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

Optimal. Leaf size=105 \[ -\frac{b \left (3 a^2+b^2\right ) \sin (c+d x)}{d}-\frac{3 a b^2 \sin ^2(c+d x)}{2 d}-\frac{(a-b)^3 \log (\sin (c+d x)+1)}{2 d}-\frac{(a+b)^3 \log (1-\sin (c+d x))}{2 d}-\frac{b^3 \sin ^3(c+d x)}{3 d} \]

[Out]

-((a + b)^3*Log[1 - Sin[c + d*x]])/(2*d) - ((a - b)^3*Log[1 + Sin[c + d*x]])/(2*d) - (b*(3*a^2 + b^2)*Sin[c +
d*x])/d - (3*a*b^2*Sin[c + d*x]^2)/(2*d) - (b^3*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.113594, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2721, 801, 633, 31} \[ -\frac{b \left (3 a^2+b^2\right ) \sin (c+d x)}{d}-\frac{3 a b^2 \sin ^2(c+d x)}{2 d}-\frac{(a-b)^3 \log (\sin (c+d x)+1)}{2 d}-\frac{(a+b)^3 \log (1-\sin (c+d x))}{2 d}-\frac{b^3 \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)^3*Log[1 - Sin[c + d*x]])/(2*d) - ((a - b)^3*Log[1 + Sin[c + d*x]])/(2*d) - (b*(3*a^2 + b^2)*Sin[c +
d*x])/d - (3*a*b^2*Sin[c + d*x]^2)/(2*d) - (b^3*Sin[c + d*x]^3)/(3*d)

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.189961, size = 90, normalized size = 0.86 \[ -\frac{6 b \left (3 a^2+b^2\right ) \sin (c+d x)+9 a b^2 \sin ^2(c+d x)+3 \left ((a-b)^3 \log (\sin (c+d x)+1)+(a+b)^3 \log (1-\sin (c+d x))\right )+2 b^3 \sin ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(3*((a + b)^3*Log[1 - Sin[c + d*x]] + (a - b)^3*Log[1 + Sin[c + d*x]]) + 6*b*(3*a^2 + b^2)*Sin[c + d*x] + 9*a
*b^2*Sin[c + d*x]^2 + 2*b^3*Sin[c + d*x]^3)/(6*d)

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Maple [A]  time = 0.046, size = 139, normalized size = 1.3 \begin{align*} -{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-3\,{\frac{{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}a{b}^{2}}{2\,d}}-3\,{\frac{a{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{3\,d}}-{\frac{{b}^{3}\sin \left ( dx+c \right ) }{d}}+{\frac{{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.70671, size = 153, normalized size = 1.46 \begin{align*} -\frac{2 \, b^{3} \sin \left (d x + c\right )^{3} + 9 \, a b^{2} \sin \left (d x + c\right )^{2} + 3 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \,{\left (3 \, a^{2} b + b^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*b^3*sin(d*x + c)^3 + 9*a*b^2*sin(d*x + c)^2 + 3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1)
+ 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*log(sin(d*x + c) - 1) + 6*(3*a^2*b + b^3)*sin(d*x + c))/d

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Fricas [A]  time = 1.60646, size = 277, normalized size = 2.64 \begin{align*} \frac{9 \, a b^{2} \cos \left (d x + c\right )^{2} - 3 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (b^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{2} b - 4 \, b^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(9*a*b^2*cos(d*x + c)^2 - 3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*log(sin(d*x + c) + 1) - 3*(a^3 + 3*a^2*b + 3*a
*b^2 + b^3)*log(-sin(d*x + c) + 1) + 2*(b^3*cos(d*x + c)^2 - 9*a^2*b - 4*b^3)*sin(d*x + c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out