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

Optimal. Leaf size=177 \[ -\frac{\left (8 a^2+24 a b+15 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 f (a+b)^2}+\frac{\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{3/2}}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f (a+b)}-\frac{(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 f (a+b)^2} \]

[Out]

((8*a^2 + 24*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(3/2)*f) - ((8*a^2 + 24
*a*b + 15*b^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^2*f) - ((8*a + 7*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2
)^(3/2))/(8*(a + b)^2*f) + (Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2))/(4*(a + b)*f)

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Rubi [A]  time = 0.210743, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3194, 89, 78, 50, 63, 208} \[ -\frac{\left (8 a^2+24 a b+15 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 f (a+b)^2}+\frac{\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{3/2}}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f (a+b)}-\frac{(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 + 24*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(3/2)*f) - ((8*a^2 + 24
*a*b + 15*b^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)^2*f) - ((8*a + 7*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2
)^(3/2))/(8*(a + b)^2*f) + (Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2))/(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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 \sqrt{a+b \sin ^2(e+f x)} \tan ^5(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \sqrt{a+b x}}{(1-x)^3} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+b x} \left (\frac{1}{2} (4 a+3 b)+2 (a+b) x\right )}{(1-x)^2} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}+\frac{\left (8 a^2+24 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^2 f}\\ &=-\frac{\left (8 a^2+24 a b+15 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac{(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}+\frac{\left (8 a^2+24 a b+15 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) f}\\ &=-\frac{\left (8 a^2+24 a b+15 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac{(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}+\frac{\left (8 a^2+24 a b+15 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) f}\\ &=\frac{\left (8 a^2+24 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 (a+b)^{3/2} f}-\frac{\left (8 a^2+24 a b+15 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b)^2 f}-\frac{(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 (a+b) f}\\ \end{align*}

Mathematica [A]  time = 0.600288, size = 143, normalized size = 0.81 \[ \frac{\left (8 a^2+24 a b+15 b^2\right ) \left (\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )-\sqrt{a+b \sin ^2(e+f x)}\right )+2 (a+b) \sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}-(8 a+7 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{8 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((8*a + 7*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(3/2)) + 2*(a + b)*Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^
(3/2) + (8*a^2 + 24*a*b + 15*b^2)*(Sqrt[a + b]*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]] - Sqrt[a + b*Si
n[e + f*x]^2]))/(8*(a + b)^2*f)

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Maple [B]  time = 4.911, size = 721, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((-16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a^2-48*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a*b-30*b^2*(a+
b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)+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+40*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+71*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+54*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+15*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+40*
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+71*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+54*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+15*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)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*(8*a+7*b)*cos(f*x+e)^2+4*(a+b)^(3/2)*(a+
b-b*cos(f*x+e)^2)^(3/2)*a+4*b*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2))/(a+b)^(3/2)/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((a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 8.29031, size = 871, normalized size = 4.92 \begin{align*} \left [\frac{{\left (8 \, a^{2} + 24 \, a b + 15 \, 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 (8 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} +{\left (8 \, a^{2} + 17 \, a b + 9 \, 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^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}, -\frac{{\left (8 \, a^{2} + 24 \, a b + 15 \, 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 (8 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} +{\left (8 \, a^{2} + 17 \, a b + 9 \, 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^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*((8*a^2 + 24*a*b + 15*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 + 2*a*b + b^2)*cos(f*x + e)^4 + (8*a^2 + 17*a*b +
9*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^2 + 2*a*b + b^2)*f*cos(f*
x + e)^4), -1/8*((8*a^2 + 24*a*b + 15*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 + 2*a*b + b^2)*cos(f*x + e)^4 + (8*a^2 + 17*a*b + 9*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^2 + 2*a*b + b^2)*f*cos(f*x + e)^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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