\(\int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [511]

Optimal result

Integrand size = 25, antiderivative size = 81 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{3/2} f}+\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3273, 79, 65, 214} \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 f (a+b)}-\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f (a+b)^{3/2}} \]

[In]

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

[Out]

-1/2*((2*a + b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/((a + b)^(3/2)*f) + (Sec[e + f*x]^2*Sqrt[a +
b*Sin[e + f*x]^2])/(2*(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 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}{(1-x)^2 \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = \frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f}-\frac {(2 a+b) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f} \\ & = \frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f}-\frac {(2 a+b) \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 )}{2 b (a+b) f} \\ & = -\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{3/2} f}+\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 (a+b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {\sec ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a+b}}{2 f} \]

[In]

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

[Out]

-1/2*(((2*a + b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(a + b)^(3/2) - (Sec[e + f*x]^2*Sqrt[a + b*S
in[e + f*x]^2])/(a + b))/f

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(352\) vs. \(2(69)=138\).

Time = 1.31 (sec) , antiderivative size = 353, normalized size of antiderivative = 4.36

method	result	size
default	\(\frac {-\left (2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{2}+2 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2}+3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a b +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {3}{2}}}{4 \left (a +b \right )^{\frac {5}{2}} \cos \left (f x +e \right )^{2} f}\)	\(353\)

[In]

int(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.72 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [\frac {{\left (2 \, a + b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{2} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}, \frac {{\left (2 \, a + b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{2} + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.53 \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\frac {2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )} {\left (a + b\right )} - a^{2} - 2 \, a b - b^{2}} - \frac {{\left (2 \, a b^{2} + b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}}}{4 \, b^{2} f} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\tan \left (f x + e\right )^{3}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

[In]

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

[Out]

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