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

Optimal. Leaf size=123 \[ -\frac{\left (15 a^2+20 a b+8 b^2\right ) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{15 a^3 f}+\frac{2 (5 a+2 b) \cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{15 a^2 f}-\frac{\cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{5 a f} \]

[Out]

-((15*a^2 + 20*a*b + 8*b^2)*Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(15*a^3*f) + (2*(5*a + 2*b)*Cos[e + f*x]^
3*Sqrt[a + b*Sec[e + f*x]^2])/(15*a^2*f) - (Cos[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2])/(5*a*f)

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Rubi [A]  time = 0.138716, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4134, 462, 453, 264} \[ -\frac{\left (15 a^2+20 a b+8 b^2\right ) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{15 a^3 f}+\frac{2 (5 a+2 b) \cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{15 a^2 f}-\frac{\cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{5 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((15*a^2 + 20*a*b + 8*b^2)*Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(15*a^3*f) + (2*(5*a + 2*b)*Cos[e + f*x]^
3*Sqrt[a + b*Sec[e + f*x]^2])/(15*a^2*f) - (Cos[e + f*x]^5*Sqrt[a + b*Sec[e + f*x]^2])/(5*a*f)

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

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 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 264

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

Rubi steps

\begin{align*} \int \frac{\sin ^5(e+f x)}{\sqrt{a+b \sec ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2}{x^6 \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{-2 (5 a+2 b)+5 a x^2}{x^4 \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac{2 (5 a+2 b) \cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{15 a^2 f}-\frac{\cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{5 a f}+\frac{\left (15 a^2+20 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x^2}} \, dx,x,\sec (e+f x)\right )}{15 a^2 f}\\ &=-\frac{\left (15 a^2+20 a b+8 b^2\right ) \cos (e+f x) \sqrt{a+b \sec ^2(e+f x)}}{15 a^3 f}+\frac{2 (5 a+2 b) \cos ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{15 a^2 f}-\frac{\cos ^5(e+f x) \sqrt{a+b \sec ^2(e+f x)}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 0.980078, size = 93, normalized size = 0.76 \[ -\frac{\sec (e+f x) (a \cos (2 (e+f x))+a+2 b) \left (3 a^2 \cos (4 (e+f x))+89 a^2-4 a (7 a+4 b) \cos (2 (e+f x))+144 a b+64 b^2\right )}{240 a^3 f \sqrt{a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + 2*b + a*Cos[2*(e + f*x)])*(89*a^2 + 144*a*b + 64*b^2 - 4*a*(7*a + 4*b)*Cos[2*(e + f*x)] + 3*a^2*Cos[4*(
e + f*x)])*Sec[e + f*x])/(240*a^3*f*Sqrt[a + b*Sec[e + f*x]^2])

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Maple [A]  time = 0.427, size = 105, normalized size = 0.9 \begin{align*} -{\frac{ \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}{a}^{2}-10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}{a}^{2}-4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}ab+15\,{a}^{2}+20\,ab+8\,{b}^{2} \right ) }{15\,f{a}^{3}\cos \left ( fx+e \right ) }{\frac{1}{\sqrt{{\frac{b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15/f/a^3*(b+a*cos(f*x+e)^2)*(3*cos(f*x+e)^4*a^2-10*cos(f*x+e)^2*a^2-4*cos(f*x+e)^2*a*b+15*a^2+20*a*b+8*b^2)
/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)

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Maxima [A]  time = 1.21052, size = 219, normalized size = 1.78 \begin{align*} -\frac{\frac{15 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a} - \frac{10 \,{\left ({\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{2}} + \frac{3 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{5}{2}} \cos \left (f x + e\right )^{5} - 10 \,{\left (a + \frac{b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac{3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt{a + \frac{b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{3}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a - 10*((a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 3*sqrt
(a + b/cos(f*x + e)^2)*b*cos(f*x + e))/a^2 + (3*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 10*(a + b/cos(f*
x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 15*sqrt(a + b/cos(f*x + e)^2)*b^2*cos(f*x + e))/a^3)/f

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Fricas [A]  time = 0.573482, size = 213, normalized size = 1.73 \begin{align*} -\frac{{\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 2 \,{\left (5 \, a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{3} +{\left (15 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt{\frac{a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*a^2*cos(f*x + e)^5 - 2*(5*a^2 + 2*a*b)*cos(f*x + e)^3 + (15*a^2 + 20*a*b + 8*b^2)*cos(f*x + e))*sqrt(
(a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^3*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*sec(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 \frac{\sin \left (f x + e\right )^{5}}{\sqrt{b \sec \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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