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[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])
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Rubi [A] time = 0.0677903, antiderivative size = 63, normalized size of antiderivative = 1.,
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 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]
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]
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 = 0.0706664, 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]))
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Maple [B] time = 8.316, size = 1317, normalized size = 20.9 \begin{align*} \text{result too large to display} \end{align*}
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)
^(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+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^
3-2*(a+b-b*cos(f*x+e)^2)^(1/2)*a*b^2-4*a^2*b*(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)+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+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/(-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)+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*si
n(f*x+e)+a))*(a+b)^(1/2)*a*b^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/(-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
*(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/(-
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+(-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
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
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]
Exception raised: ValueError
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Fricas [B] time = 2.39673, size = 672, normalized size = 10.67 \begin{align*} \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 ] \end{align*}
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)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
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)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
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]
integrate(tan(f*x + e)/(b*sin(f*x + e)^2 + a)^(3/2), x)