3.471 \(\int \frac{1}{\sqrt{b \sec (e+f x)} \sin ^{\frac{17}{2}}(e+f x)} \, dx\)

Optimal. Leaf size=121 \[ -\frac{256 b}{1155 f \sin ^{\frac{3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{64 b}{385 f \sin ^{\frac{7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{8 b}{55 f \sin ^{\frac{11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{2 b}{15 f \sin ^{\frac{15}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

[Out]

(-2*b)/(15*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(15/2)) - (8*b)/(55*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(11
/2)) - (64*b)/(385*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7/2)) - (256*b)/(1155*f*(b*Sec[e + f*x])^(3/2)*Sin[e
+ f*x]^(3/2))

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Rubi [A]  time = 0.162258, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2584, 2578} \[ -\frac{256 b}{1155 f \sin ^{\frac{3}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{64 b}{385 f \sin ^{\frac{7}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{8 b}{55 f \sin ^{\frac{11}{2}}(e+f x) (b \sec (e+f x))^{3/2}}-\frac{2 b}{15 f \sin ^{\frac{15}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(17/2)),x]

[Out]

(-2*b)/(15*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(15/2)) - (8*b)/(55*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(11
/2)) - (64*b)/(385*f*(b*Sec[e + f*x])^(3/2)*Sin[e + f*x]^(7/2)) - (256*b)/(1155*f*(b*Sec[e + f*x])^(3/2)*Sin[e
+ f*x]^(3/2))

Rule 2584

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(a*Sin[e +
f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + 1)), x] + Dist[(m - n + 2)/(a^2*(m + 1)), Int[(a*Sin[e + f*
x])^(m + 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2578

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m - n + 2,
0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{b \sec (e+f x)} \sin ^{\frac{17}{2}}(e+f x)} \, dx &=-\frac{2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac{15}{2}}(e+f x)}+\frac{4}{5} \int \frac{1}{\sqrt{b \sec (e+f x)} \sin ^{\frac{13}{2}}(e+f x)} \, dx\\ &=-\frac{2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac{15}{2}}(e+f x)}-\frac{8 b}{55 f (b \sec (e+f x))^{3/2} \sin ^{\frac{11}{2}}(e+f x)}+\frac{32}{55} \int \frac{1}{\sqrt{b \sec (e+f x)} \sin ^{\frac{9}{2}}(e+f x)} \, dx\\ &=-\frac{2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac{15}{2}}(e+f x)}-\frac{8 b}{55 f (b \sec (e+f x))^{3/2} \sin ^{\frac{11}{2}}(e+f x)}-\frac{64 b}{385 f (b \sec (e+f x))^{3/2} \sin ^{\frac{7}{2}}(e+f x)}+\frac{128}{385} \int \frac{1}{\sqrt{b \sec (e+f x)} \sin ^{\frac{5}{2}}(e+f x)} \, dx\\ &=-\frac{2 b}{15 f (b \sec (e+f x))^{3/2} \sin ^{\frac{15}{2}}(e+f x)}-\frac{8 b}{55 f (b \sec (e+f x))^{3/2} \sin ^{\frac{11}{2}}(e+f x)}-\frac{64 b}{385 f (b \sec (e+f x))^{3/2} \sin ^{\frac{7}{2}}(e+f x)}-\frac{256 b}{1155 f (b \sec (e+f x))^{3/2} \sin ^{\frac{3}{2}}(e+f x)}\\ \end{align*}

Mathematica [A]  time = 0.254319, size = 62, normalized size = 0.51 \[ \frac{2 b (150 \cos (2 (e+f x))-36 \cos (4 (e+f x))+4 \cos (6 (e+f x))-195)}{1155 f \sin ^{\frac{15}{2}}(e+f x) (b \sec (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[b*Sec[e + f*x]]*Sin[e + f*x]^(17/2)),x]

[Out]

(2*b*(-195 + 150*Cos[2*(e + f*x)] - 36*Cos[4*(e + f*x)] + 4*Cos[6*(e + f*x)]))/(1155*f*(b*Sec[e + f*x])^(3/2)*
Sin[e + f*x]^(15/2))

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Maple [A]  time = 0.15, size = 102, normalized size = 0.8 \begin{align*}{\frac{512\,\cos \left ( fx+e \right ) \left ( 128\, \left ( \cos \left ( fx+e \right ) \right ) ^{6}-480\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+660\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-385 \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{8}}{1155\,f \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{2}+ \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\cos \left ( fx+e \right ) +1 \right ) ^{8}} \left ( \sin \left ( fx+e \right ) \right ) ^{-{\frac{15}{2}}}{\frac{1}{\sqrt{{\frac{b}{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x)

[Out]

512/1155/f*cos(f*x+e)*(128*cos(f*x+e)^6-480*cos(f*x+e)^4+660*cos(f*x+e)^2-385)*(-1+cos(f*x+e))^8/sin(f*x+e)^(1
5/2)/(sin(f*x+e)^2+cos(f*x+e)^2-2*cos(f*x+e)+1)^8/(b/cos(f*x+e))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac{17}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(17/2)), x)

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Fricas [A]  time = 3.01009, size = 308, normalized size = 2.55 \begin{align*} \frac{2 \,{\left (128 \, \cos \left (f x + e\right )^{8} - 480 \, \cos \left (f x + e\right )^{6} + 660 \, \cos \left (f x + e\right )^{4} - 385 \, \cos \left (f x + e\right )^{2}\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}} \sqrt{\sin \left (f x + e\right )}}{1155 \,{\left (b f \cos \left (f x + e\right )^{8} - 4 \, b f \cos \left (f x + e\right )^{6} + 6 \, b f \cos \left (f x + e\right )^{4} - 4 \, b f \cos \left (f x + e\right )^{2} + b f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/1155*(128*cos(f*x + e)^8 - 480*cos(f*x + e)^6 + 660*cos(f*x + e)^4 - 385*cos(f*x + e)^2)*sqrt(b/cos(f*x + e)
)*sqrt(sin(f*x + e))/(b*f*cos(f*x + e)^8 - 4*b*f*cos(f*x + e)^6 + 6*b*f*cos(f*x + e)^4 - 4*b*f*cos(f*x + e)^2
+ b*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(1/sin(f*x+e)**(17/2)/(b*sec(f*x+e))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sec \left (f x + e\right )} \sin \left (f x + e\right )^{\frac{17}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(f*x+e)^(17/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(b*sec(f*x + e))*sin(f*x + e)^(17/2)), x)