3.5.68 \(\int \frac {\tan ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) [468]

Optimal. Leaf size=65 \[ \frac {a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {2 a}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}+\frac {1}{f \sqrt {a \cos ^2(e+f x)}} \]

[Out]

1/5*a^2/f/(a*cos(f*x+e)^2)^(5/2)-2/3*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 = 65, 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^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {2 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]^5/Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

a^2/(5*f*(a*Cos[e + f*x]^2)^(5/2)) - (2*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 ^5(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx &=\int \frac {\tan ^5(e+f x)}{\sqrt {a \cos ^2(e+f x)}} \, dx\\ &=-\frac {\text {Subst}\left (\int \frac {(1-x)^2}{x^3 \sqrt {a x}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^3 \text {Subst}\left (\int \frac {(1-x)^2}{(a x)^{7/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^3 \text {Subst}\left (\int \left (\frac {1}{(a x)^{7/2}}-\frac {2}{a (a x)^{5/2}}+\frac {1}{a^2 (a x)^{3/2}}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac {a^2}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}-\frac {2 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 = 0.06, size = 43, normalized size = 0.66 \begin {gather*} \frac {15-10 \sec ^2(e+f x)+3 \sec ^4(e+f x)}{15 f \sqrt {a \cos ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5/Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

(15 - 10*Sec[e + f*x]^2 + 3*Sec[e + f*x]^4)/(15*f*Sqrt[a*Cos[e + f*x]^2])

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Maple [A]
time = 14.43, size = 51, normalized size = 0.78

method	result	size
default	\(\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (15 \left (\cos ^{4}\left (f x +e \right )\right )-10 \left (\cos ^{2}\left (f x +e \right )\right )+3\right )}{15 a \cos \left (f x +e \right )^{6} f}\)	\(51\)
risch	\(\frac {2 \,{\mathrm e}^{8 i \left (f x +e \right )}+\frac {8 \,{\mathrm e}^{6 i \left (f x +e \right )}}{3}+\frac {116 \,{\mathrm e}^{4 i \left (f x +e \right )}}{15}+\frac {8 \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}+2}{\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4} f}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15/a/cos(f*x+e)^6*(a*cos(f*x+e)^2)^(1/2)*(15*cos(f*x+e)^4-10*cos(f*x+e)^2+3)/f

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Maxima [A]
time = 0.30, size = 72, normalized size = 1.11 \begin {gather*} \frac {15 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )}^{2} a^{3} + 10 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + 3 \, a^{5}}{15 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

1/15*(15*(a*sin(f*x + e)^2 - a)^2*a^3 + 10*(a*sin(f*x + e)^2 - a)*a^4 + 3*a^5)/((-a*sin(f*x + e)^2 + a)^(5/2)*
a^3*f)

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Fricas [A]
time = 0.39, size = 50, normalized size = 0.77 \begin {gather*} \frac {{\left (15 \, \cos \left (f x + e\right )^{4} - 10 \, \cos \left (f x + e\right )^{2} + 3\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{15 \, a f \cos \left (f x + e\right )^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

1/15*(15*cos(f*x + e)^4 - 10*cos(f*x + e)^2 + 3)*sqrt(a*cos(f*x + e)^2)/(a*f*cos(f*x + e)^6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{5}{\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)**5/(a-a*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(tan(e + f*x)**5/sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1)), x)

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Giac [A]
time = 1.61, size = 71, normalized size = 1.09 \begin {gather*} \frac {16 \, {\left (10 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 5 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5} \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)^5/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

16/15*(10*tan(1/2*f*x + 1/2*e)^4 - 5*tan(1/2*f*x + 1/2*e)^2 + 1)/((tan(1/2*f*x + 1/2*e)^2 - 1)^5*sqrt(a)*f*sgn
(tan(1/2*f*x + 1/2*e)^4 - 1))

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Mupad [B]
time = 23.61, size = 486, normalized size = 7.48 \begin {gather*} \frac {4\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\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}}{a\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {32\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\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\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {352\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\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}}{15\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {128\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\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}}{5\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {64\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\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}}{5\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^5\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^5/(a - a*sin(e + f*x)^2)^(1/2),x)

[Out]

(4*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(a*f*(exp(e*2
i + f*x*2i) + 1)*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (32*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x
*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(3*a*f*(exp(e*2i + f*x*2i) + 1)^2*(exp(e*1i + f*x*1i) + exp(
e*3i + f*x*3i))) + (352*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)
^(1/2))/(15*a*f*(exp(e*2i + f*x*2i) + 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (128*exp(e*3i + f*x*3i
)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(5*a*f*(exp(e*2i + f*x*2i) + 1)^4
*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) + (64*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (ex
p(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(5*a*f*(exp(e*2i + f*x*2i) + 1)^5*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))

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