3.5.9 \(\int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx\) [409]

Optimal. Leaf size=84 \[ \frac {2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac {2 a \left (a^2-b^2\right ) \tan (c+d x)}{3 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2770, 12, 2748, 3852, 8} \begin {gather*} \frac {2 a \left (a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac {2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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*} \int \sec ^4(c+d x) (a+b \sin (c+d x))^3 \, dx &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}-\frac {1}{3} \int \left (-2 a^2+2 b^2\right ) \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\\ &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac {1}{3} \left (2 \left (a^2-b^2\right )\right ) \int \sec ^2(c+d x) (a+b \sin (c+d x)) \, dx\\ &=\frac {2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac {1}{3} \left (2 a \left (a^2-b^2\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}-\frac {\left (2 a \left (a^2-b^2\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {2 b \left (a^2-b^2\right ) \sec (c+d x)}{3 d}+\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{3 d}+\frac {2 a \left (a^2-b^2\right ) \tan (c+d x)}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.43, size = 136, normalized size = 1.62 \begin {gather*} \frac {\sec ^3(c+d x) \left (24 a^2 b-4 b^3+\left (-9 a^2 b+15 b^3\right ) \cos (c+d x)-12 b^3 \cos (2 (c+d x))-3 a^2 b \cos (3 (c+d x))+5 b^3 \cos (3 (c+d x))+12 a^3 \sin (c+d x)+18 a b^2 \sin (c+d x)+4 a^3 \sin (3 (c+d x))-6 a b^2 \sin (3 (c+d x))\right )}{24 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(24*a^2*b - 4*b^3 + (-9*a^2*b + 15*b^3)*Cos[c + d*x] - 12*b^3*Cos[2*(c + d*x)] - 3*a^2*b*Cos[3
*(c + d*x)] + 5*b^3*Cos[3*(c + d*x)] + 12*a^3*Sin[c + d*x] + 18*a*b^2*Sin[c + d*x] + 4*a^3*Sin[3*(c + d*x)] -
6*a*b^2*Sin[3*(c + d*x)]))/(24*d)

________________________________________________________________________________________

Maple [A]
time = 0.36, size = 122, normalized size = 1.45

method	result	size
risch	\(-\frac {2 \left (9 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-6 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-12 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-2 i a^{3}+3 i a \,b^{2}+3 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{3}}\)	\(121\)
derivativedivides	\(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a^{2} b}{\cos \left (d x +c \right )^{3}}+\frac {a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d}\)	\(122\)
default	\(\frac {-a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a^{2} b}{\cos \left (d x +c \right )^{3}}+\frac {a \,b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{\cos \left (d x +c \right )^{3}}+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )}{d}\)	\(122\)
norman	\(\frac {-\frac {6 a^{2} b -4 b^{3}}{3 d}-\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (12 a^{2} b +8 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (18 a^{2} b +4 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (a^{2}+6 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (a^{2}+6 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a \left (7 a^{2}+12 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a \left (7 a^{2}+12 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {6 a^{2} b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {6 a^{2} b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 b \left (15 a^{2}+8 b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(318\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 80, normalized size = 0.95 \begin {gather*} \frac {3 \, a 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^{3} - \frac {{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} b^{3}}{\cos \left (d x + c\right )^{3}} + \frac {3 \, a^{2} b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.33, size = 77, normalized size = 0.92 \begin {gather*} -\frac {3 \, b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} - {\left (a^{3} + 3 \, a b^{2} + {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right )^{3} \sec ^{4}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*sin(d*x+c))**3,x)

[Out]

Integral((a + b*sin(c + d*x))**3*sec(c + d*x)**4, x)

________________________________________________________________________________________

Giac [A]
time = 5.74, size = 128, normalized size = 1.52 \begin {gather*} -\frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{2} b - 2 \, b^{3}\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-2/3*(3*a^3*tan(1/2*d*x + 1/2*c)^5 + 9*a^2*b*tan(1/2*d*x + 1/2*c)^4 - 2*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*a*b^2*
tan(1/2*d*x + 1/2*c)^3 + 6*b^3*tan(1/2*d*x + 1/2*c)^2 + 3*a^3*tan(1/2*d*x + 1/2*c) + 3*a^2*b - 2*b^3)/((tan(1/
2*d*x + 1/2*c)^2 - 1)^3*d)

________________________________________________________________________________________

Mupad [B]
time = 5.25, size = 81, normalized size = 0.96 \begin {gather*} \frac {a^2\,b+\frac {a^3\,\sin \left (c+d\,x\right )}{3}+\frac {b^3}{3}-{\cos \left (c+d\,x\right )}^2\,\left (-\frac {2\,\sin \left (c+d\,x\right )\,a^3}{3}+\sin \left (c+d\,x\right )\,a\,b^2+b^3\right )+a\,b^2\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^3/cos(c + d*x)^4,x)

[Out]

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

________________________________________________________________________________________