\(\int \frac {\tan (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [534]

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2} f}-\frac {1}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {1}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3273, 53, 65, 214} \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{5/2}}-\frac {1}{f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {1}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]]/((a + b)^(5/2)*f) - 1/(3*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2
)) - 1/((a + b)^2*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)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f} \\ & = -\frac {1}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\text {Subst}\left (\int \frac {1}{(1-x) (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 (a+b) f} \\ & = -\frac {1}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {1}{(a+b)^2 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)^2 f} \\ & = -\frac {1}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {1}{(a+b)^2 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)^2 f} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{(a+b)^{5/2} f}-\frac {1}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {1}{(a+b)^2 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 = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},1-\frac {b \cos ^2(e+f x)}{a+b}\right )}{3 (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(962\) vs. \(2(79)=158\).

Time = 2.34 (sec) , antiderivative size = 963, normalized size of antiderivative = 10.58

method	result	size
default	\(\text {Expression too large to display}\)	\(963\)

[In]

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

[Out]

1/6/b^3/(a+b)^(1/2)/a^2/(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)*(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^2*b^5*cos(f*x+e)^4+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^2*b^5*cos(f*x+e)^4+6*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^
(1/2)*(a+b)^(1/2)*a^2*b^4*cos(f*x+e)^2-6*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^4*cos(f*x+e)^2-6*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^5*cos(f*x+e)^2-6*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^4*cos(
f*x+e)^2-6*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^5*cos(f*x+e)^2-8
*a^3*b^3*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*(a+b)^(1/2)-8*a^2*b^4*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)
*(a+b)^(1/2)+3*a^4*b^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))+3*a^4*b^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))+6*a^3*b^4*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))+6*a^3*b^4*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))+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^2*b^5+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^2*b^5)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (79) = 158\).

Time = 0.37 (sec) , antiderivative size = 521, normalized size of antiderivative = 5.73 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left (3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 8 \, a b - 4 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, {\left ({\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f\right )}}, -\frac {3 \, {\left (b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) - {\left (3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, a^{2} - 8 \, a b - 4 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, {\left ({\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f\right )}}\right ] \]

[In]

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

[Out]

[1/6*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^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*(a*b + b^2)*cos(f*x +
e)^2 - 4*a^2 - 8*a*b - 4*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f*cos(f*
x + e)^4 - 2*(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 +
10*a^2*b^3 + 5*a*b^4 + b^5)*f), -1/3*(3*(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2
)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) - (3*(a*b + b^2)*cos(f*x + e)^2 -
4*a^2 - 8*a*b - 4*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f*cos(f*x + e)^
4 - 2*(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^2 + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*
b^3 + 5*a*b^4 + b^5)*f)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (79) = 158\).

Time = 0.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.23 \[ \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {2}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a + {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {6}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2} + 2 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a b + \sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}} + \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}} - \frac {3 \, \operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {5}{2}}}}{6 \, f} \]

[In]

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

[Out]

-1/6*(2/((b*sin(f*x + e)^2 + a)^(3/2)*a + (b*sin(f*x + e)^2 + a)^(3/2)*b) + 6/(sqrt(b*sin(f*x + e)^2 + a)*a^2
+ 2*sqrt(b*sin(f*x + e)^2 + a)*a*b + sqrt(b*sin(f*x + e)^2 + a)*b^2) + 3*arcsinh(b*sin(f*x + e)/(sqrt(a*b)*(si
n(f*x + e) + 1)) - a/(sqrt(a*b)*(sin(f*x + e) + 1)))/(a + b)^(5/2) - 3*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)^(5/2))/f

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[In]

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

[Out]

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