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

Optimal. Leaf size=88 \[ \frac{(2 a+3 b) \log (1-\sin (c+d x))}{4 d}+\frac{(2 a-3 b) \log (\sin (c+d x)+1)}{4 d}+\frac{\tan ^2(c+d x) (a+b \sin (c+d x))}{2 d}+\frac{3 b \sin (c+d x)}{2 d} \]

[Out]

((2*a + 3*b)*Log[1 - Sin[c + d*x]])/(4*d) + ((2*a - 3*b)*Log[1 + Sin[c + d*x]])/(4*d) + (3*b*Sin[c + d*x])/(2*
d) + ((a + b*Sin[c + d*x])*Tan[c + d*x]^2)/(2*d)

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Rubi [A]  time = 0.0771489, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2721, 819, 774, 633, 31} \[ \frac{(2 a+3 b) \log (1-\sin (c+d x))}{4 d}+\frac{(2 a-3 b) \log (\sin (c+d x)+1)}{4 d}+\frac{\tan ^2(c+d x) (a+b \sin (c+d x))}{2 d}+\frac{3 b \sin (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a + 3*b)*Log[1 - Sin[c + d*x]])/(4*d) + ((2*a - 3*b)*Log[1 + Sin[c + d*x]])/(4*d) + (3*b*Sin[c + d*x])/(2*
d) + ((a + b*Sin[c + d*x])*Tan[c + d*x]^2)/(2*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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

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

Mathematica [A]  time = 0.119188, size = 77, normalized size = 0.88 \[ \frac{a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d}-\frac{b \sin (c+d x) \tan ^2(c+d x)}{d}-\frac{3 b \left (\tanh ^{-1}(\sin (c+d x))-\tan (c+d x) \sec (c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.032, size = 96, normalized size = 1.1 \begin{align*}{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}a}{2\,d}}+{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{b \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,b\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,b\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.90057, size = 99, normalized size = 1.12 \begin{align*} \frac{{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 4 \, b \sin \left (d x + c\right ) - \frac{2 \,{\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68026, size = 236, normalized size = 2.68 \begin{align*} \frac{{\left (2 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (2 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right ) + 2 \, a}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*a - 3*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*a + 3*b)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*
(2*b*cos(d*x + c)^2 + b)*sin(d*x + c) + 2*a)/(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 \sin{\left (c + d x \right )}\right ) \tan ^{3}{\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))*tan(d*x+c)**3,x)

[Out]

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

[Out]

Timed out