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

Optimal. Leaf size=129 \[ \frac {\left (8 a^2+12 a b+5 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac {\left (8 a^2+12 a b+5 b^2\right ) \tan (e+f x) \sec (e+f x)}{16 f}+\frac {b (8 a+5 b) \tan (e+f x) \sec ^3(e+f x)}{24 f}+\frac {b \tan (e+f x) \sec ^5(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{6 f} \]

[Out]

1/16*(8*a^2+12*a*b+5*b^2)*arctanh(sin(f*x+e))/f+1/16*(8*a^2+12*a*b+5*b^2)*sec(f*x+e)*tan(f*x+e)/f+1/24*b*(8*a+
5*b)*sec(f*x+e)^3*tan(f*x+e)/f+1/6*b*sec(f*x+e)^5*(a+b-a*sin(f*x+e)^2)*tan(f*x+e)/f

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Rubi [A]  time = 0.13, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4147, 413, 385, 199, 206} \[ \frac {\left (8 a^2+12 a b+5 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac {\left (8 a^2+12 a b+5 b^2\right ) \tan (e+f x) \sec (e+f x)}{16 f}+\frac {b (8 a+5 b) \tan (e+f x) \sec ^3(e+f x)}{24 f}+\frac {b \tan (e+f x) \sec ^5(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 + 12*a*b + 5*b^2)*ArcTanh[Sin[e + f*x]])/(16*f) + ((8*a^2 + 12*a*b + 5*b^2)*Sec[e + f*x]*Tan[e + f*x])
/(16*f) + (b*(8*a + 5*b)*Sec[e + f*x]^3*Tan[e + f*x])/(24*f) + (b*Sec[e + f*x]^5*(a + b - a*Sin[e + f*x]^2)*Ta
n[e + f*x])/(6*f)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 385

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

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 4147

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

Rubi steps

\begin {align*} \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-a x^2\right )^2}{\left (1-x^2\right )^4} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {b \sec ^5(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{6 f}-\frac {\operatorname {Subst}\left (\int \frac {-(a+b) (6 a+5 b)+3 a (2 a+b) x^2}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{6 f}\\ &=\frac {b (8 a+5 b) \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac {b \sec ^5(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{6 f}+\frac {\left (8 a^2+12 a b+5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{8 f}\\ &=\frac {\left (8 a^2+12 a b+5 b^2\right ) \sec (e+f x) \tan (e+f x)}{16 f}+\frac {b (8 a+5 b) \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac {b \sec ^5(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{6 f}+\frac {\left (8 a^2+12 a b+5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{16 f}\\ &=\frac {\left (8 a^2+12 a b+5 b^2\right ) \tanh ^{-1}(\sin (e+f x))}{16 f}+\frac {\left (8 a^2+12 a b+5 b^2\right ) \sec (e+f x) \tan (e+f x)}{16 f}+\frac {b (8 a+5 b) \sec ^3(e+f x) \tan (e+f x)}{24 f}+\frac {b \sec ^5(e+f x) \left (a+b-a \sin ^2(e+f x)\right ) \tan (e+f x)}{6 f}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 94, normalized size = 0.73 \[ \frac {3 \left (8 a^2+12 a b+5 b^2\right ) \tanh ^{-1}(\sin (e+f x))+\tan (e+f x) \sec (e+f x) \left (3 \left (8 a^2+12 a b+5 b^2\right )+2 b (12 a+5 b) \sec ^2(e+f x)+8 b^2 \sec ^4(e+f x)\right )}{48 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(8*a^2 + 12*a*b + 5*b^2)*ArcTanh[Sin[e + f*x]] + Sec[e + f*x]*(3*(8*a^2 + 12*a*b + 5*b^2) + 2*b*(12*a + 5*b
)*Sec[e + f*x]^2 + 8*b^2*Sec[e + f*x]^4)*Tan[e + f*x])/(48*f)

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fricas [A]  time = 0.49, size = 143, normalized size = 1.11 \[ \frac {3 \, {\left (8 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (8 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (3 \, {\left (8 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (12 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 8 \, b^{2}\right )} \sin \left (f x + e\right )}{96 \, f \cos \left (f x + e\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(8*a^2 + 12*a*b + 5*b^2)*cos(f*x + e)^6*log(sin(f*x + e) + 1) - 3*(8*a^2 + 12*a*b + 5*b^2)*cos(f*x + e
)^6*log(-sin(f*x + e) + 1) + 2*(3*(8*a^2 + 12*a*b + 5*b^2)*cos(f*x + e)^4 + 2*(12*a*b + 5*b^2)*cos(f*x + e)^2
+ 8*b^2)*sin(f*x + e))/(f*cos(f*x + e)^6)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*(-(8*a^2+12*a*b+5*b^2)/64*ln(abs(sin(f*x+exp(1))-1))+(8*a^
2+12*a*b+5*b^2)/64*ln(abs(sin(f*x+exp(1))+1))+(-24*sin(f*x+exp(1))^5*a^2-36*sin(f*x+exp(1))^5*a*b-15*sin(f*x+e
xp(1))^5*b^2+48*sin(f*x+exp(1))^3*a^2+96*sin(f*x+exp(1))^3*a*b+40*sin(f*x+exp(1))^3*b^2-24*sin(f*x+exp(1))*a^2
-60*sin(f*x+exp(1))*a*b-33*sin(f*x+exp(1))*b^2)*1/96/(sin(f*x+exp(1))^2-1)^3)

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maple [A]  time = 1.31, size = 191, normalized size = 1.48 \[ \frac {a^{2} \tan \left (f x +e \right ) \sec \left (f x +e \right )}{2 f}+\frac {a^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2 f}+\frac {a b \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{2 f}+\frac {3 a b \tan \left (f x +e \right ) \sec \left (f x +e \right )}{4 f}+\frac {3 a b \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{4 f}+\frac {b^{2} \tan \left (f x +e \right ) \left (\sec ^{5}\left (f x +e \right )\right )}{6 f}+\frac {5 b^{2} \tan \left (f x +e \right ) \left (\sec ^{3}\left (f x +e \right )\right )}{24 f}+\frac {5 b^{2} \sec \left (f x +e \right ) \tan \left (f x +e \right )}{16 f}+\frac {5 b^{2} \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{16 f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*a^2*tan(f*x+e)*sec(f*x+e)+1/2/f*a^2*ln(sec(f*x+e)+tan(f*x+e))+1/2/f*a*b*tan(f*x+e)*sec(f*x+e)^3+3/4/f*a*
b*tan(f*x+e)*sec(f*x+e)+3/4/f*a*b*ln(sec(f*x+e)+tan(f*x+e))+1/6/f*b^2*tan(f*x+e)*sec(f*x+e)^5+5/24/f*b^2*tan(f
*x+e)*sec(f*x+e)^3+5/16*b^2*sec(f*x+e)*tan(f*x+e)/f+5/16/f*b^2*ln(sec(f*x+e)+tan(f*x+e))

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maxima [A]  time = 0.33, size = 166, normalized size = 1.29 \[ \frac {3 \, {\left (8 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 3 \, {\left (8 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} \log \left (\sin \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (8 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} \sin \left (f x + e\right )^{5} - 8 \, {\left (6 \, a^{2} + 12 \, a b + 5 \, b^{2}\right )} \sin \left (f x + e\right )^{3} + 3 \, {\left (8 \, a^{2} + 20 \, a b + 11 \, b^{2}\right )} \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{6} - 3 \, \sin \left (f x + e\right )^{4} + 3 \, \sin \left (f x + e\right )^{2} - 1}}{96 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(8*a^2 + 12*a*b + 5*b^2)*log(sin(f*x + e) + 1) - 3*(8*a^2 + 12*a*b + 5*b^2)*log(sin(f*x + e) - 1) - 2*
(3*(8*a^2 + 12*a*b + 5*b^2)*sin(f*x + e)^5 - 8*(6*a^2 + 12*a*b + 5*b^2)*sin(f*x + e)^3 + 3*(8*a^2 + 20*a*b + 1
1*b^2)*sin(f*x + e))/(sin(f*x + e)^6 - 3*sin(f*x + e)^4 + 3*sin(f*x + e)^2 - 1))/f

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mupad [B]  time = 4.55, size = 134, normalized size = 1.04 \[ \frac {\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\,\left (\frac {a^2}{2}+\frac {3\,a\,b}{4}+\frac {5\,b^2}{16}\right )}{f}-\frac {\left (\frac {a^2}{2}+\frac {3\,a\,b}{4}+\frac {5\,b^2}{16}\right )\,{\sin \left (e+f\,x\right )}^5+\left (-a^2-2\,a\,b-\frac {5\,b^2}{6}\right )\,{\sin \left (e+f\,x\right )}^3+\left (\frac {a^2}{2}+\frac {5\,a\,b}{4}+\frac {11\,b^2}{16}\right )\,\sin \left (e+f\,x\right )}{f\,\left ({\sin \left (e+f\,x\right )}^6-3\,{\sin \left (e+f\,x\right )}^4+3\,{\sin \left (e+f\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/cos(e + f*x)^2)^2/cos(e + f*x)^3,x)

[Out]

(atanh(sin(e + f*x))*((3*a*b)/4 + a^2/2 + (5*b^2)/16))/f - (sin(e + f*x)*((5*a*b)/4 + a^2/2 + (11*b^2)/16) - s
in(e + f*x)^3*(2*a*b + a^2 + (5*b^2)/6) + sin(e + f*x)^5*((3*a*b)/4 + a^2/2 + (5*b^2)/16))/(f*(3*sin(e + f*x)^
2 - 3*sin(e + f*x)^4 + sin(e + f*x)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \sec ^{3}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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