3.5.57 \(\int \sqrt {a-a \sin ^2(e+f x)} \tan ^5(e+f x) \, dx\) [457]

Optimal. Leaf size=64 \[ \frac {a^2}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {2 a}{f \sqrt {a \cos ^2(e+f x)}}-\frac {\sqrt {a \cos ^2(e+f x)}}{f} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 64, 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}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}}-\frac {2 a}{f \sqrt {a \cos ^2(e+f x)}}-\frac {\sqrt {a \cos ^2(e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.07, size = 51, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {a \cos ^2(e+f x)} \left (-1+6 \cos ^2(e+f x)+3 \cos ^4(e+f x)\right ) \sec ^4(e+f x)}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Sqrt[a*Cos[e + f*x]^2]*(-1 + 6*Cos[e + f*x]^2 + 3*Cos[e + f*x]^4)*Sec[e + f*x]^4)/f

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Maple [A]
time = 14.03, size = 48, normalized size = 0.75

method	result	size
default	\(-\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (3 \left (\cos ^{4}\left (f x +e \right )\right )+6 \left (\cos ^{2}\left (f x +e \right )\right )-1\right )}{3 \cos \left (f x +e \right )^{4} f}\)	\(48\)
risch	\(-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {4 \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (3 \,{\mathrm e}^{6 i \left (f x +e \right )}+4 \,{\mathrm e}^{4 i \left (f x +e \right )}+3 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 72, normalized size = 1.12 \begin {gather*} -\frac {3 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} a^{3} - \frac {6 \, {\left (a \sin \left (f x + e\right )^{2} - a\right )} a^{4} + a^{5}}{{\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}}{3 \, a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*sqrt(-a*sin(f*x + e)^2 + a)*a^3 - (6*(a*sin(f*x + e)^2 - a)*a^4 + a^5)/(-a*sin(f*x + e)^2 + a)^(3/2))/
(a^3*f)

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Fricas [A]
time = 0.39, size = 47, normalized size = 0.73 \begin {gather*} -\frac {{\left (3 \, \cos \left (f x + e\right )^{4} + 6 \, \cos \left (f x + e\right )^{2} - 1\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{3 \, f \cos \left (f x + e\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{5}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (59) = 118\).
time = 1.55, size = 136, normalized size = 2.12 \begin {gather*} \frac {2 \, \sqrt {a} {\left (\frac {3 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1} - \frac {3 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 12 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3}}\right )}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(a)*(3*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)/(tan(1/2*f*x + 1/2*e)^2 + 1) - (3*sgn(tan(1/2*f*x + 1/2*e)^4 -
1)*tan(1/2*f*x + 1/2*e)^4 - 12*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^2 + 5*sgn(tan(1/2*f*x + 1/
2*e)^4 - 1))/(tan(1/2*f*x + 1/2*e)^2 - 1)^3)/f

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Mupad [B]
time = 19.71, size = 326, normalized size = 5.09 \begin {gather*} -\frac {\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}}{f}-\frac {8\,{\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}}{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 {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\,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 {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\,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 )} \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]

(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*f*(exp(e*
2i + f*x*2i) + 1)^2*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (8*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))/(f*(exp(e*2i + f*x*2i) + 1)*(exp(e*1i + f*x*1i) + 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)/f - (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*f*(exp(e*2i + f*x*2i) + 1)
^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))

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