3.23 \(\int \sin ^3(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=122 \[ \frac{a^2 \cos ^3(c+d x)}{3 d}-\frac{a^2 \cos (c+d x)}{d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^2 \cos ^3(c+d x)}{3 d}+\frac{2 b^2 \cos (c+d x)}{d}+\frac{b^2 \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.130693, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3517, 2633, 2592, 302, 206, 2590, 270} \[ \frac{a^2 \cos ^3(c+d x)}{3 d}-\frac{a^2 \cos (c+d x)}{d}-\frac{2 a b \sin ^3(c+d x)}{3 d}-\frac{2 a b \sin (c+d x)}{d}+\frac{2 a b \tanh ^{-1}(\sin (c+d x))}{d}-\frac{b^2 \cos ^3(c+d x)}{3 d}+\frac{2 b^2 \cos (c+d x)}{d}+\frac{b^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

(2*a*b*ArcTanh[Sin[c + d*x]])/d - (a^2*Cos[c + d*x])/d + (2*b^2*Cos[c + d*x])/d + (a^2*Cos[c + d*x]^3)/(3*d) -
(b^2*Cos[c + d*x]^3)/(3*d) + (b^2*Sec[c + d*x])/d - (2*a*b*Sin[c + d*x])/d - (2*a*b*Sin[c + d*x]^3)/(3*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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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

Mathematica [A]  time = 1.02334, size = 152, normalized size = 1.25 \[ \frac{\sec (c+d x) \left (\left (20 b^2-8 a^2\right ) \cos (2 (c+d x))+\left (a^2-b^2\right ) \cos (4 (c+d x))-9 a^2-28 a b \sin (2 (c+d x))+2 a b \sin (4 (c+d x))-48 a b \cos (c+d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+48 a b \cos (c+d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+45 b^2\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3*(a + b*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]*(-9*a^2 + 45*b^2 + (-8*a^2 + 20*b^2)*Cos[2*(c + d*x)] + (a^2 - b^2)*Cos[4*(c + d*x)] - 48*a*b*Co
s[c + d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 48*a*b*Cos[c + d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]] - 28*a*b*Sin[2*(c + d*x)] + 2*a*b*Sin[4*(c + d*x)]))/(24*d)

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Maple [A]  time = 0.041, size = 167, normalized size = 1.4 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}+{\frac{8\,{b}^{2}\cos \left ( dx+c \right ) }{3\,d}}+{\frac{{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{4\,{b}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3\,d}}-{\frac{2\,ab \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-2\,{\frac{ab\sin \left ( dx+c \right ) }{d}}+2\,{\frac{ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}}{3\,d}}-{\frac{2\,{a}^{2}\cos \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3*(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.973981, size = 140, normalized size = 1.15 \begin{align*} \frac{{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} -{\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a b -{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b^{2}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.99257, size = 321, normalized size = 2.63 \begin{align*} \frac{{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2} + 2 \,{\left (a b \cos \left (d x + c\right )^{3} - 4 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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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 )^{2} \sin ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**3*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sin(c + d*x)**3, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^3*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError