3.31 \(\int \sin ^3(e+f x) (a+b \tan ^2(e+f x)) \, dx\)

Optimal. Leaf size=48 \[ \frac{(a-b) \cos ^3(e+f x)}{3 f}-\frac{(a-2 b) \cos (e+f x)}{f}+\frac{b \sec (e+f x)}{f} \]

[Out]

-(((a - 2*b)*Cos[e + f*x])/f) + ((a - b)*Cos[e + f*x]^3)/(3*f) + (b*Sec[e + f*x])/f

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Rubi [A]  time = 0.0470021, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3664, 448} \[ \frac{(a-b) \cos ^3(e+f x)}{3 f}-\frac{(a-2 b) \cos (e+f x)}{f}+\frac{b \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a - 2*b)*Cos[e + f*x])/f) + ((a - b)*Cos[e + f*x]^3)/(3*f) + (b*Sec[e + f*x])/f

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.0494669, size = 72, normalized size = 1.5 \[ -\frac{3 a \cos (e+f x)}{4 f}+\frac{a \cos (3 (e+f x))}{12 f}+\frac{7 b \cos (e+f x)}{4 f}-\frac{b \cos (3 (e+f x))}{12 f}+\frac{b \sec (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*Cos[e + f*x])/(4*f) + (7*b*Cos[e + f*x])/(4*f) + (a*Cos[3*(e + f*x)])/(12*f) - (b*Cos[3*(e + f*x)])/(12*
f) + (b*Sec[e + f*x])/f

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Maple [A]  time = 0.063, size = 72, normalized size = 1.5 \begin{align*}{\frac{1}{f} \left ( -{\frac{a \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+b \left ({\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{6}}{\cos \left ( fx+e \right ) }}+ \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) \cos \left ( fx+e \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^3*(a+b*tan(f*x+e)^2),x)

[Out]

1/f*(-1/3*a*(2+sin(f*x+e)^2)*cos(f*x+e)+b*(sin(f*x+e)^6/cos(f*x+e)+(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x
+e)))

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Maxima [A]  time = 1.06492, size = 59, normalized size = 1.23 \begin{align*} \frac{{\left (a - b\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (a - 2 \, b\right )} \cos \left (f x + e\right ) + \frac{3 \, b}{\cos \left (f x + e\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((a - b)*cos(f*x + e)^3 - 3*(a - 2*b)*cos(f*x + e) + 3*b/cos(f*x + e))/f

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Fricas [A]  time = 1.93285, size = 111, normalized size = 2.31 \begin{align*} \frac{{\left (a - b\right )} \cos \left (f x + e\right )^{4} - 3 \,{\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b}{3 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((a - b)*cos(f*x + e)^4 - 3*(a - 2*b)*cos(f*x + e)^2 + 3*b)/(f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right ) \sin ^{3}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**3*(a+b*tan(f*x+e)**2),x)

[Out]

Integral((a + b*tan(e + f*x)**2)*sin(e + f*x)**3, x)

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Giac [A]  time = 1.48716, size = 103, normalized size = 2.15 \begin{align*} \frac{b}{f \cos \left (f x + e\right )} + \frac{a f^{5} \cos \left (f x + e\right )^{3} - b f^{5} \cos \left (f x + e\right )^{3} - 3 \, a f^{5} \cos \left (f x + e\right ) + 6 \, b f^{5} \cos \left (f x + e\right )}{3 \, f^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b/(f*cos(f*x + e)) + 1/3*(a*f^5*cos(f*x + e)^3 - b*f^5*cos(f*x + e)^3 - 3*a*f^5*cos(f*x + e) + 6*b*f^5*cos(f*x
+ e))/f^6