\(\int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\) [397]

Optimal result

Integrand size = 21, antiderivative size = 75 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a b \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{3 d}+\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2770, 2748, 3852, 8} \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac {a b \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{3 d} \]

[In]

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

(a*b*Sec[c + d*x])/(3*d) + (Sec[c + d*x]^3*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x]))/(3*d) + ((2*a^2 - b^2)*T
an[c + d*x])/(3*d)

Rule 8

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

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2770

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*C
os[e + f*x])^(p + 1))*(a + b*Sin[e + f*x])^(m - 1)*((b + a*Sin[e + f*x])/(f*g*(p + 1))), x] + Dist[1/(g^2*(p +
1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*S
in[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (Integers
Q[2*m, 2*p] || IntegerQ[m])

Rule 3852

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]

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.40 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (8 a b+3 \left (2 a^2+b^2\right ) \sin (c+d x)+\left (2 a^2-b^2\right ) \sin (3 (c+d x))\right )}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (-1+\sin (c+d x))^2} \]

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(8*a*b + 3*(2*a^2 + b^2)*Sin[c + d*x] + (2*a^2 - b^2)*Sin[3*(c + d*x)])
)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-1 + Sin[c + d*x])^2)

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {2 a b}{3 \cos \left (d x +c \right )^{3}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(62\)
default	\(\frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {2 a b}{3 \cos \left (d x +c \right )^{3}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(62\)
risch	\(-\frac {2 i \left (8 i a b \,{\mathrm e}^{3 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 a^{2}+b^{2}\right )}{3 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3}}\)	\(71\)
parallelrisch	\(\frac {-2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}-4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b +\frac {4 \left (a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4 a b}{3}}{d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(104\)
norman	\(\frac {-\frac {4 a b}{3 d}-\frac {2 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 \left (a^{2}+b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 \left (a^{2}+b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 \left (a^{2}+4 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {8 a b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {16 a b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}\)	\(219\)

[In]

int(sec(d*x+c)^4*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.69 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {2 \, a b + {\left ({\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**4*(a+b*sin(d*x+c))**2,x)

[Out]

Integral((a + b*sin(c + d*x))**2*sec(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.68 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {b^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + \frac {2 \, a b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.36 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} d} \]

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\frac {2\,a\,b}{3}+\frac {a^2\,\sin \left (c+d\,x\right )}{3}+\frac {b^2\,\sin \left (c+d\,x\right )}{3}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {2\,a^2\,\sin \left (c+d\,x\right )}{3}-\frac {b^2\,\sin \left (c+d\,x\right )}{3}\right )}{d\,{\cos \left (c+d\,x\right )}^3} \]

[In]

int((a + b*sin(c + d*x))^2/cos(c + d*x)^4,x)

[Out]

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