\(\int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) [405]

Optimal result

Integrand size = 21, antiderivative size = 94 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 a \sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \]

[Out]

3/8*a*(a^2-b^2)*arctanh(sin(d*x+c))/d+3/8*a*sec(d*x+c)^2*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))/d+1/4*sec(d*x+c)^3*
(a+b*sin(d*x+c))^3*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 94, 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 = {2747, 743, 737, 212} \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 a \sec ^2(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec ^3(c+d x) (a+b \sin (c+d x))^3}{4 d} \]

[In]

Int[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[(2*p + 3)*((c*d^2 + a*e^2)/(2*a*c*(p + 1))), Int[(d + e*x)^(m -
2)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2, 0] && Lt
Q[p, -1]

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^m)*(2*c*x)*((a + c*x^2)
^(p + 1)/(4*a*c*(p + 1))), x] - Dist[m*((2*c*d)/(4*a*c*(p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x
], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {(a+x)^3}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\left (3 a b^3\right ) \text {Subst}\left (\int \frac {(a+x)^2}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {3 a \sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d}+\frac {\left (3 a b \left (a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {3 a \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 a \sec ^2(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^3 \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(318\) vs. \(2(94)=188\).

Time = 2.74 (sec) , antiderivative size = 318, normalized size of antiderivative = 3.38 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {-6 a \left (a^2-b^2\right )^3 (\log (1-\sin (c+d x))-\log (1+\sin (c+d x)))+a b \sec ^4(c+d x) \left (-8 a^5+8 a^3 b^2+\left (18 a^4 b-11 a^2 b^3+5 b^5\right ) \sin (3 (c+d x))\right )+a \left (8 a^6-22 a^4 b^2+29 a^2 b^4-3 b^6\right ) \sec ^3(c+d x) \tan (c+d x)+16 a^4 b \left (3 a^2-2 b^2\right ) \tan ^2(c+d x)+8 b^3 \left (4 a^4-5 a^2 b^2+b^4\right ) \tan ^4(c+d x)+4 a \sec (c+d x) \tan (c+d x) \left (3 \left (a^6-5 a^4 b^2\right )+4 b^2 \left (3 a^4-5 a^2 b^2+2 b^4\right ) \tan ^2(c+d x)\right )+16 a^2 b \sec ^2(c+d x) \left (-a^4+\left (2 a^4-5 a^2 b^2+3 b^4\right ) \tan ^2(c+d x)\right )}{32 \left (a^2-b^2\right )^2 d} \]

[In]

Integrate[Sec[c + d*x]^5*(a + b*Sin[c + d*x])^3,x]

[Out]

(-6*a*(a^2 - b^2)^3*(Log[1 - Sin[c + d*x]] - Log[1 + Sin[c + d*x]]) + a*b*Sec[c + d*x]^4*(-8*a^5 + 8*a^3*b^2 +
(18*a^4*b - 11*a^2*b^3 + 5*b^5)*Sin[3*(c + d*x)]) + a*(8*a^6 - 22*a^4*b^2 + 29*a^2*b^4 - 3*b^6)*Sec[c + d*x]^
3*Tan[c + d*x] + 16*a^4*b*(3*a^2 - 2*b^2)*Tan[c + d*x]^2 + 8*b^3*(4*a^4 - 5*a^2*b^2 + b^4)*Tan[c + d*x]^4 + 4*
a*Sec[c + d*x]*Tan[c + d*x]*(3*(a^6 - 5*a^4*b^2) + 4*b^2*(3*a^4 - 5*a^2*b^2 + 2*b^4)*Tan[c + d*x]^2) + 16*a^2*
b*Sec[c + d*x]^2*(-a^4 + (2*a^4 - 5*a^2*b^2 + 3*b^4)*Tan[c + d*x]^2))/(32*(a^2 - b^2)^2*d)

Maple [A] (verified)

Time = 1.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.66

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(156\)
default	\(\frac {a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {3 a^{2} b}{4 \cos \left (d x +c \right )^{4}}+3 a \,b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}+\frac {\sin ^{3}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{8}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 \cos \left (d x +c \right )^{4}}}{d}\)	\(156\)
parallelrisch	\(\frac {-6 \left (a +b \right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (a +b \right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \left (a -b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (-3 a^{2} b -b^{3}\right ) \cos \left (2 d x +2 c \right )+\left (-3 a^{2} b +b^{3}\right ) \cos \left (4 d x +4 c \right )+3 \left (a^{3}-a \,b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (11 a^{3}+21 a \,b^{2}\right ) \sin \left (d x +c \right )+15 a^{2} b +3 b^{3}}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(215\)
risch	\(-\frac {{\mathrm e}^{i \left (d x +c \right )} \left (3 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+11 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+21 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-11 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-21 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-48 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-3 i a^{3}+3 i a \,b^{2}+8 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{4}}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a \,b^{2}}{8 d}-\frac {3 a^{3} \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 d}+\frac {3 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right ) a \,b^{2}}{8 d}\)	\(283\)
norman	\(\frac {\frac {6 a^{2} b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} b \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (18 a^{2} b +4 b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (18 a^{2} b +4 b^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (7 a^{2}+33 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (8 a^{2} b +4 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 \left (8 a^{2} b +4 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {3 a \left (3 a^{2}+5 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (5 a^{2}+3 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (5 a^{2}+3 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {3 a \left (9 a^{2}+31 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {3 a \left (9 a^{2}+31 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {3 a \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 a \left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(431\)

[In]

int(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^3*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+3/4*a^2*b/cos(d*x+c)^4
+3*a*b^2*(1/4*sin(d*x+c)^3/cos(d*x+c)^4+1/8*sin(d*x+c)^3/cos(d*x+c)^2+1/8*sin(d*x+c)-1/8*ln(sec(d*x+c)+tan(d*x
+c)))+1/4*b^3*sin(d*x+c)^4/cos(d*x+c)^4)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.47 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 8 \, b^{3} \cos \left (d x + c\right )^{2} + 12 \, a^{2} b + 4 \, b^{3} + 2 \, {\left (2 \, a^{3} + 6 \, a b^{2} + 3 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/16*(3*(a^3 - a*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(a^3 - a*b^2)*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) - 8*b^3*cos(d*x + c)^2 + 12*a^2*b + 4*b^3 + 2*(2*a^3 + 6*a*b^2 + 3*(a^3 - a*b^2)*cos(d*x + c)^2)*sin(d*x
+ c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.45 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} - a b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} - a b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (4 \, b^{3} \sin \left (d x + c\right )^{2} - 3 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b - 2 \, b^{3} + {\left (5 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/16*(3*(a^3 - a*b^2)*log(sin(d*x + c) + 1) - 3*(a^3 - a*b^2)*log(sin(d*x + c) - 1) + 2*(4*b^3*sin(d*x + c)^2
- 3*(a^3 - a*b^2)*sin(d*x + c)^3 + 6*a^2*b - 2*b^3 + (5*a^3 + 3*a*b^2)*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d
*x + c)^2 + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.48 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 3 \, {\left (a^{3} - a b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{3} \sin \left (d x + c\right )^{3} - 3 \, a b^{2} \sin \left (d x + c\right )^{3} - 4 \, b^{3} \sin \left (d x + c\right )^{2} - 5 \, a^{3} \sin \left (d x + c\right ) - 3 \, a b^{2} \sin \left (d x + c\right ) - 6 \, a^{2} b + 2 \, b^{3}\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]

[In]

integrate(sec(d*x+c)^5*(a+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/16*(3*(a^3 - a*b^2)*log(abs(sin(d*x + c) + 1)) - 3*(a^3 - a*b^2)*log(abs(sin(d*x + c) - 1)) - 2*(3*a^3*sin(d
*x + c)^3 - 3*a*b^2*sin(d*x + c)^3 - 4*b^3*sin(d*x + c)^2 - 5*a^3*sin(d*x + c) - 3*a*b^2*sin(d*x + c) - 6*a^2*
b + 2*b^3)/(sin(d*x + c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 4.58 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \sec ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {3\,a\,b^2}{8}-\frac {3\,a^3}{8}\right )+\frac {3\,a^2\,b}{4}-\frac {b^3}{4}+\sin \left (c+d\,x\right )\,\left (\frac {5\,a^3}{8}+\frac {3\,a\,b^2}{8}\right )+\frac {b^3\,{\sin \left (c+d\,x\right )}^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (a^2-b^2\right )}{8\,d} \]

[In]

int((a + b*sin(c + d*x))^3/cos(c + d*x)^5,x)

[Out]

(sin(c + d*x)^3*((3*a*b^2)/8 - (3*a^3)/8) + (3*a^2*b)/4 - b^3/4 + sin(c + d*x)*((3*a*b^2)/8 + (5*a^3)/8) + (b^
3*sin(c + d*x)^2)/2)/(d*(sin(c + d*x)^4 - 2*sin(c + d*x)^2 + 1)) + (3*a*atanh(sin(c + d*x))*(a^2 - b^2))/(8*d)