\(\int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx\) [821]

Optimal result

Integrand size = 29, antiderivative size = 117 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x}{a}-\frac {\cos (c+d x)}{a d}-\frac {3 \sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\tan (c+d x)}{a d}-\frac {\tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d} \]

[Out]

-x/a-cos(d*x+c)/a/d-3*sec(d*x+c)/a/d+sec(d*x+c)^3/a/d-1/5*sec(d*x+c)^5/a/d+tan(d*x+c)/a/d-1/3*tan(d*x+c)^3/a/d
+1/5*tan(d*x+c)^5/a/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2918, 3554, 8, 2670, 276} \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\cos (c+d x)}{a d}+\frac {\tan ^5(c+d x)}{5 a d}-\frac {\tan ^3(c+d x)}{3 a d}+\frac {\tan (c+d x)}{a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\sec ^3(c+d x)}{a d}-\frac {3 \sec (c+d x)}{a d}-\frac {x}{a} \]

[In]

Int[(Sin[c + d*x]^2*Tan[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

-(x/a) - Cos[c + d*x]/(a*d) - (3*Sec[c + d*x])/(a*d) + Sec[c + d*x]^3/(a*d) - Sec[c + d*x]^5/(5*a*d) + Tan[c +
d*x]/(a*d) - Tan[c + d*x]^3/(3*a*d) + Tan[c + d*x]^5/(5*a*d)

Rule 8

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

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \tan ^6(c+d x) \, dx}{a}-\frac {\int \sin (c+d x) \tan ^6(c+d x) \, dx}{a} \\ & = \frac {\tan ^5(c+d x)}{5 a d}-\frac {\int \tan ^4(c+d x) \, dx}{a}+\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d}+\frac {\int \tan ^2(c+d x) \, dx}{a}+\frac {\text {Subst}\left (\int \left (-1+\frac {1}{x^6}-\frac {3}{x^4}+\frac {3}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\cos (c+d x)}{a d}-\frac {3 \sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\tan (c+d x)}{a d}-\frac {\tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d}-\frac {\int 1 \, dx}{a} \\ & = -\frac {x}{a}-\frac {\cos (c+d x)}{a d}-\frac {3 \sec (c+d x)}{a d}+\frac {\sec ^3(c+d x)}{a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\tan (c+d x)}{a d}-\frac {\tan ^3(c+d x)}{3 a d}+\frac {\tan ^5(c+d x)}{5 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.18 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.91 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {1200+18 (-103+40 c+40 d x) \cos (c+d x)+1568 \cos (2 (c+d x))-618 \cos (3 (c+d x))+240 c \cos (3 (c+d x))+240 d x \cos (3 (c+d x))+304 \cos (4 (c+d x))+216 \sin (c+d x)-618 \sin (2 (c+d x))+240 c \sin (2 (c+d x))+240 d x \sin (2 (c+d x))+532 \sin (3 (c+d x))-309 \sin (4 (c+d x))+120 c \sin (4 (c+d x))+120 d x \sin (4 (c+d x))+60 \sin (5 (c+d x))}{960 a d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

[In]

Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^4)/(a + a*Sin[c + d*x]),x]

[Out]

-1/960*(1200 + 18*(-103 + 40*c + 40*d*x)*Cos[c + d*x] + 1568*Cos[2*(c + d*x)] - 618*Cos[3*(c + d*x)] + 240*c*C
os[3*(c + d*x)] + 240*d*x*Cos[3*(c + d*x)] + 304*Cos[4*(c + d*x)] + 216*Sin[c + d*x] - 618*Sin[2*(c + d*x)] +
240*c*Sin[2*(c + d*x)] + 240*d*x*Sin[2*(c + d*x)] + 532*Sin[3*(c + d*x)] - 309*Sin[4*(c + d*x)] + 120*c*Sin[4*
(c + d*x)] + 120*d*x*Sin[4*(c + d*x)] + 60*Sin[5*(c + d*x)])/(a*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos
[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.36

method	result	size
derivativedivides	\(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {23}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\)	\(159\)
default	\(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {7}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {23}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\)	\(159\)
risch	\(-\frac {x}{a}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a d}-\frac {2 \left (63 \,{\mathrm e}^{3 i \left (d x +c \right )}+105 i {\mathrm e}^{4 i \left (d x +c \right )}+91 i {\mathrm e}^{2 i \left (d x +c \right )}+75 \,{\mathrm e}^{5 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}+23 i+45 i {\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} d a}\)	\(159\)
parallelrisch	\(\frac {\left (-150 d x -480\right ) \cos \left (3 d x +3 c \right )-300 d x \cos \left (d x +c \right )-30 d x \cos \left (5 d x +5 c \right )-15 \cos \left (6 d x +6 c \right )+100 \sin \left (d x +c \right )+50 \sin \left (3 d x +3 c \right )+46 \sin \left (5 d x +5 c \right )-960 \cos \left (d x +c \right )-705 \cos \left (2 d x +2 c \right )-270 \cos \left (4 d x +4 c \right )-96 \cos \left (5 d x +5 c \right )-546}{30 a d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(163\)
norman	\(\frac {\frac {x}{a}-\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {88 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {4 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {2 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {4 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {4 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {3 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {32}{5 a d}-\frac {20 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {54 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 d a}+\frac {28 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}+\frac {16 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {182 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {208 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(415\)

[In]

int(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

128/d/a*(-1/768/(tan(1/2*d*x+1/2*c)-1)^3-1/512/(tan(1/2*d*x+1/2*c)-1)^2+7/1024/(tan(1/2*d*x+1/2*c)-1)-1/64/(1+
tan(1/2*d*x+1/2*c)^2)-1/64*arctan(tan(1/2*d*x+1/2*c))-1/320/(tan(1/2*d*x+1/2*c)+1)^5+1/128/(tan(1/2*d*x+1/2*c)
+1)^4+1/384/(tan(1/2*d*x+1/2*c)+1)^3-3/256/(tan(1/2*d*x+1/2*c)+1)^2-23/1024/(tan(1/2*d*x+1/2*c)+1))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {15 \, d x \cos \left (d x + c\right )^{3} + 38 \, \cos \left (d x + c\right )^{4} + 11 \, \cos \left (d x + c\right )^{2} + {\left (15 \, d x \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )^{4} + 22 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 1}{15 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/15*(15*d*x*cos(d*x + c)^3 + 38*cos(d*x + c)^4 + 11*cos(d*x + c)^2 + (15*d*x*cos(d*x + c)^3 + 15*cos(d*x + c
)^4 + 22*cos(d*x + c)^2 - 4)*sin(d*x + c) - 1)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**4*sin(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 400 vs. \(2 (111) = 222\).

Time = 0.31 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.42 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \, {\left (\frac {\frac {81 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {78 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {172 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {26 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {22 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {70 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {20 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {30 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 48}{a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {2 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {4 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}\right )}}{15 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-2/15*((81*sin(d*x + c)/(cos(d*x + c) + 1) - 78*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 172*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 26*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 22*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 70*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + 20*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 30*sin(d*x + c)^8/(cos(d*x + c) + 1)^8
- 15*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 48)/(a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - a*sin(d*x + c)^2/(c
os(d*x + c) + 1)^2 - 4*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 2*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 - 2*a*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) + 15*arctan(sin(d*
x + c)/(cos(d*x + c) + 1))/a)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {120 \, {\left (d x + c\right )}}{a} + \frac {240}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a} - \frac {5 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 48 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 23\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {345 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1560 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2570 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1720 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 413}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \]

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/120*(120*(d*x + c)/a + 240/((tan(1/2*d*x + 1/2*c)^2 + 1)*a) - 5*(21*tan(1/2*d*x + 1/2*c)^2 - 48*tan(1/2*d*x
+ 1/2*c) + 23)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) + (345*tan(1/2*d*x + 1/2*c)^4 + 1560*tan(1/2*d*x + 1/2*c)^3 +
2570*tan(1/2*d*x + 1/2*c)^2 + 1720*tan(1/2*d*x + 1/2*c) + 413)/(a*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

Mupad [B] (verification not implemented)

Time = 20.42 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.47 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {52\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}-\frac {344\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}-\frac {52\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {54\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {32}{5}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {x}{a} \]

[In]

int(sin(c + d*x)^6/(cos(c + d*x)^4*(a + a*sin(c + d*x))),x)

[Out]

((54*tan(c/2 + (d*x)/2))/5 - (52*tan(c/2 + (d*x)/2)^2)/5 - (344*tan(c/2 + (d*x)/2)^3)/15 - (52*tan(c/2 + (d*x)
/2)^4)/15 + (44*tan(c/2 + (d*x)/2)^5)/15 + (28*tan(c/2 + (d*x)/2)^6)/3 + (8*tan(c/2 + (d*x)/2)^7)/3 - 4*tan(c/
2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)^9 + 32/5)/(a*d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^5*(ta
n(c/2 + (d*x)/2)^2 + 1)) - x/a