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

Optimal. Leaf size=79 \[ \frac {2 b \left (a^2+b^2\right ) \cos (c+d x)}{d}+\frac {a b^2 \sin (c+d x) \cos (c+d x)}{d}-3 a b^2 x+\frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2691, 2734} \[ \frac {2 b \left (a^2+b^2\right ) \cos (c+d x)}{d}+\frac {a b^2 \sin (c+d x) \cos (c+d x)}{d}-3 a b^2 x+\frac {\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2691

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)*Sin
[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[
2*m, 2*p] || IntegerQ[m])

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.31, size = 68, normalized size = 0.86 \[ \frac {\sec (c+d x) \left (6 a^2 b+b^3 \cos (2 (c+d x))+3 b^3\right )+2 a \left (a^2+3 b^2\right ) \tan (c+d x)-6 a b^2 (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 70, normalized size = 0.89 \[ -\frac {3 \, a b^{2} d x \cos \left (d x + c\right ) - b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} - {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.73, size = 123, normalized size = 1.56 \[ -\frac {3 \, {\left (d x + c\right )} a b^{2} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b + 2 \, b^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.40, size = 89, normalized size = 1.13 \[ \frac {a^{3} \tan \left (d x +c \right )+\frac {3 a^{2} b}{\cos \left (d x +c \right )}+3 a \,b^{2} \left (\tan \left (d x +c \right )-d x -c \right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 70, normalized size = 0.89 \[ -\frac {3 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2} - b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - a^{3} \tan \left (d x + c\right ) - \frac {3 \, a^{2} b}{\cos \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.81, size = 103, normalized size = 1.30 \[ -\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^3+6\,a\,b^2\right )+6\,a^2\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,a^3+6\,a\,b^2\right )+4\,b^3+6\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-1\right )}-3\,a\,b^2\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________