3.5.69 \(\int \frac {\tan ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) [469]

Optimal. Leaf size=42 \[ \frac {a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3255, 3284, 16, 45} \begin {gather*} \frac {a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a/(3*f*(a*Cos[e + f*x]^2)^(3/2)) - 1/(f*Sqrt[a*Cos[e + f*x]^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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3255

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 3284

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

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Mathematica [A]
time = 0.05, size = 31, normalized size = 0.74 \begin {gather*} \frac {-3+\sec ^2(e+f x)}{3 f \sqrt {a \cos ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 + Sec[e + f*x]^2)/(3*f*Sqrt[a*Cos[e + f*x]^2])

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Maple [A]
time = 11.01, size = 41, normalized size = 0.98

method	result	size
default	\(-\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (3 \left (\cos ^{2}\left (f x +e \right )\right )-1\right )}{3 a \cos \left (f x +e \right )^{4} f}\)	\(41\)
risch	\(-\frac {2 \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )}+3\right )}{3 \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} f}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 48, normalized size = 1.14 \begin {gather*} \frac {3 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{2} + a^{3}}{3 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.41, size = 40, normalized size = 0.95 \begin {gather*} -\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (3 \, \cos \left (f x + e\right )^{2} - 1\right )}}{3 \, a f \cos \left (f x + e\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{3}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.85, size = 57, normalized size = 1.36 \begin {gather*} \frac {4 \, {\left (3 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} \sqrt {a} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

4/3*(3*tan(1/2*f*x + 1/2*e)^2 - 1)/((tan(1/2*f*x + 1/2*e)^2 - 1)^3*sqrt(a)*f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))

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Mupad [B]
time = 19.50, size = 100, normalized size = 2.38 \begin {gather*} -\frac {4\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\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}\,\left (2\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+3\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}+3\right )}{3\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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