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

Optimal. Leaf size=21 \[ \frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

[Out]

1/3/f/(a*cos(f*x+e)^2)^(3/2)

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Rubi [A]  time = 0.07, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3176, 3205, 16, 32} \[ \frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(3*f*(a*Cos[e + f*x]^2)^(3/2))

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rubi steps

\begin {align*} \int \frac {\tan (e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac {\tan (e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}} \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x (a x)^{3/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {a \operatorname {Subst}\left (\int \frac {1}{(a x)^{5/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.00 \[ \frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(3*f*(a*Cos[e + f*x]^2)^(3/2))

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fricas [A]  time = 0.45, size = 28, normalized size = 1.33 \[ \frac {\sqrt {a \cos \left (f x + e\right )^{2}}}{3 \, a^{2} f \cos \left (f x + e\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(a*cos(f*x + e)^2)/(a^2*f*cos(f*x + e)^4)

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giac [B]  time = 0.55, size = 64, normalized size = 3.05 \[ \frac {3 \, \sqrt {a \tan \left (f x + e\right )^{2} + a} + \frac {{\left (a \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {a \tan \left (f x + e\right )^{2} + a} a}{a}}{3 \, a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*sqrt(a*tan(f*x + e)^2 + a) + ((a*tan(f*x + e)^2 + a)^(3/2) - 3*sqrt(a*tan(f*x + e)^2 + a)*a)/a)/(a^2*f)

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maple [A]  time = 0.11, size = 21, normalized size = 1.00 \[ \frac {1}{3 f \left (a -a \left (\sin ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.44, size = 95, normalized size = 4.52 \[ \frac {\frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right ) + \sqrt {-a \sin \left (f x + e\right )^{2} + a} a} - \frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right ) - \sqrt {-a \sin \left (f x + e\right )^{2} + a} a}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(1/(sqrt(-a*sin(f*x + e)^2 + a)*a*sin(f*x + e) + sqrt(-a*sin(f*x + e)^2 + a)*a) - 1/(sqrt(-a*sin(f*x + e)^
2 + a)*a*sin(f*x + e) - sqrt(-a*sin(f*x + e)^2 + a)*a))/f

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mupad [B]  time = 18.31, size = 72, normalized size = 3.43 \[ \frac {16\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(16*exp(e*4i + f*x*4i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(3*a^2*f*(ex
p(e*2i + f*x*2i) + 1)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\left (e + f x \right )}}{\left (- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(e + f*x)/(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))**(3/2), x)

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