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

Optimal. Leaf size=111 \[ \frac{\left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

[Out]

-((Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/f) + Sqrt[a + b*Sec[e + f*x]^2]/f - ((a + 2*b)*(a + b*
Sec[e + f*x]^2)^(3/2))/(3*b^2*f) + (a + b*Sec[e + f*x]^2)^(5/2)/(5*b^2*f)

________________________________________________________________________________________

Rubi [A]  time = 0.141112, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4139, 446, 88, 50, 63, 208} \[ \frac{\left (a+b \sec ^2(e+f x)\right )^{5/2}}{5 b^2 f}-\frac{(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b^2 f}+\frac{\sqrt{a+b \sec ^2(e+f x)}}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[e + f*x]^2]/Sqrt[a]])/f) + Sqrt[a + b*Sec[e + f*x]^2]/f - ((a + 2*b)*(a + b*
Sec[e + f*x]^2)^(3/2))/(3*b^2*f) + (a + b*Sec[e + f*x]^2)^(5/2)/(5*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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [F]  time = 2.41833, size = 0, normalized size = 0. \[ \int \sqrt{a+b \sec ^2(e+f x)} \tan ^5(e+f x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.439, size = 924, normalized size = 8.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30/f/(a+b)^(1/2)/b^2*4^(1/2)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)*(-1+cos(f*x+e))*(15*cos(f*x+e)^5*(a+b)^
(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))*a^(1/2)*b^2+15*cos(f*x+e)^5*ln(-2/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)
*(a+b)^(1/2)+b)/sin(f*x+e)^2)*a*b^2-15*cos(f*x+e)^5*ln(-4/(a+b)^(1/2)*(-1+cos(f*x+e))*(cos(f*x+e)*((b+a*cos(f*
x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/
2)+b)/sin(f*x+e)^2)*a*b^2+2*cos(f*x+e)^5*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+10*cos(f*
x+e)^5*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-15*cos(f*x+e)^5*(a+b)^(1/2)*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2+2*cos(f*x+e)^4*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a^2+10
*cos(f*x+e)^4*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b-15*cos(f*x+e)^4*(a+b)^(1/2)*((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2-cos(f*x+e)^3*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a
*b+10*cos(f*x+e)^3*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2-cos(f*x+e)^2*(a+b)^(1/2)*((b+a*
cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a*b+10*cos(f*x+e)^2*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1
/2)*b^2-3*cos(f*x+e)*(a+b)^(1/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2-3*(a+b)^(1/2)*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(1/2)*b^2)/sin(f*x+e)^2/cos(f*x+e)^4/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 6.67986, size = 1123, normalized size = 10.12 \begin{align*} \left [\frac{15 \, \sqrt{a} b^{2} \cos \left (f x + e\right )^{4} \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 ({\left (2 \, a^{2} + 10 \, a b - 15 \, b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, b^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{120 \, b^{2} f \cos \left (f x + e\right )^{4}}, \frac{15 \, \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 )^{4} - 4 \,{\left ({\left (2 \, a^{2} + 10 \, a b - 15 \, b^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, b^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{60 \, b^{2} f \cos \left (f x + e\right )^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/120*(15*sqrt(a)*b^2*cos(f*x + e)^4*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 + 10*a*b -
15*b^2)*cos(f*x + e)^4 - (a*b - 10*b^2)*cos(f*x + e)^2 - 3*b^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(
b^2*f*cos(f*x + e)^4), 1/60*(15*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)*sq
rt(-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))*co
s(f*x + e)^4 - 4*((2*a^2 + 10*a*b - 15*b^2)*cos(f*x + e)^4 - (a*b - 10*b^2)*cos(f*x + e)^2 - 3*b^2)*sqrt((a*co
s(f*x + e)^2 + b)/cos(f*x + e)^2))/(b^2*f*cos(f*x + e)^4)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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