3.510 \(\int \frac{\tan ^5(e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=134 \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{5/2}}+\frac{\sec ^4(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{4 f (a+b)}-\frac{(8 a+5 b) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{8 f (a+b)^2} \]

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(5/2)*f) - ((8*a + 5*b)*S
ec[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^2*f) + (Sec[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2])/(4*(a
+ b)*f)

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Rubi [A]  time = 0.173842, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3194, 89, 78, 63, 208} \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{5/2}}+\frac{\sec ^4(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{4 f (a+b)}-\frac{(8 a+5 b) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{8 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(5/2)*f) - ((8*a + 5*b)*S
ec[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^2*f) + (Sec[e + f*x]^4*Sqrt[a + b*Sin[e + f*x]^2])/(4*(a
+ b)*f)

Rule 3194

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

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tan ^5(e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1-x)^3 \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^4(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{4 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (4 a+b)+2 (a+b) x}{(1-x)^2 \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(8 a+5 b) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{4 (a+b) f}+\frac{\left (8 a^2+8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^2 f}\\ &=-\frac{(8 a+5 b) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{4 (a+b) f}+\frac{\left (8 a^2+8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{8 b (a+b)^2 f}\\ &=\frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 (a+b)^{5/2} f}-\frac{(8 a+5 b) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{4 (a+b) f}\\ \end{align*}

Mathematica [A]  time = 0.49333, size = 108, normalized size = 0.81 \[ \frac{\left (8 a^2+8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )+\sqrt{a+b} \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)} \left (2 (a+b) \sec ^2(e+f x)-8 a-5 b\right )}{8 f (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 + 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] + Sqrt[a + b]*Sec[e + f*x]^2*(-8*a -
5*b + 2*(a + b)*Sec[e + f*x]^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^(5/2)*f)

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Maple [B]  time = 3.356, size = 644, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((8*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^4+24*ln(2/(1+sin(f*x+e
))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b+27*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*c
os(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^2+14*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*s
in(f*x+e)+a))*a*b^3+3*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^4+8*ln(2/
(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^4+24*ln(2/(-1+sin(f*x+e))*((a+b)^(1
/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b+27*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2
)^(1/2)+b*sin(f*x+e)+a))*a^2*b^2+14*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+
a))*a*b^3+3*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^4)*cos(f*x+e)^4-2*
(a+b)^(5/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*(8*a+5*b)*cos(f*x+e)^2+4*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*a+4*(a+
b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*b)/(a+b)^(5/2)/cos(f*x+e)^4/(a^2+2*a*b+b^2)/f

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.03885, size = 792, normalized size = 5.91 \begin{align*} \left [\frac{{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt{a + b} \cos \left (f x + e\right )^{4} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \,{\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{16 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}, -\frac{{\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{a + b}\right ) \cos \left (f x + e\right )^{4} +{\left ({\left (8 \, a^{2} + 13 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a^{2} - 4 \, a b - 2 \, b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{8 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((8*a^2 + 8*a*b + 3*b^2)*sqrt(a + b)*cos(f*x + e)^4*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a
+ b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) - 2*((8*a^2 + 13*a*b + 5*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b -
2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4), -1/8*((8*a^2 + 8*a
*b + 3*b^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^4 + ((8*a^2
+ 13*a*b + 5*b^2)*cos(f*x + e)^2 - 2*a^2 - 4*a*b - 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^3 + 3*a^2*b +
3*a*b^2 + b^3)*f*cos(f*x + e)^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (e + f x \right )}}{\sqrt{a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(e + f*x)**5/sqrt(a + b*sin(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{5}}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(f*x + e)^5/sqrt(b*sin(f*x + e)^2 + a), x)