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

Optimal. Leaf size=161 \[ -\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 f (a-b)}+\frac{2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{15 f (a-b)^2}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{f} \]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a - b + b*Sec[
e + f*x]^2])/f + (2*(5*a - 4*b)*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(3/2))/(15*(a - b)^2*f) - (Cos[e + f
*x]^5*(a - b + b*Sec[e + f*x]^2)^(3/2))/(5*(a - b)*f)

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Rubi [A]  time = 0.165315, antiderivative size = 161, 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 = {3664, 462, 451, 277, 217, 206} \[ -\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 f (a-b)}+\frac{2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{15 f (a-b)^2}-\frac{\cos (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a - b + b*Sec[
e + f*x]^2])/f + (2*(5*a - 4*b)*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(3/2))/(15*(a - b)^2*f) - (Cos[e + f
*x]^5*(a - b + b*Sec[e + f*x]^2)^(3/2))/(5*(a - b)*f)

Rule 3664

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

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 451

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sin ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2 \sqrt{a-b+b x^2}}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{\left (-2 (5 a-4 b)+5 (a-b) x^2\right ) \sqrt{a-b+b x^2}}{x^4} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac{2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a-b+b x^2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{f}+\frac{2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{f}+\frac{2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{f}\\ &=\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac{\cos (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{f}+\frac{2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}\\ \end{align*}

Mathematica [A]  time = 3.25176, size = 208, normalized size = 1.29 \[ \frac{\cos (e+f x) \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt{(a-b) \cos (2 (e+f x))+a+b} \left (4 \left (7 a^2-15 a b+8 b^2\right ) \cos (2 (e+f x))-89 a^2-3 (a-b)^2 \cos (4 (e+f x))+254 a b-149 b^2\right )+120 \sqrt{2} \sqrt{b} (a-b)^2 \tanh ^{-1}\left (\frac{\sqrt{(a-b) \cos (2 (e+f x))+a+b}}{\sqrt{2} \sqrt{b}}\right )\right )}{120 \sqrt{2} f (a-b)^2 \sqrt{(a-b) \cos (2 (e+f x))+a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(120*Sqrt[2]*(a - b)^2*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])]
+ Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*(-89*a^2 + 254*a*b - 149*b^2 + 4*(7*a^2 - 15*a*b + 8*b^2)*Cos[2*(e +
f*x)] - 3*(a - b)^2*Cos[4*(e + f*x)]))*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(120*Sqrt[2]*
(a - b)^2*f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])

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Maple [B]  time = 0.752, size = 7044, normalized size = 43.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.36526, size = 941, normalized size = 5.84 \begin{align*} \left [\frac{15 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt{b} \log \left (-\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt{b} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \,{\left (3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} -{\left (10 \, a^{2} - 21 \, a b + 11 \, b^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} - 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} f}, -\frac{15 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) +{\left (3 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} -{\left (10 \, a^{2} - 21 \, a b + 11 \, b^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} - 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \,{\left (a^{2} - 2 \, a b + b^{2}\right )} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*(a^2 - 2*a*b + b^2)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) - 2*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - (10*a^2 - 2
1*a*b + 11*b^2)*cos(f*x + e)^3 + (15*a^2 - 40*a*b + 23*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/co
s(f*x + e)^2))/((a^2 - 2*a*b + b^2)*f), -1/15*(15*(a^2 - 2*a*b + b^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*c
os(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + (3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - (10*a^2 - 21*a*b
+ 11*b^2)*cos(f*x + e)^3 + (15*a^2 - 40*a*b + 23*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x
+ e)^2))/((a^2 - 2*a*b + b^2)*f)]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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