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

Optimal. Leaf size=56 \[ \frac {\sqrt {a+b \sec ^2(e+f x)}}{b f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \]

[Out]

arctanh((a+b*sec(f*x+e)^2)^(1/2)/a^(1/2))/f/a^(1/2)+(a+b*sec(f*x+e)^2)^(1/2)/b/f

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Rubi [A]  time = 0.10, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4139, 446, 80, 63, 208} \[ \frac {\sqrt {a+b \sec ^2(e+f x)}}{b f}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sec ^2(e+f x)}}{\sqrt {a}}\right )}{\sqrt {a} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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 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])

Rubi steps

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

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Mathematica [F]  time = 1.50, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.69, size = 328, normalized size = 5.86 \[ \left [\frac {\sqrt {a} b \log \left (128 \, a^{4} \cos \left (f x + e\right )^{8} + 256 \, a^{3} b \cos \left (f x + e\right )^{6} + 160 \, a^{2} b^{2} \cos \left (f x + e\right )^{4} + 32 \, a b^{3} \cos \left (f x + e\right )^{2} + b^{4} + 8 \, {\left (16 \, a^{3} \cos \left (f x + e\right )^{8} + 24 \, a^{2} b \cos \left (f x + e\right )^{6} + 10 \, a b^{2} \cos \left (f x + e\right )^{4} + b^{3} \cos \left (f x + e\right )^{2}\right )} \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}\right ) + 8 \, a \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{8 \, a b f}, -\frac {\sqrt {-a} b \arctan \left (\frac {{\left (8 \, a^{2} \cos \left (f x + e\right )^{4} + 8 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, a^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b \cos \left (f x + e\right )^{2} + a b^{2}\right )}}\right ) - 4 \, a \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, a b f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(a)*b*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*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e
)^2))/(a*b*f), -1/4*(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*cos(f*x + e)^4 + 3*a^2*b*cos(f*x + e)^2 + a*b^2)) - 4*a*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*b*f)]

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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(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)-2/f*8*(1/4*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)-sqrt(a+b)+sqrt(a
*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))/(-2*s
qrt(a+b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+e
xp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))-(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*t
an((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^2-a+3*b)/sign(tan((f*x+exp(1))/
2)^2-1)-1/8*atan(1/2*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)-sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1
))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))/sqrt(-a))/sqrt(-a)/sign(tan((f*x+exp(1))/2)^
2-1))

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maple [B]  time = 1.50, size = 303, normalized size = 5.41 \[ -\frac {\left (\sin ^{2}\left (f x +e \right )\right ) \left (a^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \cos \left (f x +e \right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+4 a \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) b +\cos \left (f x +e \right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \ln \left (4 \cos \left (f x +e \right ) \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+4 a \cos \left (f x +e \right )+4 \sqrt {a}\, \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\right ) b +\sqrt {a}\, b \right )}{f \cos \left (f x +e \right )^{2} \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )-1\right ) b \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^3/(a + b/cos(e + f*x)^2)^(1/2),x)

[Out]

int(tan(e + f*x)^3/(a + b/cos(e + f*x)^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**3/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

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

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