3.894 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=91 \[ \frac{2 a^2 \tan ^7(c+d x)}{7 d}+\frac{2 a^2 \tan ^5(c+d x)}{5 d}+\frac{2 a^2 \sec ^7(c+d x)}{7 d}-\frac{3 a^2 \sec ^5(c+d x)}{5 d}+\frac{a^2 \sec ^3(c+d x)}{3 d} \]

[Out]

(a^2*Sec[c + d*x]^3)/(3*d) - (3*a^2*Sec[c + d*x]^5)/(5*d) + (2*a^2*Sec[c + d*x]^7)/(7*d) + (2*a^2*Tan[c + d*x]
^5)/(5*d) + (2*a^2*Tan[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.21374, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2873, 2606, 14, 2607, 270} \[ \frac{2 a^2 \tan ^7(c+d x)}{7 d}+\frac{2 a^2 \tan ^5(c+d x)}{5 d}+\frac{2 a^2 \sec ^7(c+d x)}{7 d}-\frac{3 a^2 \sec ^5(c+d x)}{5 d}+\frac{a^2 \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

(a^2*Sec[c + d*x]^3)/(3*d) - (3*a^2*Sec[c + d*x]^5)/(5*d) + (2*a^2*Sec[c + d*x]^7)/(7*d) + (2*a^2*Tan[c + d*x]
^5)/(5*d) + (2*a^2*Tan[c + d*x]^7)/(7*d)

Rule 2873

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 270

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]

Rubi steps

\begin{align*} \int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx &=\int \left (a^2 \sec ^5(c+d x) \tan ^3(c+d x)+2 a^2 \sec ^4(c+d x) \tan ^4(c+d x)+a^2 \sec ^3(c+d x) \tan ^5(c+d x)\right ) \, dx\\ &=a^2 \int \sec ^5(c+d x) \tan ^3(c+d x) \, dx+a^2 \int \sec ^3(c+d x) \tan ^5(c+d x) \, dx+\left (2 a^2\right ) \int \sec ^4(c+d x) \tan ^4(c+d x) \, dx\\ &=\frac{a^2 \operatorname{Subst}\left (\int x^4 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac{a^2 \operatorname{Subst}\left (\int x^2 \left (-1+x^2\right )^2 \, dx,x,\sec (c+d x)\right )}{d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int x^4 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a^2 \operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac{a^2 \operatorname{Subst}\left (\int \left (-x^4+x^6\right ) \, dx,x,\sec (c+d x)\right )}{d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \left (x^4+x^6\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{a^2 \sec ^3(c+d x)}{3 d}-\frac{3 a^2 \sec ^5(c+d x)}{5 d}+\frac{2 a^2 \sec ^7(c+d x)}{7 d}+\frac{2 a^2 \tan ^5(c+d x)}{5 d}+\frac{2 a^2 \tan ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.984646, size = 139, normalized size = 1.53 \[ -\frac{a^2 \sec ^7(c+d x) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 (448 \sin (c+d x)-104 \sin (2 (c+d x))-144 \sin (3 (c+d x))-52 \sin (4 (c+d x))+48 \sin (5 (c+d x))+182 \cos (c+d x)+736 \cos (2 (c+d x))+39 \cos (3 (c+d x))-192 \cos (4 (c+d x))-13 \cos (5 (c+d x))-672)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

-(a^2*Sec[c + d*x]^7*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*(-672 + 182*Cos[c + d*x] + 736*Cos[2*(c + d*x)] +
39*Cos[3*(c + d*x)] - 192*Cos[4*(c + d*x)] - 13*Cos[5*(c + d*x)] + 448*Sin[c + d*x] - 104*Sin[2*(c + d*x)] -
144*Sin[3*(c + d*x)] - 52*Sin[4*(c + d*x)] + 48*Sin[5*(c + d*x)]))/(6720*d)

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Maple [B]  time = 0.089, size = 248, normalized size = 2.7 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{35\,\cos \left ( dx+c \right ) }}+{\frac{\cos \left ( dx+c \right ) }{35} \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) +2\,{a}^{2} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) +{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{35}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(a^2*(1/7*sin(d*x+c)^6/cos(d*x+c)^7+1/35*sin(d*x+c)^6/cos(d*x+c)^5-1/105*sin(d*x+c)^6/cos(d*x+c)^3+1/35*si
n(d*x+c)^6/cos(d*x+c)+1/35*(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))+2*a^2*(1/7*sin(d*x+c)^5/cos(d*x+c)^
7+2/35*sin(d*x+c)^5/cos(d*x+c)^5)+a^2*(1/7*sin(d*x+c)^4/cos(d*x+c)^7+3/35*sin(d*x+c)^4/cos(d*x+c)^5+1/35*sin(d
*x+c)^4/cos(d*x+c)^3-1/35*sin(d*x+c)^4/cos(d*x+c)-1/35*(2+sin(d*x+c)^2)*cos(d*x+c)))

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Maxima [A]  time = 1.03654, size = 123, normalized size = 1.35 \begin{align*} \frac{6 \,{\left (5 \, \tan \left (d x + c\right )^{7} + 7 \, \tan \left (d x + c\right )^{5}\right )} a^{2} + \frac{{\left (35 \, \cos \left (d x + c\right )^{4} - 42 \, \cos \left (d x + c\right )^{2} + 15\right )} a^{2}}{\cos \left (d x + c\right )^{7}} - \frac{3 \,{\left (7 \, \cos \left (d x + c\right )^{2} - 5\right )} a^{2}}{\cos \left (d x + c\right )^{7}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/105*(6*(5*tan(d*x + c)^7 + 7*tan(d*x + c)^5)*a^2 + (35*cos(d*x + c)^4 - 42*cos(d*x + c)^2 + 15)*a^2/cos(d*x
+ c)^7 - 3*(7*cos(d*x + c)^2 - 5)*a^2/cos(d*x + c)^7)/d

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Fricas [A]  time = 1.3343, size = 284, normalized size = 3.12 \begin{align*} -\frac{24 \, a^{2} \cos \left (d x + c\right )^{4} - 47 \, a^{2} \cos \left (d x + c\right )^{2} + 25 \, a^{2} - 2 \,{\left (6 \, a^{2} \cos \left (d x + c\right )^{4} - 9 \, a^{2} \cos \left (d x + c\right )^{2} + 5 \, a^{2}\right )} \sin \left (d x + c\right )}{105 \,{\left (d \cos \left (d x + c\right )^{5} + 2 \, d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/105*(24*a^2*cos(d*x + c)^4 - 47*a^2*cos(d*x + c)^2 + 25*a^2 - 2*(6*a^2*cos(d*x + c)^4 - 9*a^2*cos(d*x + c)^
2 + 5*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5 + 2*d*cos(d*x + c)^3*sin(d*x + c) - 2*d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.28235, size = 186, normalized size = 2.04 \begin{align*} -\frac{\frac{35 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}} - \frac{105 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1015 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1330 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1302 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 469 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 67 \, a^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{7}}}{840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/840*(35*(3*a^2*tan(1/2*d*x + 1/2*c) + a^2)/(tan(1/2*d*x + 1/2*c) + 1)^3 - (105*a^2*tan(1/2*d*x + 1/2*c)^5 -
1015*a^2*tan(1/2*d*x + 1/2*c)^4 + 1330*a^2*tan(1/2*d*x + 1/2*c)^3 - 1302*a^2*tan(1/2*d*x + 1/2*c)^2 + 469*a^2
*tan(1/2*d*x + 1/2*c) - 67*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^7)/d