3.402 \(\int \frac{\tan ^5(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=89 \[ \frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}-\frac{(a+2 b) \sqrt{a+b \sec ^2(e+f x)}}{b^2 f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f} \]

[Out]

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

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Rubi [A]  time = 0.133072, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4139, 446, 88, 63, 208} \[ \frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}-\frac{(a+2 b) \sqrt{a+b \sec ^2(e+f x)}}{b^2 f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

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

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, 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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\tan ^5(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-1+x)^2}{x \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{-a-2 b}{b \sqrt{a+b x}}+\frac{1}{x \sqrt{a+b x}}+\frac{\sqrt{a+b x}}{b}\right ) \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a+2 b) \sqrt{a+b \sec ^2(e+f x)}}{b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a+2 b) \sqrt{a+b \sec ^2(e+f x)}}{b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sec ^2(e+f x)}\right )}{b f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} f}-\frac{(a+2 b) \sqrt{a+b \sec ^2(e+f x)}}{b^2 f}+\frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}\\ \end{align*}

Mathematica [F]  time = 2.1216, size = 0, normalized size = 0. \[ \int \frac{\tan ^5(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[e + f*x]^5/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

Integrate[Tan[e + f*x]^5/Sqrt[a + b*Sec[e + f*x]^2], x]

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Maple [B]  time = 0.44, size = 358, normalized size = 4. \begin{align*}{\frac{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3\,f{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{4} \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}-1 \right ) } \left ( 2\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{5/2}+3\,\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\ln \left ( 4\,\cos \left ( fx+e \right ) \sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\sqrt{a}+4\,a\cos \left ( fx+e \right ) +4\,\sqrt{a}\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}{b}^{2}+6\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{3/2}b+3\,\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\ln \left ( 4\,\cos \left ( fx+e \right ) \sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}}\sqrt{a}+4\,a\cos \left ( fx+e \right ) +4\,\sqrt{a}\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( 1+\cos \left ( fx+e \right ) \right ) ^{2}}}} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}{b}^{2}+ \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{{\frac{3}{2}}}b+6\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sqrt{a}{b}^{2}-\sqrt{a}{b}^{2} \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

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

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Maxima [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(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.07901, size = 1006, normalized size = 11.3 \begin{align*} \left [\frac{3 \, \sqrt{a} b^{2} \cos \left (f x + e\right )^{2} \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} - 8 \,{\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) - 8 \,{\left (2 \,{\left (a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, a b^{2} f \cos \left (f x + e\right )^{2}}, \frac{3 \, \sqrt{-a} b^{2} \arctan \left (\frac{{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \,{\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) \cos \left (f x + e\right )^{2} - 4 \,{\left (2 \,{\left (a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} - a b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, a b^{2} f \cos \left (f x + e\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/24*(3*sqrt(a)*b^2*cos(f*x + e)^2*log(128*a^4*cos(f*x + e)^8 + 256*a^3*b*cos(f*x + e)^6 + 160*a^2*b^2*cos(f*
x + e)^4 + 32*a*b^3*cos(f*x + e)^2 + b^4 - 8*(16*a^3*cos(f*x + e)^8 + 24*a^2*b*cos(f*x + e)^6 + 10*a*b^2*cos(f
*x + e)^4 + b^3*cos(f*x + e)^2)*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)) - 8*(2*(a^2 + 3*a*b)*cos(
f*x + e)^2 - a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*b^2*f*cos(f*x + e)^2), 1/12*(3*sqrt(-a)*b^2*
arctan(1/4*(8*a^2*cos(f*x + e)^4 + 8*a*b*cos(f*x + e)^2 + b^2)*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x +
e)^2)/(2*a^3*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2))*cos(f*x + e)^2 - 4*(2*(a^2 + 3*a*b)*cos(f*x + e
)^2 - a*b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*b^2*f*cos(f*x + e)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (e + f x \right )}}{\sqrt{a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(tan(e + f*x)**5/sqrt(a + b*sec(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{5}}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^5/sqrt(b*sec(f*x + e)^2 + a), x)