3.4.85 \(\int (b \sec (e+f x))^{3/2} \sin ^7(e+f x) \, dx\) [385]

Optimal. Leaf size=83 \[ \frac {2 b^7}{11 f (b \sec (e+f x))^{11/2}}-\frac {6 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {2 b^3}{f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[Out]

2/11*b^7/f/(b*sec(f*x+e))^(11/2)-6/7*b^5/f/(b*sec(f*x+e))^(7/2)+2*b^3/f/(b*sec(f*x+e))^(3/2)+2*b*(b*sec(f*x+e)
)^(1/2)/f

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Rubi [A]
time = 0.04, antiderivative size = 83, 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 = {2702, 276} \begin {gather*} \frac {2 b^7}{11 f (b \sec (e+f x))^{11/2}}-\frac {6 b^5}{7 f (b \sec (e+f x))^{7/2}}+\frac {2 b^3}{f (b \sec (e+f x))^{3/2}}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^7,x]

[Out]

(2*b^7)/(11*f*(b*Sec[e + f*x])^(11/2)) - (6*b^5)/(7*f*(b*Sec[e + f*x])^(7/2)) + (2*b^3)/(f*(b*Sec[e + f*x])^(3
/2)) + (2*b*Sqrt[b*Sec[e + f*x]])/f

Rule 276

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 2702

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

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Mathematica [A]
time = 0.10, size = 52, normalized size = 0.63 \begin {gather*} \frac {b (3370+809 \cos (2 (e+f x))-90 \cos (4 (e+f x))+7 \cos (6 (e+f x))) \sqrt {b \sec (e+f x)}}{1232 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^7,x]

[Out]

(b*(3370 + 809*Cos[2*(e + f*x)] - 90*Cos[4*(e + f*x)] + 7*Cos[6*(e + f*x)])*Sqrt[b*Sec[e + f*x]])/(1232*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(968\) vs. \(2(71)=142\).
time = 1.63, size = 969, normalized size = 11.67

method	result	size
default	\(\text {Expression too large to display}\)	\(969\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x,method=_RETURNVERBOSE)

[Out]

1/154/f*(cos(f*x+e)+1)^2*(-1+cos(f*x+e))^2*(77*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)-77*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)+28*cos(f*x
+e)^7+231*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)-231*cos(f*x+e)^2*(-
cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)+231*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^
(3/2)*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)-231*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*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)-132*cos(f*x+e)^5+77*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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)-77*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)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+308*cos(f*x+e)^3+308*cos(f*x+e))*(b/cos(f*x+e))^(3/2)/s
in(f*x+e)^4

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Maxima [A]
time = 0.31, size = 76, normalized size = 0.92 \begin {gather*} \frac {2 \, b {\left (\frac {7 \, b^{6}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {11}{2}}} - \frac {33 \, b^{4}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {7}{2}}} + \frac {77 \, b^{2}}{\left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}} + 77 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )}}{77 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x, algorithm="maxima")

[Out]

2/77*b*(7*b^6/(b/cos(f*x + e))^(11/2) - 33*b^4/(b/cos(f*x + e))^(7/2) + 77*b^2/(b/cos(f*x + e))^(3/2) + 77*sqr
t(b/cos(f*x + e)))/f

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Fricas [A]
time = 0.38, size = 58, normalized size = 0.70 \begin {gather*} \frac {2 \, {\left (7 \, b \cos \left (f x + e\right )^{6} - 33 \, b \cos \left (f x + e\right )^{4} + 77 \, b \cos \left (f x + e\right )^{2} + 77 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{77 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x, algorithm="fricas")

[Out]

2/77*(7*b*cos(f*x + e)^6 - 33*b*cos(f*x + e)^4 + 77*b*cos(f*x + e)^2 + 77*b)*sqrt(b/cos(f*x + e))/f

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**(3/2)*sin(f*x+e)**7,x)

[Out]

Timed out

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Giac [A]
time = 4.00, size = 106, normalized size = 1.28 \begin {gather*} \frac {2 \, {\left (7 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{5} - 33 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )^{3} + 77 \, \sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right ) + \frac {77 \, b^{6}}{\sqrt {b \cos \left (f x + e\right )}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{77 \, b^{4} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^(3/2)*sin(f*x+e)^7,x, algorithm="giac")

[Out]

2/77*(7*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^5 - 33*sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)^3 + 77*sqrt(b*cos(f
*x + e))*b^5*cos(f*x + e) + 77*b^6/sqrt(b*cos(f*x + e)))*sgn(cos(f*x + e))/(b^4*f)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (e+f\,x\right )}^7\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^7*(b/cos(e + f*x))^(3/2),x)

[Out]

int(sin(e + f*x)^7*(b/cos(e + f*x))^(3/2), x)

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