\(\int \frac {\tan (e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [523]

Optimal result

Integrand size = 23, antiderivative size = 63 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2} f}-\frac {1}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3273, 53, 65, 214} \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{3/2}}-\frac {1}{f (a+b) \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]]/((a + b)^(3/2)*f) - 1/((a + b)*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 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 {1}{(1-x) (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = -\frac {1}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 (a+b) f} \\ & = -\frac {1}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\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 )}{b (a+b) f} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{3/2} f}-\frac {1}{(a+b) 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 = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.86 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\frac {b \cos ^2(e+f x)}{a+b}\right )}{(a+b) f \sqrt {a+b-b \cos ^2(e+f x)}} \]

[In]

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

[Out]

-(Hypergeometric2F1[-1/2, 1, 1/2, 1 - (b*Cos[e + f*x]^2)/(a + b)]/((a + b)*f*Sqrt[a + b - b*Cos[e + f*x]^2]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1316\) vs. \(2(55)=110\).

Time = 3.62 (sec) , antiderivative size = 1317, normalized size of antiderivative = 20.90

method	result	size
default	\(\text {Expression too large to display}\)	\(1317\)

[In]

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

[Out]

1/2/a/(cos(f*x+e)^4*a^2*b^2+2*a*b^3*cos(f*x+e)^4+b^4*cos(f*x+e)^4-2*cos(f*x+e)^2*a^3*b-6*cos(f*x+e)^2*a^2*b^2-
6*a*b^3*cos(f*x+e)^2-2*b^4*cos(f*x+e)^2+a^4+4*a^3*b+6*a^2*b^2+4*a*b^3+b^4)*(-(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)
^(1/2)*b^3+(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*b^2-2*(a+b-b*cos(f*x+e)^2)^(1/2)*a^3-(-b*cos(f*x+e)^2+(a*b^
2+b^3)/b^2)^(3/2)*a^2+(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^3-(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^2*
a+(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b*a^2-4*b*(a+b-b*cos(f*x+e)^2)^(1/2)*a^2-2*b^2*(a+b-b*cos(f*x+e)^2)^
(1/2)*a+(a+b)^(1/2)*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^3+(a+b)^(1/
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^3+b^2*(a+b)^(1/2)*ln(2/(1+si
n(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*b*(a+b)^(1/2)*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+b^2*(a+b)^(1/2)*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*b*(a+b)^(1/2)*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+b^2*((a+b)^(1/2)*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+(a+b)^(1/2)*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+(-b*c
os(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a-(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b-2*(a+b-b*cos(f*x+e)^2)^(1/2)*a)
*cos(f*x+e)^4-b*((-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*b-2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*b^2+2*b*(
a+b)^(1/2)*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*b*(a+b)^(1/2)*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+(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(
3/2)*a-4*(a+b-b*cos(f*x+e)^2)^(1/2)*a*b+2*(a+b)^(1/2)*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+2*(a+b)^(1/2)*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+2*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^2-4*(a+b-b*cos(f*x+e)^2)^(1/2)*a^2)*cos(f*x+e)^2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (55) = 110\).

Time = 0.35 (sec) , antiderivative size = 281, normalized size of antiderivative = 4.46 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left ({\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}, -\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) - \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (55) = 110\).

Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.27 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} - \frac {\operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} + \frac {2}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a + \sqrt {b \sin \left (f x + e\right )^{2} + a} b}}{2 \, f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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