3.1482 \(\int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=135 \[ -\frac{15 a \sin (c+d x)}{8 d}+\frac{a \sin (c+d x) \tan ^4(c+d x)}{4 d}-\frac{5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac{15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \cos ^2(c+d x)}{2 d}+\frac{b \sec ^4(c+d x)}{4 d}-\frac{3 b \sec ^2(c+d x)}{2 d}-\frac{3 b \log (\cos (c+d x))}{d} \]

[Out]

(15*a*ArcTanh[Sin[c + d*x]])/(8*d) + (b*Cos[c + d*x]^2)/(2*d) - (3*b*Log[Cos[c + d*x]])/d - (3*b*Sec[c + d*x]^
2)/(2*d) + (b*Sec[c + d*x]^4)/(4*d) - (15*a*Sin[c + d*x])/(8*d) - (5*a*Sin[c + d*x]*Tan[c + d*x]^2)/(8*d) + (a
*Sin[c + d*x]*Tan[c + d*x]^4)/(4*d)

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Rubi [A]  time = 0.133346, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2834, 2592, 288, 321, 206, 2590, 266, 43} \[ -\frac{15 a \sin (c+d x)}{8 d}+\frac{a \sin (c+d x) \tan ^4(c+d x)}{4 d}-\frac{5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac{15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \cos ^2(c+d x)}{2 d}+\frac{b \sec ^4(c+d x)}{4 d}-\frac{3 b \sec ^2(c+d x)}{2 d}-\frac{3 b \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a*ArcTanh[Sin[c + d*x]])/(8*d) + (b*Cos[c + d*x]^2)/(2*d) - (3*b*Log[Cos[c + d*x]])/d - (3*b*Sec[c + d*x]^
2)/(2*d) + (b*Sec[c + d*x]^4)/(4*d) - (15*a*Sin[c + d*x])/(8*d) - (5*a*Sin[c + d*x]*Tan[c + d*x]^2)/(8*d) + (a
*Sin[c + d*x]*Tan[c + d*x]^4)/(4*d)

Rule 2834

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

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sin (c+d x) (a+b \sin (c+d x)) \tan ^5(c+d x) \, dx &=a \int \sin (c+d x) \tan ^5(c+d x) \, dx+b \int \sin ^2(c+d x) \tan ^5(c+d x) \, dx\\ &=\frac{a \operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3} \, dx,x,\sin (c+d x)\right )}{d}-\frac{b \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^5} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{a \sin (c+d x) \tan ^4(c+d x)}{4 d}-\frac{(5 a) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (c+d x)\right )}{4 d}-\frac{b \operatorname{Subst}\left (\int \frac{(1-x)^3}{x^3} \, dx,x,\cos ^2(c+d x)\right )}{2 d}\\ &=-\frac{5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac{a \sin (c+d x) \tan ^4(c+d x)}{4 d}+\frac{(15 a) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\sin (c+d x)\right )}{8 d}-\frac{b \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^3}-\frac{3}{x^2}+\frac{3}{x}\right ) \, dx,x,\cos ^2(c+d x)\right )}{2 d}\\ &=\frac{b \cos ^2(c+d x)}{2 d}-\frac{3 b \log (\cos (c+d x))}{d}-\frac{3 b \sec ^2(c+d x)}{2 d}+\frac{b \sec ^4(c+d x)}{4 d}-\frac{15 a \sin (c+d x)}{8 d}-\frac{5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac{a \sin (c+d x) \tan ^4(c+d x)}{4 d}+\frac{(15 a) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (c+d x)\right )}{8 d}\\ &=\frac{15 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b \cos ^2(c+d x)}{2 d}-\frac{3 b \log (\cos (c+d x))}{d}-\frac{3 b \sec ^2(c+d x)}{2 d}+\frac{b \sec ^4(c+d x)}{4 d}-\frac{15 a \sin (c+d x)}{8 d}-\frac{5 a \sin (c+d x) \tan ^2(c+d x)}{8 d}+\frac{a \sin (c+d x) \tan ^4(c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.352517, size = 133, normalized size = 0.99 \[ -\frac{a \sin (c+d x) \tan ^4(c+d x)}{d}-\frac{5 a \left (6 \tan (c+d x) \sec ^3(c+d x)-8 \tan ^3(c+d x) \sec (c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d}-\frac{b \left (2 \sin ^2(c+d x)-\sec ^4(c+d x)+6 \sec ^2(c+d x)+12 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(12*Log[Cos[c + d*x]] + 6*Sec[c + d*x]^2 - Sec[c + d*x]^4 + 2*Sin[c + d*x]^2))/(4*d) - (a*Sin[c + d*x]*Tan
[c + d*x]^4)/d - (5*a*(6*Sec[c + d*x]^3*Tan[c + d*x] - 8*Sec[c + d*x]*Tan[c + d*x]^3 - 3*(ArcTanh[Sin[c + d*x]
] + Sec[c + d*x]*Tan[c + d*x])))/(8*d)

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Maple [A]  time = 0.05, size = 205, normalized size = 1.5 \begin{align*}{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,a \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}-{\frac{5\,a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{15\,a\sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{3\,b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}b}{2\,d}}-3\,{\frac{b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.996085, size = 163, normalized size = 1.21 \begin{align*} -\frac{8 \, b \sin \left (d x + c\right )^{2} - 3 \,{\left (5 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (5 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \sin \left (d x + c\right ) - \frac{2 \,{\left (9 \, a \sin \left (d x + c\right )^{3} + 12 \, b \sin \left (d x + c\right )^{2} - 7 \, a \sin \left (d x + c\right ) - 10 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(8*b*sin(d*x + c)^2 - 3*(5*a - 8*b)*log(sin(d*x + c) + 1) + 3*(5*a + 8*b)*log(sin(d*x + c) - 1) + 16*a*s
in(d*x + c) - 2*(9*a*sin(d*x + c)^3 + 12*b*sin(d*x + c)^2 - 7*a*sin(d*x + c) - 10*b)/(sin(d*x + c)^4 - 2*sin(d
*x + c)^2 + 1))/d

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Fricas [A]  time = 2.06085, size = 360, normalized size = 2.67 \begin{align*} \frac{8 \, b \cos \left (d x + c\right )^{6} + 3 \,{\left (5 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (5 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 4 \, b \cos \left (d x + c\right )^{4} - 24 \, b \cos \left (d x + c\right )^{2} - 2 \,{\left (8 \, a \cos \left (d x + c\right )^{4} + 9 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(8*b*cos(d*x + c)^6 + 3*(5*a - 8*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(5*a + 8*b)*cos(d*x + c)^4*l
og(-sin(d*x + c) + 1) - 4*b*cos(d*x + c)^4 - 24*b*cos(d*x + c)^2 - 2*(8*a*cos(d*x + c)^4 + 9*a*cos(d*x + c)^2
- 2*a)*sin(d*x + c) + 4*b)/(d*cos(d*x + c)^4)

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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)**5*sin(d*x+c)**6*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.25774, size = 167, normalized size = 1.24 \begin{align*} -\frac{8 \, b \sin \left (d x + c\right )^{2} - 3 \,{\left (5 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 3 \,{\left (5 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 16 \, a \sin \left (d x + c\right ) - \frac{2 \,{\left (18 \, b \sin \left (d x + c\right )^{4} + 9 \, a \sin \left (d x + c\right )^{3} - 24 \, b \sin \left (d x + c\right )^{2} - 7 \, a \sin \left (d x + c\right ) + 8 \, b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/16*(8*b*sin(d*x + c)^2 - 3*(5*a - 8*b)*log(abs(sin(d*x + c) + 1)) + 3*(5*a + 8*b)*log(abs(sin(d*x + c) - 1)
) + 16*a*sin(d*x + c) - 2*(18*b*sin(d*x + c)^4 + 9*a*sin(d*x + c)^3 - 24*b*sin(d*x + c)^2 - 7*a*sin(d*x + c) +
8*b)/(sin(d*x + c)^2 - 1)^2)/d