3 + b^3*Sec[c + d*x]^4)/(4*d)

Maple [A] (verified)

Time = 8.52 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {\frac {b^{3} \sec \left (d x +c \right )^{4}}{4}+a \,b^{2} \sec \left (d x +c \right )^{3}+\frac {3 a^{2} b \sec \left (d x +c \right )^{2}}{2}-\frac {b^{3} \sec \left (d x +c \right )^{2}}{2}+\sec \left (d x +c \right ) a^{3}-3 \sec \left (d x +c \right ) a \,b^{2}+\frac {a^{3}}{\sec \left (d x +c \right )}-3 a^{2} b \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(106\)
default	\(\frac {\frac {b^{3} \sec \left (d x +c \right )^{4}}{4}+a \,b^{2} \sec \left (d x +c \right )^{3}+\frac {3 a^{2} b \sec \left (d x +c \right )^{2}}{2}-\frac {b^{3} \sec \left (d x +c \right )^{2}}{2}+\sec \left (d x +c \right ) a^{3}-3 \sec \left (d x +c \right ) a \,b^{2}+\frac {a^{3}}{\sec \left (d x +c \right )}-3 a^{2} b \ln \left (\sec \left (d x +c \right )\right )}{d}\)	\(106\)
risch	\(-3 i a^{2} b x +\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {6 i b \,a^{2} c}{d}+\frac {2 a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-6 a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-2 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+6 a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-10 a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-10 a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 a^{3} {\mathrm e}^{i \left (d x +c \right )}-6 a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}\)	\(282\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {4 \, a^{3} \cos \left (d x + c\right )^{5} + 12 \, a^{2} b \cos \left (d x + c\right )^{4} \log \left (-\cos \left (d x + c\right )\right ) + 4 \, a b^{2} \cos \left (d x + c\right ) + 4 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + b^{3} + 2 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{4 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/4*(4*a^3*cos(d*x + c)^5 + 12*a^2*b*cos(d*x + c)^4*log(-cos(d*x + c)) + 4*a*b^2*cos(d*x + c) + 4*(a^3 - 3*a*b
^2)*cos(d*x + c)^3 + b^3 + 2*(3*a^2*b - b^3)*cos(d*x + c)^2)/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {b^{3} \tan \left (d x + c\right )^{4} - 6 \, a^{2} b {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} + 4 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac {4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a b^{2}}{\cos \left (d x + c\right )^{3}}}{4 \, d} \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Modgcd: no suitable evaluation pointindex.cc index_m operator + Error: Bad Argument ValueDone

Mupad [B] (verification not implemented)

Time = 26.51 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.01 \[ \int \sec ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-12\,a^3+6\,a^2\,b+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^3-6\,a^2\,b+4\,a\,b^2+4\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (4\,a^3+6\,a^2\,b+12\,a\,b^2-4\,b^3\right )-4\,a\,b^2+4\,a^3+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {6\,a^2\,b\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^2*(12*a*b^2 + 6*a^2*b - 12*a^3) + tan(c/2 + (d*x)/2)^4*(4*a*b^2 - 6*a^2*b + 12*a^3 + 4*b^3
) - tan(c/2 + (d*x)/2)^6*(12*a*b^2 + 6*a^2*b + 4*a^3 - 4*b^3) - 4*a*b^2 + 4*a^3 + 6*a^2*b*tan(c/2 + (d*x)/2)^8
)/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + tan(
c/2 + (d*x)/2)^10 + 1)) - (6*a^2*b*atanh(tan(c/2 + (d*x)/2)^2))/d