3.34 \(\int \sin (c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=133 \[ -\frac{3 a^2 b \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cos (c+d x)}{d}+\frac{3 a b^2 \cos (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{3 b^3 \sin (c+d x)}{2 d}+\frac{b^3 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac{3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.116482, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {3517, 2638, 2592, 321, 206, 2590, 14, 288} \[ -\frac{3 a^2 b \sin (c+d x)}{d}+\frac{3 a^2 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^3 \cos (c+d x)}{d}+\frac{3 a b^2 \cos (c+d x)}{d}+\frac{3 a b^2 \sec (c+d x)}{d}+\frac{3 b^3 \sin (c+d x)}{2 d}+\frac{b^3 \sin (c+d x) \tan ^2(c+d x)}{2 d}-\frac{3 b^3 \tanh ^{-1}(\sin (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [B]  time = 6.1559, size = 637, normalized size = 4.79 \[ -\frac{a \left (a^2-3 b^2\right ) \cos ^4(c+d x) (a+b \tan (c+d x))^3}{d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{b \left (3 a^2-b^2\right ) \sin (c+d x) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{3 \left (2 a^2 b-b^3\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 \left (2 a^2 b-b^3\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{2 d (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{3 a b^2 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{3 a b^2 \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^3}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (a \cos (c+d x)+b \sin (c+d x))^3}+\frac{b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3}-\frac{b^3 \cos ^3(c+d x) (a+b \tan (c+d x))^3}{4 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 (a \cos (c+d x)+b \sin (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*b^2*Cos[c + d*x]^3*(a + b*Tan[c + d*x])^3)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (a*(a^2 - 3*b^2)*Cos
[c + d*x]^4*(a + b*Tan[c + d*x])^3)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*(2*a^2*b - b^3)*Cos[c + d*x]^
3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) +
(3*(2*a^2*b - b^3)*Cos[c + d*x]^3*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Tan[c + d*x])^3)/(2*d*(a*Co
s[c + d*x] + b*Sin[c + d*x])^3) + (b^3*Cos[c + d*x]^3*(a + b*Tan[c + d*x])^3)/(4*d*(Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) + (3*a*b^2*Cos[c + d*x]^3*Sin[(c + d*x)/2]*(a + b*Tan[c + d*
x])^3)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (b^3*Cos[c + d*x]^3*(a
+ b*Tan[c + d*x])^3)/(4*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2*(a*Cos[c + d*x] + b*Sin[c + d*x])^3) - (3*a*
b^2*Cos[c + d*x]^3*Sin[(c + d*x)/2]*(a + b*Tan[c + d*x])^3)/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(a*Cos[c
+ d*x] + b*Sin[c + d*x])^3) - (b*(3*a^2 - b^2)*Cos[c + d*x]^3*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(d*(a*Cos[c
+ d*x] + b*Sin[c + d*x])^3)

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Maple [A]  time = 0.043, size = 193, normalized size = 1.5 \begin{align*}{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{b}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,{b}^{3}\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{a{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d\cos \left ( dx+c \right ) }}+3\,{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}a{b}^{2}}{d}}+6\,{\frac{a{b}^{2}\cos \left ( dx+c \right ) }{d}}-3\,{\frac{b{a}^{2}\sin \left ( dx+c \right ) }{d}}+3\,{\frac{b{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{3}\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))^3,x)

[Out]

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

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Maxima [A]  time = 1.11825, size = 173, normalized size = 1.3 \begin{align*} -\frac{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 )} - 12 \, a b^{2}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 6 \, a^{2} 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 )} + 4 \, a^{3} \cos \left (d x + c\right )}{4 \, 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))^3,x, algorithm="maxima")

[Out]

-1/4*(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)) - 12*a*b^2*(1/cos(d*x + c) + cos(d*x + c)) - 6*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2
*sin(d*x + c)) + 4*a^3*cos(d*x + c))/d

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Fricas [A]  time = 2.12391, size = 346, normalized size = 2.6 \begin{align*} \frac{12 \, a b^{2} \cos \left (d x + c\right ) - 4 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (b^{3} - 2 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/4*(12*a*b^2*cos(d*x + c) - 4*(a^3 - 3*a*b^2)*cos(d*x + c)^3 + 3*(2*a^2*b - b^3)*cos(d*x + c)^2*log(sin(d*x +
c) + 1) - 3*(2*a^2*b - b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(b^3 - 2*(3*a^2*b - b^3)*cos(d*x + c)^2
)*sin(d*x + c))/(d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{3} \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))**3,x)

[Out]

Integral((a + b*tan(c + d*x))**3*sin(c + d*x), x)

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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))^3,x, algorithm="giac")

[Out]

Timed out