\(\int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [510]

Optimal result

Integrand size = 25, antiderivative size = 134 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{5/2} f}-\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f} \]

[Out]

1/8*(8*a^2+8*a*b+3*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/f-1/8*(8*a+5*b)*sec(f*x+e)^2
*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)^2/f+1/4*sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)/f

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 91, 79, 65, 214} \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 f (a+b)^{5/2}}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 f (a+b)}-\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f (a+b)^2} \]

[In]

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

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(5/2)*f) - ((8*a + 5*b)*S
ec[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^2*f) + (Sec[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2])/(4*(a
+ b)*f)

Rule 65

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 79

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

Rule 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 214

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

Rule 3273

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2}{(1-x)^3 \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = \frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f}-\frac {\text {Subst}\left (\int \frac {\frac {1}{2} (4 a+b)+2 (a+b) x}{(1-x)^2 \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f} \\ & = -\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^2 f} \\ & = -\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f}+\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^2(e+f x)}\right )}{8 b (a+b)^2 f} \\ & = \frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{5/2} f}-\frac {(8 a+5 b) \sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac {\sec ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{4 (a+b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )+\sqrt {a+b} \sec ^2(e+f x) \left (-8 a-5 b+2 (a+b) \sec ^2(e+f x)\right ) \sqrt {a+b \sin ^2(e+f x)}}{8 (a+b)^{5/2} f} \]

[In]

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

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] + Sqrt[a + b]*Sec[e + f*x]^2*(-8*a -
5*b + 2*(a + b)*Sec[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^(5/2)*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(643\) vs. \(2(118)=236\).

Time = 1.58 (sec) , antiderivative size = 644, normalized size of antiderivative = 4.81

method	result	size
default	\(\frac {\left (8 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{4}+24 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3} b +27 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b^{2}+14 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{3}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{4}+8 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{4}+24 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3} b +27 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b^{2}+14 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{3}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{4}\right ) \left (\cos ^{4}\left (f x +e \right )\right )-2 \left (a +b \right )^{\frac {5}{2}} \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (8 a +5 b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+4 a \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}}+4 b \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {5}{2}}}{16 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f}\)	\(644\)

[In]

int(tan(f*x+e)^5/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16*((8*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^4+24*ln(2/(1+sin(f*x+e
))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b+27*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*c
os(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2+14*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*s
in(f*x+e)+a))*a*b^3+3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^4+8*ln(2/
(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^4+24*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2
)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b+27*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(
1/2)+b*sin(f*x+e)+a))*a^2*b^2+14*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*
a*b^3+3*ln(2/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^4)*cos(f*x+e)^4-2*(a+b)
^(5/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(8*a+5*b)*cos(f*x+e)^2+4*a*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)+4*b*(a+b-b
*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2))/(a+b)^(5/2)/cos(f*x+e)^4/(a^2+2*a*b+b^2)/f

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.45 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}, -\frac {{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{4} + {\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}\right ] \]

[In]

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

[Out]

[1/16*((8*a^2 + 8*a*b + 3*b^2)*sqrt(a + b)*cos(f*x + e)^4*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a
+ b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 2*((8*a^2 + 13*a*b + 5*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b -
2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4), -1/8*((8*a^2 + 8*a
*b + 3*b^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^4 + ((8*a^2
+ 13*a*b + 5*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b - 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3 + 3*a^2*b +
3*a*b^2 + b^3)*f*cos(f*x + e)^4)]

Sympy [F]

\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (118) = 236\).

Time = 0.30 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.85 \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {{\left (8 \, a^{2} b^{3} + 8 \, a b^{4} + 3 \, b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b}} - \frac {2 \, {\left ({\left (8 \, a b^{4} + 5 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - {\left (8 \, a^{2} b^{4} + 11 \, a b^{5} + 3 \, b^{6}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}\right )}}{a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}}}{16 \, b^{3} f} \]

[In]

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

[Out]

-1/16*((8*a^2*b^3 + 8*a*b^4 + 3*b^5)*log((sqrt(b*sin(f*x + e)^2 + a) - sqrt(a + b))/(sqrt(b*sin(f*x + e)^2 + a
) + sqrt(a + b)))/((a^2 + 2*a*b + b^2)*sqrt(a + b)) - 2*((8*a*b^4 + 5*b^5)*(b*sin(f*x + e)^2 + a)^(3/2) - (8*a
^2*b^4 + 11*a*b^5 + 3*b^6)*sqrt(b*sin(f*x + e)^2 + a))/(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 + (b*sin(f*x
+ e)^2 + a)^2*(a^2 + 2*a*b + b^2) - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(b*sin(f*x + e)^2 + a)))/(b^3*f)

Giac [F]

\[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^5}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

[In]

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

[Out]

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