3.377 \(\int \sqrt{a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx\)

Optimal. Leaf size=80 \[ \frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b f}-\frac{\sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

[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 + b*Sec[e + f*x]^2
)^(3/2)/(3*b*f)

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Rubi [A]  time = 0.104652, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {4139, 446, 80, 50, 63, 208} \[ \frac{\left (a+b \sec ^2(e+f x)\right )^{3/2}}{3 b f}-\frac{\sqrt{a+b \sec ^2(e+f x)}}{f}+\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sec[e + f*x]^2]*Tan[e + f*x]^3,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 + b*Sec[e + f*x]^2
)^(3/2)/(3*b*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 80

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

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

Mathematica [F]  time = 1.07293, size = 0, normalized size = 0. \[ \int \sqrt{a+b \sec ^2(e+f x)} \tan ^3(e+f x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 0.36, size = 648, normalized size = 8.1 \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)^3,x)

[Out]

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

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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 )^{3}\,{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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.888, size = 969, normalized size = 12.11 \begin{align*} \left [\frac{3 \, \sqrt{a} b \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 ({\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, b f \cos \left (f x + e\right )^{2}}, -\frac{3 \, \sqrt{-a} b \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 ({\left (a - 3 \, b\right )} \cos \left (f x + e\right )^{2} + b\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, b f \cos \left (f x + e\right )^{2}}\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)^3,x, algorithm="fricas")

[Out]

[1/24*(3*sqrt(a)*b*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*((a - 3*b)*cos(f*x + e)
^2 + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^2), -1/12*(3*sqrt(-a)*b*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*co
s(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2))*cos(f*x + e)^2 - 4*((a - 3*b)*cos(f*x + e)^2 + b)*sqrt((a*cos(
f*x + e)^2 + b)/cos(f*x + e)^2))/(b*f*cos(f*x + e)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec ^{2}{\left (e + f x \right )}} \tan ^{3}{\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)**3,x)

[Out]

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

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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 )^{3}\,{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)^3,x, algorithm="giac")

[Out]

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