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

Optimal. Leaf size=77 \[ \frac{a^2 \log (\cos (e+f x))}{f}+\frac{b (2 a-b) \sec ^4(e+f x)}{4 f}+\frac{a (a-2 b) \sec ^2(e+f x)}{2 f}+\frac{b^2 \sec ^6(e+f x)}{6 f} \]

[Out]

(a^2*Log[Cos[e + f*x]])/f + (a*(a - 2*b)*Sec[e + f*x]^2)/(2*f) + ((2*a - b)*b*Sec[e + f*x]^4)/(4*f) + (b^2*Sec
[e + f*x]^6)/(6*f)

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Rubi [A]  time = 0.0840397, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4138, 446, 76} \[ \frac{a^2 \log (\cos (e+f x))}{f}+\frac{b (2 a-b) \sec ^4(e+f x)}{4 f}+\frac{a (a-2 b) \sec ^2(e+f x)}{2 f}+\frac{b^2 \sec ^6(e+f x)}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Log[Cos[e + f*x]])/f + (a*(a - 2*b)*Sec[e + f*x]^2)/(2*f) + ((2*a - b)*b*Sec[e + f*x]^4)/(4*f) + (b^2*Sec
[e + f*x]^6)/(6*f)

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^3(e+f x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right ) \left (b+a x^2\right )^2}{x^7} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(1-x) (b+a x)^2}{x^4} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b^2}{x^4}+\frac{(2 a-b) b}{x^3}+\frac{a (a-2 b)}{x^2}-\frac{a^2}{x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{a^2 \log (\cos (e+f x))}{f}+\frac{a (a-2 b) \sec ^2(e+f x)}{2 f}+\frac{(2 a-b) b \sec ^4(e+f x)}{4 f}+\frac{b^2 \sec ^6(e+f x)}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.277409, size = 107, normalized size = 1.39 \[ \frac{\cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \left (12 a^2 \log (\cos (e+f x))+3 b (2 a-b) \sec ^4(e+f x)+6 a (a-2 b) \sec ^2(e+f x)+2 b^2 \sec ^6(e+f x)\right )}{3 f (a \cos (2 e+2 f x)+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2*(12*a^2*Log[Cos[e + f*x]] + 6*a*(a - 2*b)*Sec[e + f*x]^2 + 3*(2*a - b
)*b*Sec[e + f*x]^4 + 2*b^2*Sec[e + f*x]^6))/(3*f*(a + 2*b + a*Cos[2*e + 2*f*x])^2)

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Maple [A]  time = 0.056, size = 103, normalized size = 1.3 \begin{align*}{\frac{{a}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{2\,f}}+{\frac{{a}^{2}\ln \left ( \cos \left ( fx+e \right ) \right ) }{f}}+{\frac{ab \left ( \sin \left ( fx+e \right ) \right ) ^{4}}{2\,f \left ( \cos \left ( fx+e \right ) \right ) ^{4}}}+{\frac{{b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{4}}{6\,f \left ( \cos \left ( fx+e \right ) \right ) ^{6}}}+{\frac{{b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{4}}{12\,f \left ( \cos \left ( fx+e \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*a^2*tan(f*x+e)^2+a^2*ln(cos(f*x+e))/f+1/2/f*a*b*sin(f*x+e)^4/cos(f*x+e)^4+1/6/f*b^2*sin(f*x+e)^4/cos(f*x
+e)^6+1/12/f*b^2*sin(f*x+e)^4/cos(f*x+e)^4

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Maxima [A]  time = 1.00919, size = 154, normalized size = 2. \begin{align*} \frac{6 \, a^{2} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac{6 \,{\left (a^{2} - 2 \, a b\right )} \sin \left (f x + e\right )^{4} - 3 \,{\left (4 \, a^{2} - 6 \, a b - b^{2}\right )} \sin \left (f x + e\right )^{2} + 6 \, a^{2} - 6 \, a b - b^{2}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a^2*log(sin(f*x + e)^2 - 1) - (6*(a^2 - 2*a*b)*sin(f*x + e)^4 - 3*(4*a^2 - 6*a*b - b^2)*sin(f*x + e)^2
+ 6*a^2 - 6*a*b - b^2)/(sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(f*x + e)^2 - 1))/f

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Fricas [A]  time = 0.541965, size = 192, normalized size = 2.49 \begin{align*} \frac{12 \, a^{2} \cos \left (f x + e\right )^{6} \log \left (-\cos \left (f x + e\right )\right ) + 6 \,{\left (a^{2} - 2 \, a b\right )} \cos \left (f x + e\right )^{4} + 3 \,{\left (2 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, b^{2}}{12 \, f \cos \left (f x + e\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*a^2*cos(f*x + e)^6*log(-cos(f*x + e)) + 6*(a^2 - 2*a*b)*cos(f*x + e)^4 + 3*(2*a*b - b^2)*cos(f*x + e)
^2 + 2*b^2)/(f*cos(f*x + e)^6)

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Sympy [A]  time = 7.92297, size = 128, normalized size = 1.66 \begin{align*} \begin{cases} - \frac{a^{2} \log{\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac{a^{2} \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac{a b \tan ^{2}{\left (e + f x \right )} \sec ^{2}{\left (e + f x \right )}}{2 f} - \frac{a b \sec ^{2}{\left (e + f x \right )}}{2 f} + \frac{b^{2} \tan ^{2}{\left (e + f x \right )} \sec ^{4}{\left (e + f x \right )}}{6 f} - \frac{b^{2} \sec ^{4}{\left (e + f x \right )}}{12 f} & \text{for}\: f \neq 0 \\x \left (a + b \sec ^{2}{\left (e \right )}\right )^{2} \tan ^{3}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*log(tan(e + f*x)**2 + 1)/(2*f) + a**2*tan(e + f*x)**2/(2*f) + a*b*tan(e + f*x)**2*sec(e + f*x
)**2/(2*f) - a*b*sec(e + f*x)**2/(2*f) + b**2*tan(e + f*x)**2*sec(e + f*x)**4/(6*f) - b**2*sec(e + f*x)**4/(12
*f), Ne(f, 0)), (x*(a + b*sec(e)**2)**2*tan(e)**3, True))

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Giac [B]  time = 1.85568, size = 560, normalized size = 7.27 \begin{align*} -\frac{6 \, a^{2} \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right ) - 6 \, a^{2} \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2\right ) + \frac{11 \, a^{2}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{3} + 90 \, a^{2}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )}^{2} + 228 \, a^{2}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 96 \, a b{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 48 \, b^{2}{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 184 \, a^{2} - 192 \, a b + 32 \, b^{2}}{{\left (\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}^{3}}}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*a^2*log(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2) - 6*a^2*l
og(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2) + (11*a^2*((cos(f*x + e
) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^3 + 90*a^2*((cos(f*x + e) + 1)/(cos(f*x + e
) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1))^2 + 228*a^2*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x
+ e) - 1)/(cos(f*x + e) + 1)) - 96*a*b*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x +
e) + 1)) - 48*b^2*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1)) + 184*a^2 -
192*a*b + 32*b^2)/((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 2)^3)/f