3.371 \(\int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx\)

Optimal. Leaf size=85 \[ \frac {2 b^7}{13 f (b \sec (e+f x))^{13/2}}-\frac {2 b^5}{3 f (b \sec (e+f x))^{9/2}}+\frac {6 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

[Out]

2/13*b^7/f/(b*sec(f*x+e))^(13/2)-2/3*b^5/f/(b*sec(f*x+e))^(9/2)+6/5*b^3/f/(b*sec(f*x+e))^(5/2)-2*b/f/(b*sec(f*
x+e))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2622, 270} \[ \frac {2 b^7}{13 f (b \sec (e+f x))^{13/2}}-\frac {2 b^5}{3 f (b \sec (e+f x))^{9/2}}+\frac {6 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^7,x]

[Out]

(2*b^7)/(13*f*(b*Sec[e + f*x])^(13/2)) - (2*b^5)/(3*f*(b*Sec[e + f*x])^(9/2)) + (6*b^3)/(5*f*(b*Sec[e + f*x])^
(5/2)) - (2*b)/(f*Sqrt[b*Sec[e + f*x]])

Rule 270

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

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \sqrt {b \sec (e+f x)} \sin ^7(e+f x) \, dx &=\frac {b^7 \operatorname {Subst}\left (\int \frac {\left (-1+\frac {x^2}{b^2}\right )^3}{x^{15/2}} \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {b^7 \operatorname {Subst}\left (\int \left (-\frac {1}{x^{15/2}}+\frac {3}{b^2 x^{11/2}}-\frac {3}{b^4 x^{7/2}}+\frac {1}{b^6 x^{3/2}}\right ) \, dx,x,b \sec (e+f x)\right )}{f}\\ &=\frac {2 b^7}{13 f (b \sec (e+f x))^{13/2}}-\frac {2 b^5}{3 f (b \sec (e+f x))^{9/2}}+\frac {6 b^3}{5 f (b \sec (e+f x))^{5/2}}-\frac {2 b}{f \sqrt {b \sec (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 58, normalized size = 0.68 \[ \frac {(-8939 \cos (e+f x)+887 \cos (3 (e+f x))-155 \cos (5 (e+f x))+15 \cos (7 (e+f x))) \sqrt {b \sec (e+f x)}}{6240 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^7,x]

[Out]

((-8939*Cos[e + f*x] + 887*Cos[3*(e + f*x)] - 155*Cos[5*(e + f*x)] + 15*Cos[7*(e + f*x)])*Sqrt[b*Sec[e + f*x]]
)/(6240*f)

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fricas [A]  time = 0.81, size = 56, normalized size = 0.66 \[ \frac {2 \, {\left (15 \, \cos \left (f x + e\right )^{7} - 65 \, \cos \left (f x + e\right )^{5} + 117 \, \cos \left (f x + e\right )^{3} - 195 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{195 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7*(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/195*(15*cos(f*x + e)^7 - 65*cos(f*x + e)^5 + 117*cos(f*x + e)^3 - 195*cos(f*x + e))*sqrt(b/cos(f*x + e))/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)^7*(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)-256/195*(390*b^2*sqrt(-b)*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^9+702*b^
3*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^8-1716*b^3*sqrt(-b)*(-sqrt(-b)*tan(1/
2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^7-1716*b^4*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*ta
n(1/2*(f*x+exp(1)))^4+b))^6+1872*b^4*sqrt(-b)*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))
^4+b))^5+1040*b^5*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^4+26*b^6*sqrt(-b)*(-s
qrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))-572*b^5*sqrt(-b)*(-sqrt(-b)*tan(1/2*(f*x+e
xp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b))^3-156*b^6*(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*
x+exp(1)))^4+b))^2+2*b^7)*sign(cos(f*x+exp(1)))/(-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)
))^4+b)-sqrt(-b))^13/f

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maple [B]  time = 0.38, size = 517, normalized size = 6.08 \[ \frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (60 \left (\cos ^{7}\left (f x +e \right )\right )-260 \left (\cos ^{5}\left (f x +e \right )\right )+468 \left (\cos ^{3}\left (f x +e \right )\right )+195 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )-195 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )+195 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )-195 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )-780 \cos \left (f x +e \right )\right ) \left (\cos \left (f x +e \right )+1\right )^{2} \sqrt {\frac {b}{\cos \left (f x +e \right )}}}{390 f \sin \left (f x +e \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^7*(b*sec(f*x+e))^(1/2),x)

[Out]

1/390/f*(-1+cos(f*x+e))^2*(60*cos(f*x+e)^7-260*cos(f*x+e)^5+468*cos(f*x+e)^3+195*cos(f*x+e)*(-cos(f*x+e)/(cos(
f*x+e)+1)^2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-co
s(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-195*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-(2*c
os(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1
/2)-1)/sin(f*x+e)^2)+195*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-195*(-cos(f*x+e)/
(cos(f*x+e)+1)^2)^(1/2)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)-780*cos(f*x+e))*(cos(f*x+e)+1)^2*(b/cos(f*x+e))^(1/2)/sin
(f*x+e)^4

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maxima [A]  time = 0.32, size = 63, normalized size = 0.74 \[ \frac {2 \, {\left (15 \, b^{6} - \frac {65 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {117 \, b^{6}}{\cos \left (f x + e\right )^{4}} - \frac {195 \, b^{6}}{\cos \left (f x + e\right )^{6}}\right )} b}{195 \, f \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^7*(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

2/195*(15*b^6 - 65*b^6/cos(f*x + e)^2 + 117*b^6/cos(f*x + e)^4 - 195*b^6/cos(f*x + e)^6)*b/(f*(b/cos(f*x + e))
^(13/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^7\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^7*(b/cos(e + f*x))^(1/2),x)

[Out]

int(sin(e + f*x)^7*(b/cos(e + f*x))^(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)**7*(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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