3.9.13 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx\) [813]

3.9.13.1 Optimal result
3.9.13.2 Mathematica [A] (verified)
3.9.13.3 Rubi [A] (verified)
3.9.13.4 Maple [C] (verified)
3.9.13.5 Fricas [B] (verification not implemented)
3.9.13.6 Sympy [F(-1)]
3.9.13.7 Maxima [A] (verification not implemented)
3.9.13.8 Giac [A] (verification not implemented)
3.9.13.9 Mupad [B] (verification not implemented)

3.9.13.1 Optimal result

Integrand size = 29, antiderivative size = 77 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=3 a^3 x-\frac {3 a^3 \cos (c+d x)}{d}-\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d} \]

output
3*a^3*x-3*a^3*cos(d*x+c)/d-2*a^5*cos(d*x+c)^3/d/(a-a*sin(d*x+c))^2+1/3*sec 
(d*x+c)^3*(a+a*sin(d*x+c))^3/d
 
3.9.13.2 Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.34 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {a^3 (1+\sin (c+d x))^3 \left (9 \arctan (\tan (c+d x))-15 \sec (c+d x)+4 \sec ^3(c+d x)+6 \sin ^2\left (\frac {1}{2} (c+d x)\right )-9 \tan (c+d x)+4 \tan ^3(c+d x)\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

input
Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^2,x]
 
output
(a^3*(1 + Sin[c + d*x])^3*(9*ArcTan[Tan[c + d*x]] - 15*Sec[c + d*x] + 4*Se 
c[c + d*x]^3 + 6*Sin[(c + d*x)/2]^2 - 9*Tan[c + d*x] + 4*Tan[c + d*x]^3))/ 
(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)
 
3.9.13.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 3350, 3042, 3149, 3042, 3159, 3042, 3161, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^2 (a \sin (c+d x)+a)^3}{\cos (c+d x)^4}dx\)

\(\Big \downarrow \) 3350

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-\int \sec ^2(c+d x) (\sin (c+d x) a+a)^3dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-\int \frac {(\sin (c+d x) a+a)^3}{\cos (c+d x)^2}dx\)

\(\Big \downarrow \) 3149

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^6 \int \frac {\cos ^4(c+d x)}{(a-a \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^6 \int \frac {\cos (c+d x)^4}{(a-a \sin (c+d x))^3}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \int \frac {\cos ^2(c+d x)}{a-a \sin (c+d x)}dx}{a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \int \frac {\cos (c+d x)^2}{a-a \sin (c+d x)}dx}{a^2}\right )\)

\(\Big \downarrow \) 3161

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \left (\frac {\int 1dx}{a}-\frac {\cos (c+d x)}{a d}\right )}{a^2}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d}-a^6 \left (\frac {2 \cos ^3(c+d x)}{a d (a-a \sin (c+d x))^2}-\frac {3 \left (\frac {x}{a}-\frac {\cos (c+d x)}{a d}\right )}{a^2}\right )\)

input
Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*Tan[c + d*x]^2,x]
 
output
(Sec[c + d*x]^3*(a + a*Sin[c + d*x])^3)/(3*d) - a^6*((-3*(x/a - Cos[c + d* 
x]/(a*d)))/a^2 + (2*Cos[c + d*x]^3)/(a*d*(a - a*Sin[c + d*x])^2))
 

3.9.13.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3149
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(a/g)^(2*m)   Int[(g*Cos[e + f*x])^(2*m + p)/( 
a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2 
, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]
 

rule 3159
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

rule 3161
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si 
mp[g^2/a   Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x 
] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
 

rule 3350
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + 
(b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^( 
p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*m)), x] - Simp[1/g^2   Int[(g*Cos[e + 
 f*x])^(p + 2)*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, m, p 
}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + p + 1, 0]
 
3.9.13.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25

method result size
risch \(3 a^{3} x -\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 \left (-24 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-13 a^{3}+15 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d}\) \(96\)
parallelrisch \(\frac {a^{3} \left (9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d x -27 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) x d +36 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) d x +18 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) x d -54 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d x +58 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 d x -66 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+28\right )}{3 d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\) \(166\)
derivativedivides \(\frac {a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\) \(184\)
default \(\frac {a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d}\) \(184\)
norman \(\frac {-3 a^{3} x +\frac {28 a^{3}}{3 d}+\frac {6 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {14 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {44 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {128 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {44 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {14 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {6 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+9 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{3} x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {12 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {40 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\) \(267\)

input
int(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
3*a^3*x-1/2*a^3/d*exp(I*(d*x+c))-1/2*a^3/d*exp(-I*(d*x+c))-2/3*(-24*I*a^3* 
exp(I*(d*x+c))-13*a^3+15*a^3*exp(2*I*(d*x+c)))/(exp(I*(d*x+c))-I)^3/d
 
3.9.13.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (76) = 152\).

Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.19 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {3 \, a^{3} \cos \left (d x + c\right )^{3} + 18 \, a^{3} d x + 2 \, a^{3} - {\left (9 \, a^{3} d x + 16 \, a^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, a^{3} d x - 17 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (18 \, a^{3} d x - 3 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} + {\left (9 \, a^{3} d x - 19 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

input
integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="frica 
s")
 
output
-1/3*(3*a^3*cos(d*x + c)^3 + 18*a^3*d*x + 2*a^3 - (9*a^3*d*x + 16*a^3)*cos 
(d*x + c)^2 + (9*a^3*d*x - 17*a^3)*cos(d*x + c) - (18*a^3*d*x - 3*a^3*cos( 
d*x + c)^2 - 2*a^3 + (9*a^3*d*x - 19*a^3)*cos(d*x + c))*sin(d*x + c))/(d*c 
os(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d 
)
 
3.9.13.6 Sympy [F(-1)]

Timed out. \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**4*sin(d*x+c)**2*(a+a*sin(d*x+c))**3,x)
 
output
Timed out
 
3.9.13.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.39 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {a^{3} \tan \left (d x + c\right )^{3} + 3 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - a^{3} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - \frac {3 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

input
integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxim 
a")
 
output
1/3*(a^3*tan(d*x + c)^3 + 3*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c) 
)*a^3 - a^3*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)) - 3*( 
3*cos(d*x + c)^2 - 1)*a^3/cos(d*x + c)^3)/d
 
3.9.13.8 Giac [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.13 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {9 \, {\left (d x + c\right )} a^{3} - \frac {6 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

input
integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac" 
)
 
output
1/3*(9*(d*x + c)*a^3 - 6*a^3/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(9*a^3*tan(1 
/2*d*x + 1/2*c)^2 - 24*a^3*tan(1/2*d*x + 1/2*c) + 11*a^3)/(tan(1/2*d*x + 1 
/2*c) - 1)^3)/d
 
3.9.13.9 Mupad [B] (verification not implemented)

Time = 12.68 (sec) , antiderivative size = 182, normalized size of antiderivative = 2.36 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx=3\,a^3\,x+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3\,\left (27\,d\,x-66\right )}{3}-9\,a^3\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^3\,\left (27\,d\,x-18\right )}{3}-9\,a^3\,d\,x\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^3\,\left (36\,d\,x-54\right )}{3}-12\,a^3\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (36\,d\,x-58\right )}{3}-12\,a^3\,d\,x\right )-\frac {a^3\,\left (9\,d\,x-28\right )}{3}+3\,a^3\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

input
int((sin(c + d*x)^2*(a + a*sin(c + d*x))^3)/cos(c + d*x)^4,x)
 
output
3*a^3*x + (tan(c/2 + (d*x)/2)*((a^3*(27*d*x - 66))/3 - 9*a^3*d*x) - tan(c/ 
2 + (d*x)/2)^4*((a^3*(27*d*x - 18))/3 - 9*a^3*d*x) + tan(c/2 + (d*x)/2)^3* 
((a^3*(36*d*x - 54))/3 - 12*a^3*d*x) - tan(c/2 + (d*x)/2)^2*((a^3*(36*d*x 
- 58))/3 - 12*a^3*d*x) - (a^3*(9*d*x - 28))/3 + 3*a^3*d*x)/(d*(tan(c/2 + ( 
d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2)^2 + 1))