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

Optimal. Leaf size=180 \[ \frac{6 a^2 b^2 \cos (c+d x)}{d}+\frac{6 a^2 b^2 \sec (c+d x)}{d}-\frac{4 a^3 b \sin (c+d x)}{d}+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^4 \cos (c+d x)}{d}+\frac{6 a b^3 \sin (c+d x)}{d}+\frac{2 a b^3 \sin (c+d x) \tan ^2(c+d x)}{d}-\frac{6 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^4 \cos (c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 d}-\frac{2 b^4 \sec (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.15661, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {3517, 2638, 2592, 321, 206, 2590, 14, 288, 270} \[ \frac{6 a^2 b^2 \cos (c+d x)}{d}+\frac{6 a^2 b^2 \sec (c+d x)}{d}-\frac{4 a^3 b \sin (c+d x)}{d}+\frac{4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^4 \cos (c+d x)}{d}+\frac{6 a b^3 \sin (c+d x)}{d}+\frac{2 a b^3 \sin (c+d x) \tan ^2(c+d x)}{d}-\frac{6 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^4 \cos (c+d x)}{d}+\frac{b^4 \sec ^3(c+d x)}{3 d}-\frac{2 b^4 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3517

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[Expand[Sin[e
+ f*x]^m*(a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [B]  time = 5.45856, size = 383, normalized size = 2.13 \[ \frac{-48 a b \left (a^2-b^2\right ) \sin (c+d x)-12 \left (-6 a^2 b^2+a^4+b^4\right ) \cos (c+d x)+\frac{2 b^2 \left (36 a^2-11 b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{2 b^2 \left (11 b^2-36 a^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-24 a b \left (2 a^2-3 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+24 a b \left (2 a^2-3 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+72 a^2 b^2+\frac{b^3 (12 a+b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{b^3 (b-12 a)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 b^4 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}-\frac{2 b^4 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}-22 b^4}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(72*a^2*b^2 - 22*b^4 - 12*(a^4 - 6*a^2*b^2 + b^4)*Cos[c + d*x] - 24*a*b*(2*a^2 - 3*b^2)*Log[Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] + 24*a*b*(2*a^2 - 3*b^2)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^3*(12*a + b))/(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^4*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (2*b^2*
(36*a^2 - 11*b^2)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (2*b^4*Sin[(c + d*x)/2])/(Cos[(c +
d*x)/2] + Sin[(c + d*x)/2])^3 + (b^3*(-12*a + b))/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (2*b^2*(-36*a^2 +
11*b^2)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 48*a*b*(a^2 - b^2)*Sin[c + d*x])/(12*d)

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 309, normalized size = 1.7 \begin{align*}{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{b}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}-{\frac{8\,{b}^{4}\cos \left ( dx+c \right ) }{3\,d}}-{\frac{{b}^{4}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}-{\frac{4\,{b}^{4}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{{b}^{3}a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{{b}^{3}a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d}}+6\,{\frac{{b}^{3}a\sin \left ( dx+c \right ) }{d}}-6\,{\frac{{b}^{3}a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,{\frac{{a}^{2}{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}+6\,{\frac{{a}^{2}{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}+12\,{\frac{{a}^{2}{b}^{2}\cos \left ( dx+c \right ) }{d}}-4\,{\frac{b{a}^{3}\sin \left ( dx+c \right ) }{d}}+4\,{\frac{b{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{4}\cos \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)*(a+b*tan(d*x+c))^4,x)

[Out]

1/3/d*b^4*sin(d*x+c)^6/cos(d*x+c)^3-1/d*b^4*sin(d*x+c)^6/cos(d*x+c)-8/3*b^4*cos(d*x+c)/d-1/d*b^4*cos(d*x+c)*si
n(d*x+c)^4-4/3/d*b^4*cos(d*x+c)*sin(d*x+c)^2+2/d*b^3*a*sin(d*x+c)^5/cos(d*x+c)^2+2*a*b^3*sin(d*x+c)^3/d+6*a*b^
3*sin(d*x+c)/d-6/d*b^3*a*ln(sec(d*x+c)+tan(d*x+c))+6/d*a^2*b^2*sin(d*x+c)^4/cos(d*x+c)+6/d*a^2*b^2*cos(d*x+c)*
sin(d*x+c)^2+12*a^2*b^2*cos(d*x+c)/d-4*a^3*b*sin(d*x+c)/d+4/d*b*a^3*ln(sec(d*x+c)+tan(d*x+c))-a^4*cos(d*x+c)/d

________________________________________________________________________________________

Maxima [A]  time = 1.09334, size = 224, normalized size = 1.24 \begin{align*} -\frac{3 \, a b^{3}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} - 18 \, a^{2} b^{2}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + b^{4}{\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 \, a^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 3 \, a^{4} \cos \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*a*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*sin
(d*x + c)) - 18*a^2*b^2*(1/cos(d*x + c) + cos(d*x + c)) + b^4*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d
*x + c)) - 6*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c)) + 3*a^4*cos(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 2.06641, size = 414, normalized size = 2.3 \begin{align*} -\frac{3 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} - 3 \,{\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (2 \, a^{3} b - 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - b^{4} - 6 \,{\left (3 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (a b^{3} \cos \left (d x + c\right ) - 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{4} \sin{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+b*tan(d*x+c))**4,x)

[Out]

Integral((a + b*tan(c + d*x))**4*sin(c + d*x), x)

________________________________________________________________________________________

Giac [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(sin(d*x+c)*(a+b*tan(d*x+c))^4,x, algorithm="giac")

[Out]

Timed out