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

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

[Out]

-1/15*(15*a^2+20*a*b+8*b^2)*cos(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/a^3/f+2/15*(5*a+2*b)*cos(f*x+e)^3*(a+b*sec(f*x
+e)^2)^(1/2)/a^2/f-1/5*cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2)/a/f

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Rubi [A]  time = 0.14, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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]

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 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 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])

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*}

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Mathematica [A]  time = 0.93, 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]

-1/240*((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*C
os[4*(e + f*x)])*Sec[e + f*x])/(a^3*f*Sqrt[a + b*Sec[e + f*x]^2])

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fricas [A]  time = 0.54, size = 87, normalized size = 0.71 \[ -\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 2 \, {\left (5 \, a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, a^{3} f} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)2/f*2/15*((320*a+320*b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/
2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^7+sqrt(a+b)*(640*a-320*b)*(-tan((f*x+exp(1))/2)
^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1
))/2)^2+a+b))^6+(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan
((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^5*(-832*a^2-960*b^2-2560*a*b)+(-tan((f*x+exp(1))/2)^2*sqrt(
a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+
a+b))*(2880*a^4-320*b^4-1920*a*b^3+4480*a^3*b)+sqrt(a+b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp
(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^4*(-2560*a^2+960*b
^2-1600*a*b)+(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f
*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^3*(-320*a^3+960*b^3+4160*a*b^2+2880*a^2*b)+sqrt(a+b)*(-tan((f*
x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*ta
n((f*x+exp(1))/2)^2+a+b))^2*(3200*a^3-960*b^3+1280*a*b^2+5440*a^2*b)+sqrt(a+b)*(768*a^4+320*b^4-320*a*b^3-832*
a^2*b^2+576*a^3*b))/(2*sqrt(a+b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp
(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))-(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*t
an((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^2+3*a-b
)^5/sign(tan((f*x+exp(1))/2)^2-1)

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maple [A]  time = 1.96, size = 105, normalized size = 0.85 \[ -\frac {\left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right ) \left (3 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2}-10 a^{2} \left (\cos ^{2}\left (f x +e \right )\right )-4 \left (\cos ^{2}\left (f x +e \right )\right ) a b +15 a^{2}+20 a b +8 b^{2}\right )}{15 f \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) a^{3}} \]

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

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maxima [A]  time = 0.35, size = 162, normalized size = 1.32 \[ -\frac {\frac {15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a} - \frac {10 \, {\left ({\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 3 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{2}} + \frac {3 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 10 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{3}}}{15 \, f} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (e+f\,x\right )}^5}{\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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