3.796 \(\int \sin (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=101 \[ -\frac{a \cos (c+d x)}{d}+\frac{5 a \tan ^3(c+d x)}{6 d}-\frac{5 a \tan (c+d x)}{2 d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{2 a \sec (c+d x)}{d}-\frac{a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{5 a x}{2} \]

[Out]

(5*a*x)/2 - (a*Cos[c + d*x])/d - (2*a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (5*a*Tan[c + d*x])/(2*d) +
(5*a*Tan[c + d*x]^3)/(6*d) - (a*Sin[c + d*x]^2*Tan[c + d*x]^3)/(2*d)

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Rubi [A]  time = 0.130055, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2838, 2590, 270, 2591, 288, 302, 203} \[ -\frac{a \cos (c+d x)}{d}+\frac{5 a \tan ^3(c+d x)}{6 d}-\frac{5 a \tan (c+d x)}{2 d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{2 a \sec (c+d x)}{d}-\frac{a \sin ^2(c+d x) \tan ^3(c+d x)}{2 d}+\frac{5 a x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*x)/2 - (a*Cos[c + d*x])/d - (2*a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (5*a*Tan[c + d*x])/(2*d) +
(5*a*Tan[c + d*x]^3)/(6*d) - (a*Sin[c + d*x]^2*Tan[c + d*x]^3)/(2*d)

Rule 2838

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

Rule 2590

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 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]

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.245321, size = 76, normalized size = 0.75 \[ \frac{a \left (-3 \sin (2 (c+d x))-12 \cos (c+d x)-28 \tan (c+d x)+4 \sec ^3(c+d x)-24 \sec (c+d x)+4 \tan (c+d x) \sec ^2(c+d x)+30 c+30 d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(30*c + 30*d*x - 12*Cos[c + d*x] - 24*Sec[c + d*x] + 4*Sec[c + d*x]^3 - 3*Sin[2*(c + d*x)] - 28*Tan[c + d*x
] + 4*Sec[c + d*x]^2*Tan[c + d*x]))/(12*d)

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Maple [A]  time = 0.069, size = 154, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{3\,\cos \left ( dx+c \right ) }}-{\frac{4\,\cos \left ( dx+c \right ) }{3} \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{2}}+{\frac{5\,c}{2}} \right ) +a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}- \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cos \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x)

[Out]

1/d*(a*(1/3*sin(d*x+c)^7/cos(d*x+c)^3-4/3*sin(d*x+c)^7/cos(d*x+c)-4/3*(sin(d*x+c)^5+5/4*sin(d*x+c)^3+15/8*sin(
d*x+c))*cos(d*x+c)+5/2*d*x+5/2*c)+a*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+c)^4+4
/3*sin(d*x+c)^2)*cos(d*x+c)))

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Maxima [A]  time = 1.69724, size = 117, normalized size = 1.16 \begin{align*} \frac{{\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac{3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a - 2 \, a{\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 )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/6*((2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a - 2*a*((6*co
s(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)))/d

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Fricas [A]  time = 1.40778, size = 258, normalized size = 2.55 \begin{align*} \frac{3 \, a \cos \left (d x + c\right )^{4} - 15 \, a d x \cos \left (d x + c\right ) + 17 \, a \cos \left (d x + c\right )^{2} +{\left (15 \, a d x \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right ) - 4 \, a}{6 \,{\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*a*cos(d*x + c)^4 - 15*a*d*x*cos(d*x + c) + 17*a*cos(d*x + c)^2 + (15*a*d*x*cos(d*x + c) - 3*a*cos(d*x +
c)^2 + 2*a)*sin(d*x + c) - 4*a)/(d*cos(d*x + c)*sin(d*x + c) - d*cos(d*x + c))

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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)**4*sin(d*x+c)**5*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.26608, size = 181, normalized size = 1.79 \begin{align*} \frac{15 \,{\left (d x + c\right )} a + \frac{3 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} + \frac{6 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}} + \frac{21 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 48 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 23 \, a}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(15*(d*x + c)*a + 3*a/(tan(1/2*d*x + 1/2*c) + 1) + 6*(a*tan(1/2*d*x + 1/2*c)^3 - 2*a*tan(1/2*d*x + 1/2*c)^
2 - a*tan(1/2*d*x + 1/2*c) - 2*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 + (21*a*tan(1/2*d*x + 1/2*c)^2 - 48*a*tan(1/2
*d*x + 1/2*c) + 23*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d