3.478 \(\int \frac {\tan ^5(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=68 \[ \frac {a^2}{7 f \left (a \cos ^2(e+f x)\right )^{7/2}}-\frac {2 a}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}+\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

[Out]

1/7*a^2/f/(a*cos(f*x+e)^2)^(7/2)-2/5*a/f/(a*cos(f*x+e)^2)^(5/2)+1/3/f/(a*cos(f*x+e)^2)^(3/2)

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Rubi [A]  time = 0.13, antiderivative size = 68, 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 = {3176, 3205, 16, 43} \[ \frac {a^2}{7 f \left (a \cos ^2(e+f x)\right )^{7/2}}-\frac {2 a}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}+\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5/(a - a*Sin[e + f*x]^2)^(3/2),x]

[Out]

a^2/(7*f*(a*Cos[e + f*x]^2)^(7/2)) - (2*a)/(5*f*(a*Cos[e + f*x]^2)^(5/2)) + 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 43

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 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 ^5(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac {\tan ^5(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}} \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(1-x)^2}{x^3 (a x)^{3/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \frac {(1-x)^2}{(a x)^{9/2}} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac {a^3 \operatorname {Subst}\left (\int \left (\frac {1}{(a x)^{9/2}}-\frac {2}{a (a x)^{7/2}}+\frac {1}{a^2 (a x)^{5/2}}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac {a^2}{7 f \left (a \cos ^2(e+f x)\right )^{7/2}}-\frac {2 a}{5 f \left (a \cos ^2(e+f x)\right )^{5/2}}+\frac {1}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 51, normalized size = 0.75 \[ \frac {\left (35 \cos ^4(e+f x)-42 \cos ^2(e+f x)+15\right ) \sec ^4(e+f x)}{105 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5/(a - a*Sin[e + f*x]^2)^(3/2),x]

[Out]

((15 - 42*Cos[e + f*x]^2 + 35*Cos[e + f*x]^4)*Sec[e + f*x]^4)/(105*f*(a*Cos[e + f*x]^2)^(3/2))

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fricas [A]  time = 0.50, size = 50, normalized size = 0.74 \[ \frac {{\left (35 \, \cos \left (f x + e\right )^{4} - 42 \, \cos \left (f x + e\right )^{2} + 15\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{105 \, a^{2} f \cos \left (f x + e\right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

1/105*(35*cos(f*x + e)^4 - 42*cos(f*x + e)^2 + 15)*sqrt(a*cos(f*x + e)^2)/(a^2*f*cos(f*x + e)^8)

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giac [B]  time = 0.77, size = 151, normalized size = 2.22 \[ \frac {\frac {7 \, {\left (3 \, {\left (a \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} - 10 \, {\left (a \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {a \tan \left (f x + e\right )^{2} + a} a^{2}\right )}}{a^{2}} + \frac {3 \, {\left (5 \, {\left (a \tan \left (f x + e\right )^{2} + a\right )}^{\frac {7}{2}} - 21 \, {\left (a \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} a + 35 \, {\left (a \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {a \tan \left (f x + e\right )^{2} + a} a^{3}\right )}}{a^{3}}}{105 \, a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

1/105*(7*(3*(a*tan(f*x + e)^2 + a)^(5/2) - 10*(a*tan(f*x + e)^2 + a)^(3/2)*a + 15*sqrt(a*tan(f*x + e)^2 + a)*a
^2)/a^2 + 3*(5*(a*tan(f*x + e)^2 + a)^(7/2) - 21*(a*tan(f*x + e)^2 + a)^(5/2)*a + 35*(a*tan(f*x + e)^2 + a)^(3
/2)*a^2 - 35*sqrt(a*tan(f*x + e)^2 + a)*a^3)/a^3)/(a^2*f)

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maple [A]  time = 2.13, size = 51, normalized size = 0.75 \[ \frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (35 \left (\cos ^{4}\left (f x +e \right )\right )-42 \left (\cos ^{2}\left (f x +e \right )\right )+15\right )}{105 a^{2} \cos \left (f x +e \right )^{8} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(3/2),x)

[Out]

1/105/a^2/cos(f*x+e)^8*(a*cos(f*x+e)^2)^(1/2)*(35*cos(f*x+e)^4-42*cos(f*x+e)^2+15)/f

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maxima [A]  time = 0.33, size = 69, normalized size = 1.01 \[ \frac {35 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )}^{2} a^{3} + 42 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + 15 \, a^{5}}{105 \, {\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {7}{2}} a^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

1/105*(35*(a*sin(f*x + e)^2 - a)^2*a^3 + 42*(a*sin(f*x + e)^2 - a)*a^4 + 15*a^5)/((-a*sin(f*x + e)^2 + a)^(7/2
)*a^3*f)

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mupad [B]  time = 33.91, size = 583, normalized size = 8.57 \[ \frac {16\,{\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^2\,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 {464\,{\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^2\,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 {3072\,{\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}}{35\,a^2\,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 {4736\,{\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}}{35\,a^2\,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 )}+\frac {768\,{\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}}{7\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^6\,\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 {256\,{\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}}{7\,a^2\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^7\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^5/(a - a*sin(e + f*x)^2)^(3/2),x)

[Out]

(16*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^2*f*(ex
p(e*2i + f*x*2i) + 1)^2*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (464*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^2*f*(exp(e*2i + f*x*2i) + 1)^3*(exp(e*1i + f*x
*1i) + exp(e*3i + f*x*3i))) + (3072*exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1
i)*1i)/2)^2)^(1/2))/(35*a^2*f*(exp(e*2i + f*x*2i) + 1)^4*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (4736*ex
p(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))/(35*a^2*f*(exp(e*2
i + f*x*2i) + 1)^5*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) + (768*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))/(7*a^2*f*(exp(e*2i + f*x*2i) + 1)^6*(exp(e*1i + f*x*1i) +
exp(e*3i + f*x*3i))) - (256*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))/(7*a^2*f*(exp(e*2i + f*x*2i) + 1)^7*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5/(a-a*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral(tan(e + f*x)**5/(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))**(3/2), x)

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