3.1459 \(\int \sec (c+d x) (a+b \sin (c+d x))^3 \tan (c+d x) \, dx\)

Optimal. Leaf size=75 \[ -\frac {3}{2} b x \left (2 a^2+b^2\right )+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {\sec (c+d x) (a+b \sin (c+d x))^3}{d}+\frac {3 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2861, 12, 2644} \[ -\frac {3}{2} b x \left (2 a^2+b^2\right )+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {\sec (c+d x) (a+b \sin (c+d x))^3}{d}+\frac {3 b^3 \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(2*a^2 + b^2)*x)/2 + (6*a*b^2*Cos[c + d*x])/d + (3*b^3*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (Sec[c + d*x]*
(a + b*Sin[c + d*x])^3)/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 2644

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

Rule 2861

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*(d + c*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 - 1)*Simp[a*c*(p +
2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x
])

Rubi steps

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

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Mathematica [A]  time = 0.55, size = 91, normalized size = 1.21 \[ \frac {\sec (c+d x) \left (8 a^3+12 a b^2 \cos (2 (c+d x))+36 a b^2+b^3 \sin (3 (c+d x))\right )+3 b \left (\left (8 a^2+3 b^2\right ) \tan (c+d x)-4 \left (2 a^2+b^2\right ) (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(8*a^3 + 36*a*b^2 + 12*a*b^2*Cos[2*(c + d*x)] + b^3*Sin[3*(c + d*x)]) + 3*b*(-4*(2*a^2 + b^2)*(c
+ d*x) + (8*a^2 + 3*b^2)*Tan[c + d*x]))/(8*d)

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fricas [A]  time = 0.42, size = 90, normalized size = 1.20 \[ \frac {6 \, a b^{2} \cos \left (d x + c\right )^{2} - 3 \, {\left (2 \, a^{2} b + b^{3}\right )} d x \cos \left (d x + c\right ) + 2 \, a^{3} + 6 \, a b^{2} + {\left (b^{3} \cos \left (d x + c\right )^{2} + 6 \, a^{2} b + 2 \, b^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.20, size = 148, normalized size = 1.97 \[ -\frac {3 \, {\left (2 \, a^{2} b + b^{3}\right )} {\left (d x + c\right )} + \frac {4 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.45, size = 132, normalized size = 1.76 \[ \frac {\frac {a^{3}}{\cos \left (d x +c \right )}+3 a^{2} b \left (\tan \left (d x +c \right )-d x -c \right )+3 a \,b^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 99, normalized size = 1.32 \[ -\frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b + {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b^{3} - 6 \, a b^{2} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac {2 \, a^{3}}{\cos \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 16.29, size = 219, normalized size = 2.92 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,b+3\,b^3\right )+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+12\,a\,b^2+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^2\,b+3\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (12\,a^2\,b+2\,b^3\right )+2\,a^3}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,b\,\mathrm {atan}\left (\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+b^2\right )}{6\,a^2\,b+3\,b^3}\right )\,\left (2\,a^2+b^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(6*a^2*b + 3*b^3) + 2*a^3*tan(c/2 + (d*x)/2)^4 + 12*a*b^2 + tan(c/2 + (d*x)/2)^2*(12*a*b^2
+ 4*a^3) + tan(c/2 + (d*x)/2)^5*(6*a^2*b + 3*b^3) + tan(c/2 + (d*x)/2)^3*(12*a^2*b + 2*b^3) + 2*a^3)/(d*(tan(
c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)^6 + 1)) - (3*b*atan((3*b*tan(c/2 + (d*x)/2)*(2*a^
2 + b^2))/(6*a^2*b + 3*b^3))*(2*a^2 + b^2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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