\(\int \frac {\tan ^3(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [533]

Optimal result

Integrand size = 25, antiderivative size = 153 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{7/2} f}+\frac {2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 a-3 b}{2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

-1/2*(2*a-3*b)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(7/2)/f+1/6*(2*a-3*b)/(a+b)^2/f/(a+b*sin(f*
x+e)^2)^(3/2)+1/2*sec(f*x+e)^2/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)+1/2*(2*a-3*b)/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(1/
2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3273, 79, 53, 65, 214} \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 f (a+b)^{7/2}}+\frac {2 a-3 b}{2 f (a+b)^3 \sqrt {a+b \sin ^2(e+f x)}}+\frac {2 a-3 b}{6 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x)}{2 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[In]

Int[Tan[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

-1/2*((2*a - 3*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/((a + b)^(7/2)*f) + (2*a - 3*b)/(6*(a + b)^
2*f*(a + b*Sin[e + f*x]^2)^(3/2)) + Sec[e + f*x]^2/(2*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*a - 3*b)/(2
*(a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 53

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

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 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = \frac {\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(2 a-3 b) \text {Subst}\left (\int \frac {1}{(1-x) (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f} \\ & = \frac {2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(2 a-3 b) \text {Subst}\left (\int \frac {1}{(1-x) (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b)^2 f} \\ & = \frac {2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 a-3 b}{2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a-3 b) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b)^3 f} \\ & = \frac {2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 a-3 b}{2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(2 a-3 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)^3 f} \\ & = -\frac {(2 a-3 b) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{2 (a+b)^{7/2} f}+\frac {2 a-3 b}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^2(e+f x)}{2 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 a-3 b}{2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.50 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {(2 a-3 b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin ^2(e+f x)}{a+b}\right )+3 (a+b) \sec ^2(e+f x)}{6 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[In]

Integrate[Tan[e + f*x]^3/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

((2*a - 3*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[e + f*x]^2)/(a + b)] + 3*(a + b)*Sec[e + f*x]^2)/(6*(
a + b)^2*f*(a + b*Sin[e + f*x]^2)^(3/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(833\) vs. \(2(133)=266\).

Time = 3.28 (sec) , antiderivative size = 834, normalized size of antiderivative = 5.45

method	result	size
default	\(\frac {\frac {b^{2} \left (-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}{\left (a +b \right ) \left (\sin \left (f x +e \right )-1\right )}+\frac {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 )}{\left (a +b \right )^{\frac {3}{2}}}\right )}{4 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2}}-\frac {b^{2} \left (-\frac {\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}{\left (a +b \right ) \left (1+\sin \left (f x +e \right )\right )}-\frac {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 )}{\left (a +b \right )^{\frac {3}{2}}}\right )}{4 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2}}+\frac {b^{3} \left (a -b \right ) \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 )}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {a +b}}+\frac {b^{3} \left (a -b \right ) \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 )}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {a +b}}+\frac {b^{3} \left (a -b \right ) \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {-a b}\, \left (\sin \left (f x +e \right )+\frac {\sqrt {-a b}}{b}\right )}-\frac {b^{3} \left (a -b \right ) \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a b +b^{2}}{b}}}{2 \left (b +\sqrt {-a b}\right )^{3} \left (-b +\sqrt {-a b}\right )^{3} \sqrt {-a b}\, \left (\sin \left (f x +e \right )-\frac {\sqrt {-a b}}{b}\right )}-\frac {b^{3} \sqrt {-a b}\, \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \left (\sqrt {-a b}\, \sin \left (f x +e \right )+2 a \right )}{12 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2} a \left (-\sqrt {-a b}\, b^{2} \left (\cos ^{2}\left (f x +e \right )\right )+2 a \,b^{2} \sin \left (f x +e \right )+\left (-a b \right )^{\frac {3}{2}}+b^{2} \sqrt {-a b}\right )}+\frac {b^{3} \sqrt {-a b}\, \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, \left (\sqrt {-a b}\, \sin \left (f x +e \right )-2 a \right )}{12 \left (b +\sqrt {-a b}\right )^{2} \left (-b +\sqrt {-a b}\right )^{2} a \left (-\sqrt {-a b}\, b^{2} \left (\cos ^{2}\left (f x +e \right )\right )-2 a \,b^{2} \sin \left (f x +e \right )+\left (-a b \right )^{\frac {3}{2}}+b^{2} \sqrt {-a b}\right )}}{f}\)	\(834\)

[In]

int(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

(1/4*b^2/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2*(-1/(a+b)/(sin(f*x+e)-1)*(a+b-b*cos(f*x+e)^2)^(1/2)+b/(a+b)^(3
/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1)))-1/4*b^2/(b+(-a*b)^(1/2))
^2/(-b+(-a*b)^(1/2))^2*(-1/(a+b)/(1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2)^(1/2)-b/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+
b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e))))+1/2*b^3*(a-b)/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2)
)^3/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))+1/2*b^3*(a-b)
/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e
)+2*a)/(sin(f*x+e)-1))+1/2*b^3*(a-b)/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1
/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/2*b^3*(a-b)/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(-a*b)^(1/2)/(
sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12*b^3*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^2/(-b+(-
a*b)^(1/2))^2/a/(-(-a*b)^(1/2)*b^2*cos(f*x+e)^2+2*a*b^2*sin(f*x+e)+(-a*b)^(3/2)+b^2*(-a*b)^(1/2))*(-b*cos(f*x+
e)^2+(a*b^2+b^3)/b^2)^(1/2)*((-a*b)^(1/2)*sin(f*x+e)+2*a)+1/12*b^3*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^
(1/2))^2/a/(-(-a*b)^(1/2)*b^2*cos(f*x+e)^2-2*a*b^2*sin(f*x+e)+(-a*b)^(3/2)+b^2*(-a*b)^(1/2))*(-b*cos(f*x+e)^2+
(a*b^2+b^3)/b^2)^(1/2)*((-a*b)^(1/2)*sin(f*x+e)-2*a))/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 377 vs. \(2 (133) = 266\).

Time = 0.44 (sec) , antiderivative size = 769, normalized size of antiderivative = 5.03 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left ({\left (2 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (2 \, a^{2} b - a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + {\left (2 \, a^{3} + a^{2} b - 4 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a + b} \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 (3 \, {\left (2 \, a^{2} b - a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 3 \, a^{3} - 9 \, a^{2} b - 9 \, a b^{2} - 3 \, b^{3} - 4 \, {\left (2 \, a^{3} + a^{2} b - 4 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{6} - 2 \, {\left (a^{5} b + 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 10 \, a^{2} b^{4} + 5 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 6 \, a^{5} b + 15 \, a^{4} b^{2} + 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} + 6 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac {3 \, {\left ({\left (2 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{6} - 2 \, {\left (2 \, a^{2} b - a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + {\left (2 \, a^{3} + a^{2} b - 4 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{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 ) - {\left (3 \, {\left (2 \, a^{2} b - a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - 3 \, a^{3} - 9 \, a^{2} b - 9 \, a b^{2} - 3 \, b^{3} - 4 \, {\left (2 \, a^{3} + a^{2} b - 4 \, a b^{2} - 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left ({\left (a^{4} b^{2} + 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} + 4 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{6} - 2 \, {\left (a^{5} b + 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} + 10 \, a^{2} b^{4} + 5 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{4} + {\left (a^{6} + 6 \, a^{5} b + 15 \, a^{4} b^{2} + 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} + 6 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \]

[In]

integrate(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(3*((2*a*b^2 - 3*b^3)*cos(f*x + e)^6 - 2*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 + (2*a^3 + a^2*b - 4*
a*b^2 - 3*b^3)*cos(f*x + e)^2)*sqrt(a + b)*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*(3*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 - 3*a^3 - 9*a^2*b - 9*a*b^2 -
3*b^3 - 4*(2*a^3 + a^2*b - 4*a*b^2 - 3*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b^2 + 4*a^3
*b^3 + 6*a^2*b^4 + 4*a*b^5 + b^6)*f*cos(f*x + e)^6 - 2*(a^5*b + 5*a^4*b^2 + 10*a^3*b^3 + 10*a^2*b^4 + 5*a*b^5
+ b^6)*f*cos(f*x + e)^4 + (a^6 + 6*a^5*b + 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)*f*cos(f*x + e
)^2), 1/6*(3*((2*a*b^2 - 3*b^3)*cos(f*x + e)^6 - 2*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 + (2*a^3 + a^2*b -
4*a*b^2 - 3*b^3)*cos(f*x + e)^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) -
(3*(2*a^2*b - a*b^2 - 3*b^3)*cos(f*x + e)^4 - 3*a^3 - 9*a^2*b - 9*a*b^2 - 3*b^3 - 4*(2*a^3 + a^2*b - 4*a*b^2 -
3*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b^2 + 4*a^3*b^3 + 6*a^2*b^4 + 4*a*b^5 + b^6)*f*
cos(f*x + e)^6 - 2*(a^5*b + 5*a^4*b^2 + 10*a^3*b^3 + 10*a^2*b^4 + 5*a*b^5 + b^6)*f*cos(f*x + e)^4 + (a^6 + 6*a
^5*b + 15*a^4*b^2 + 20*a^3*b^3 + 15*a^2*b^4 + 6*a*b^5 + b^6)*f*cos(f*x + e)^2)]

Sympy [F]

\[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(tan(f*x+e)**3/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Integral(tan(e + f*x)**3/(a + b*sin(e + f*x)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.71 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {3 \, {\left (2 \, a b^{2} - 3 \, b^{3}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a + b}} - \frac {2 \, {\left (2 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + 2 \, a b^{4} - 3 \, {\left (2 \, a b^{2} - 3 \, b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} + 2 \, {\left (2 \, a^{2} b^{2} - a b^{3} - 3 \, b^{4}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} - {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}}{12 \, b^{2} f} \]

[In]

integrate(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

1/12*(3*(2*a*b^2 - 3*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^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a + b)) - 2*(2*a^3*b^2 + 4*a^2*b^3 + 2*a*b^4 - 3*(2*a*b^2 - 3*b^3
)*(b*sin(f*x + e)^2 + a)^2 + 2*(2*a^2*b^2 - a*b^3 - 3*b^4)*(b*sin(f*x + e)^2 + a))/((a^3 + 3*a^2*b + 3*a*b^2 +
b^3)*(b*sin(f*x + e)^2 + a)^(5/2) - (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(b*sin(f*x + e)^2 + a)^(3/2))
)/(b^2*f)

Giac [F]

\[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(f*x+e)^3/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

[In]

int(tan(e + f*x)^3/(a + b*sin(e + f*x)^2)^(5/2),x)

[Out]

int(tan(e + f*x)^3/(a + b*sin(e + f*x)^2)^(5/2), x)