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

Optimal. Leaf size=63 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{3/2}}-\frac {1}{f (a+b) \sqrt {a+b \sin ^2(e+f x)}} \]

[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]  time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3194, 51, 63, 208} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{f (a+b)^{3/2}}-\frac {1}{f (a+b) \sqrt {a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[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 51

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 63

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 208

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

Rule 3194

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*} \int \frac {\tan (e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {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 {\operatorname {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 {\operatorname {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 {\tanh ^{-1}\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]  time = 0.07, size = 54, normalized size = 0.86 \[ -\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};1-\frac {b \cos ^2(e+f x)}{a+b}\right )}{f (a+b) \sqrt {a-b \cos ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[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]))

________________________________________________________________________________________

fricas [B]  time = 0.56, size = 281, normalized size = 4.46 \[ \left [\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left ({\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )}}, -\frac {{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) - \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

________________________________________________________________________________________

giac [B]  time = 0.89, size = 250, normalized size = 3.97 \[ -\frac {\frac {\frac {{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}} + \frac {a^{2} b + 2 \, a b^{2} + b^{3}}{a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}}}{\sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}} + \frac {2 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} - \sqrt {a}}{2 \, \sqrt {-a - b}}\right )}{{\left (a + b\right )} \sqrt {-a - b}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(((a^2*b + 2*a*b^2 + b^3)*tan(1/2*f*x + 1/2*e)^2/(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4) + (a^2*b + 2*a*b^2 + b^3
)/(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4))/sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/
2*f*x + 1/2*e)^2 + a) + 2*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*ta
n(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/((a + b)*sqrt(-a - b)))/f

________________________________________________________________________________________

maple [B]  time = 8.49, size = 1317, normalized size = 20.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.66, size = 143, normalized size = 2.27 \[ -\frac {\frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) + 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} - \frac {\operatorname {arsinh}\left (-\frac {b \sin \left (f x + e\right )}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}} - \frac {a}{\sqrt {a b} {\left (\sin \left (f x + e\right ) - 1\right )}}\right )}{{\left (a + b\right )}^{\frac {3}{2}}} + \frac {2}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a + \sqrt {b \sin \left (f x + e\right )^{2} + a} b}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {tan}\left (e+f\,x\right )}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________