3.69 \(\int \frac{\sin ^3(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=133 \[ \frac{\cos ^3(e+f x)}{3 f (a-b)^2}-\frac{(a+b) \cos (e+f x)}{f (a-b)^3}-\frac{a b \sec (e+f x)}{2 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 f (a-b)^{7/2}} \]

[Out]

-(Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*(a - b)^(7/2)*f) - ((a + b)*Cos[e + f*x])
/((a - b)^3*f) + Cos[e + f*x]^3/(3*(a - b)^2*f) - (a*b*Sec[e + f*x])/(2*(a - b)^3*f*(a - b + b*Sec[e + f*x]^2)
)

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Rubi [A]  time = 0.180449, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3664, 456, 1261, 205} \[ \frac{\cos ^3(e+f x)}{3 f (a-b)^2}-\frac{(a+b) \cos (e+f x)}{f (a-b)^3}-\frac{a b \sec (e+f x)}{2 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 f (a-b)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(2*(a - b)^(7/2)*f) - ((a + b)*Cos[e + f*x])
/((a - b)^3*f) + Cos[e + f*x]^3/(3*(a - b)^2*f) - (a*b*Sec[e + f*x])/(2*(a - b)^3*f*(a - b + b*Sec[e + f*x]^2)
)

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 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{2}{(a-b) b}-\frac{2 a x^2}{(a-b)^2 b}+\frac{a x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{b \operatorname{Subst}\left (\int \left (\frac{2}{(a-b)^2 b x^4}+\frac{2 (a+b)}{b (-a+b)^3 x^2}+\frac{3 a+2 b}{(a-b)^3 \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=-\frac{(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac{\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{(b (3 a+2 b)) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 (a-b)^3 f}\\ &=-\frac{\sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{2 (a-b)^{7/2} f}-\frac{(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac{\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac{a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.98622, size = 182, normalized size = 1.37 \[ \frac{-\frac{\cos (e+f x) \left (\frac{12 a b}{(a-b) \cos (2 (e+f x))+a+b}+9 a+15 b\right )+(b-a) \cos (3 (e+f x))}{(a-b)^3}+\frac{6 \sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{a-b}-\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{7/2}}+\frac{6 \sqrt{b} (3 a+2 b) \tan ^{-1}\left (\frac{\sqrt{a-b}+\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{7/2}}}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) + (6*Sqrt[b]*(
3*a + 2*b)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) - (Cos[e + f*x]*(9*a + 15*b
+ (12*a*b)/(a + b + (a - b)*Cos[2*(e + f*x)])) + (-a + b)*Cos[3*(e + f*x)])/(a - b)^3)/(12*f)

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Maple [B]  time = 0.078, size = 269, normalized size = 2. \begin{align*}{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{\cos \left ( fx+e \right ) a}{f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{\cos \left ( fx+e \right ) b}{f \left ({a}^{2}-2\,ab+{b}^{2} \right ) \left ( a-b \right ) }}-{\frac{ab\cos \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{3} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) }}+{\frac{3\,ab}{2\,f \left ( a-b \right ) ^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}}+{\frac{{b}^{2}}{f \left ( a-b \right ) ^{3}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \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)^2,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.20797, size = 1015, normalized size = 7.63 \begin{align*} \left [\frac{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \,{\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt{-\frac{b}{a - b}} \log \left (-\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (a - b\right )} \sqrt{-\frac{b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 6 \,{\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \,{\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}, \frac{2 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 2 \,{\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt{\frac{b}{a - b}} \arctan \left (-\frac{{\left (a - b\right )} \sqrt{\frac{b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - 3 \,{\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{6 \,{\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} +{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}\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)^2,x, algorithm="fricas")

[Out]

[1/12*(4*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 4*(3*a^2 - a*b - 2*b^2)*cos(f*x + e)^3 - 3*((3*a^2 - a*b - 2*b^2
)*cos(f*x + e)^2 + 3*a*b + 2*b^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*c
os(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) - 6*(3*a*b + 2*b^2)*cos(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 -
4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f), 1/6*(2*(a^2 - 2*a*b + b^2)*cos(f*x
+ e)^5 - 2*(3*a^2 - a*b - 2*b^2)*cos(f*x + e)^3 - 3*((3*a^2 - a*b - 2*b^2)*cos(f*x + e)^2 + 3*a*b + 2*b^2)*sqr
t(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) - 3*(3*a*b + 2*b^2)*cos(f*x + e))/((a^4 - 4*a^3*b
+ 6*a^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f)]

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Sympy [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(f*x+e)**3/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.38939, size = 497, normalized size = 3.74 \begin{align*} \frac{a^{4} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a^{3} b f^{11} \cos \left (f x + e\right )^{3} + 6 \, a^{2} b^{2} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a b^{3} f^{11} \cos \left (f x + e\right )^{3} + b^{4} f^{11} \cos \left (f x + e\right )^{3} - 3 \, a^{4} f^{11} \cos \left (f x + e\right ) + 6 \, a^{3} b f^{11} \cos \left (f x + e\right ) - 6 \, a b^{3} f^{11} \cos \left (f x + e\right ) + 3 \, b^{4} f^{11} \cos \left (f x + e\right )}{3 \,{\left (a^{6} f^{12} - 6 \, a^{5} b f^{12} + 15 \, a^{4} b^{2} f^{12} - 20 \, a^{3} b^{3} f^{12} + 15 \, a^{2} b^{4} f^{12} - 6 \, a b^{5} f^{12} + b^{6} f^{12}\right )}} - \frac{a b \cos \left (f x + e\right )}{2 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}{\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )} f} + \frac{{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac{a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt{a b - b^{2}}}\right )}{2 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt{a b - b^{2}} 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)^2,x, algorithm="giac")

[Out]

1/3*(a^4*f^11*cos(f*x + e)^3 - 4*a^3*b*f^11*cos(f*x + e)^3 + 6*a^2*b^2*f^11*cos(f*x + e)^3 - 4*a*b^3*f^11*cos(
f*x + e)^3 + b^4*f^11*cos(f*x + e)^3 - 3*a^4*f^11*cos(f*x + e) + 6*a^3*b*f^11*cos(f*x + e) - 6*a*b^3*f^11*cos(
f*x + e) + 3*b^4*f^11*cos(f*x + e))/(a^6*f^12 - 6*a^5*b*f^12 + 15*a^4*b^2*f^12 - 20*a^3*b^3*f^12 + 15*a^2*b^4*
f^12 - 6*a*b^5*f^12 + b^6*f^12) - 1/2*a*b*cos(f*x + e)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a*cos(f*x + e)^2 - b*
cos(f*x + e)^2 + b)*f) + 1/2*(3*a*b + 2*b^2)*arctan((a*cos(f*x + e) - b*cos(f*x + e))/sqrt(a*b - b^2))/((a^3 -
3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b - b^2)*f)