3.466 \(\int \frac{\sin ^{\frac{3}{2}}(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx\)

Optimal. Leaf size=363 \[ -\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}+\frac{\sqrt{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}+1\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}} \]

[Out]

(Sqrt[b]*ArcTan[1 - (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/(Sqrt[b]*Sqrt[Sin[e + f*x]])])/(4*Sqrt[2]*f*Sqrt[b*Cos[e +
f*x]]*Sqrt[b*Sec[e + f*x]]) - (Sqrt[b]*ArcTan[1 + (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/(Sqrt[b]*Sqrt[Sin[e + f*x]])]
)/(4*Sqrt[2]*f*Sqrt[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (Sqrt[b]*Log[Sqrt[b] + Sqrt[b]*Cot[e + f*x] - (Sqr
t[2]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[e + f*x]]])/(8*Sqrt[2]*f*Sqrt[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (Sqr
t[b]*Log[Sqrt[b] + Sqrt[b]*Cot[e + f*x] + (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[e + f*x]]])/(8*Sqrt[2]*f*Sqr
t[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (b*Sqrt[Sin[e + f*x]])/(2*f*(b*Sec[e + f*x])^(3/2))

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Rubi [A]  time = 0.267997, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {2583, 2585, 2575, 297, 1162, 617, 204, 1165, 628} \[ -\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}+\frac{\sqrt{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}+1\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}+\sqrt{b}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTan[1 - (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/(Sqrt[b]*Sqrt[Sin[e + f*x]])])/(4*Sqrt[2]*f*Sqrt[b*Cos[e +
f*x]]*Sqrt[b*Sec[e + f*x]]) - (Sqrt[b]*ArcTan[1 + (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/(Sqrt[b]*Sqrt[Sin[e + f*x]])]
)/(4*Sqrt[2]*f*Sqrt[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (Sqrt[b]*Log[Sqrt[b] + Sqrt[b]*Cot[e + f*x] - (Sqr
t[2]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[e + f*x]]])/(8*Sqrt[2]*f*Sqrt[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) + (Sqr
t[b]*Log[Sqrt[b] + Sqrt[b]*Cot[e + f*x] + (Sqrt[2]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[e + f*x]]])/(8*Sqrt[2]*f*Sqr
t[b*Cos[e + f*x]]*Sqrt[b*Sec[e + f*x]]) - (b*Sqrt[Sin[e + f*x]])/(2*f*(b*Sec[e + f*x])^(3/2))

Rule 2583

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

Rule 2585

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

Rule 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{k = Denomina
tor[m]}, -Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^{\frac{3}{2}}(e+f x)}{\sqrt{b \sec (e+f x)}} \, dx &=-\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}+\frac{1}{4} \int \frac{1}{\sqrt{b \sec (e+f x)} \sqrt{\sin (e+f x)}} \, dx\\ &=-\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}+\frac{\int \frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}} \, dx}{4 \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}-\frac{b \operatorname{Subst}\left (\int \frac{x^2}{b^2+x^4} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{2 f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}+\frac{b \operatorname{Subst}\left (\int \frac{b-x^2}{b^2+x^4} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{4 f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \operatorname{Subst}\left (\int \frac{b+x^2}{b^2+x^4} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{4 f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{b}+2 x}{-b-\sqrt{2} \sqrt{b} x-x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{b}-2 x}{-b+\sqrt{2} \sqrt{b} x-x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \operatorname{Subst}\left (\int \frac{1}{b-\sqrt{2} \sqrt{b} x+x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \operatorname{Subst}\left (\int \frac{1}{b+\sqrt{2} \sqrt{b} x+x^2} \, dx,x,\frac{\sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=-\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}-\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}\\ &=\frac{\sqrt{b} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{b} \sqrt{\sin (e+f x)}}\right )}{4 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)-\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}+\frac{\sqrt{b} \log \left (\sqrt{b}+\sqrt{b} \cot (e+f x)+\frac{\sqrt{2} \sqrt{b \cos (e+f x)}}{\sqrt{\sin (e+f x)}}\right )}{8 \sqrt{2} f \sqrt{b \cos (e+f x)} \sqrt{b \sec (e+f x)}}-\frac{b \sqrt{\sin (e+f x)}}{2 f (b \sec (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.54861, size = 218, normalized size = 0.6 \[ -\frac{\sqrt{\sin (e+f x)} \sqrt{b \sec (e+f x)} \left (2 \sqrt{2} \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}\right )-2 \sqrt{2} \tan ^{-1}\left (\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}+1\right )+4 \sqrt [4]{\tan ^2(e+f x)}+\sqrt{2} \log \left (\sqrt{\tan ^2(e+f x)}-\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}+1\right )-\sqrt{2} \log \left (\sqrt{\tan ^2(e+f x)}+\sqrt{2} \sqrt [4]{\tan ^2(e+f x)}+1\right )+4 \cos (2 (e+f x)) \sqrt [4]{\tan ^2(e+f x)}\right )}{16 b f \sqrt [4]{\tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[b*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]*(2*Sqrt[2]*ArcTan[1 - Sqrt[2]*(Tan[e + f*x]^2)^(1/4)] - 2*Sqrt[2]*Ar
cTan[1 + Sqrt[2]*(Tan[e + f*x]^2)^(1/4)] + Sqrt[2]*Log[1 - Sqrt[2]*(Tan[e + f*x]^2)^(1/4) + Sqrt[Tan[e + f*x]^
2]] - Sqrt[2]*Log[1 + Sqrt[2]*(Tan[e + f*x]^2)^(1/4) + Sqrt[Tan[e + f*x]^2]] + 4*(Tan[e + f*x]^2)^(1/4) + 4*Co
s[2*(e + f*x)]*(Tan[e + f*x]^2)^(1/4)))/(16*b*f*(Tan[e + f*x]^2)^(1/4))

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Maple [C]  time = 0.129, size = 648, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^(3/2)/(b*sec(f*x+e))^(1/2),x)

[Out]

1/8/f*2^(1/2)*(I*sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e
))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,
1/2*2^(1/2))-I*sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))
^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/
2*2^(1/2))-sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/
2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^
(1/2))-sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(
(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticPi(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2
))+2*sin(f*x+e)*((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-
1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF(((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))-2*2^(1/2)*
cos(f*x+e)^3+2*2^(1/2)*cos(f*x+e)^2)*sin(f*x+e)^(1/2)/(-1+cos(f*x+e))/(b/cos(f*x+e))^(1/2)/cos(f*x+e)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{\frac{3}{2}}}{\sqrt{b \sec \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(3/2)/(b*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [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)^(3/2)/(b*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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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)**(3/2)/(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{\sin \left (f x + e\right )^{\frac{3}{2}}}{\sqrt{b \sec \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^(3/2)/(b*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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