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

Optimal. Leaf size=82 \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{2 f \sqrt{a+b}}+\frac{\tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{2 f} \]

[Out]

(a*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2*Sqrt[a + b]*f) + (Sec[e + f*x]*Sqrt[a +
b*Sin[e + f*x]^2]*Tan[e + f*x])/(2*f)

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Rubi [A]  time = 0.0941853, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3190, 378, 377, 206} \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{a+b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{2 f \sqrt{a+b}}+\frac{\tan (e+f x) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(2*Sqrt[a + b]*f) + (Sec[e + f*x]*Sqrt[a +
b*Sin[e + f*x]^2]*Tan[e + f*x])/(2*f)

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 2.329, size = 164, normalized size = 2. \[ \frac{\sin (e+f x) \left (\sqrt{2} a \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} \tanh ^{-1}\left (\frac{\sqrt{\frac{(a+b) \sin ^2(e+f x)}{a}}}{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}\right )+\sec ^2(e+f x) \sqrt{\frac{(a+b) \sin ^2(e+f x)}{a}} (2 a-b \cos (2 (e+f x))+b)\right )}{4 f \sqrt{\frac{(a+b) \sin ^2(e+f x)}{a}} \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sin[e + f*x]*(Sqrt[2]*a*ArcTanh[Sqrt[((a + b)*Sin[e + f*x]^2)/a]/Sqrt[1 + (b*Sin[e + f*x]^2)/a]]*Sqrt[(2*a +
b - b*Cos[2*(e + f*x)])/a] + (2*a + b - b*Cos[2*(e + f*x)])*Sec[e + f*x]^2*Sqrt[((a + b)*Sin[e + f*x]^2)/a]))/
(4*f*Sqrt[((a + b)*Sin[e + f*x]^2)/a]*Sqrt[a + b*Sin[e + f*x]^2])

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Maple [B]  time = 3.897, size = 290, normalized size = 3.5 \begin{align*}{\frac{1}{4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}f} \left ( 2\,\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}b\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\, \left ( a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{3/2}\sqrt{a+b}\sin \left ( fx+e \right ) +a \left ( \ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ) a+\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ) b-\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ) a-\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ) b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \left ( a+b \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b*sin(f*x+e)*cos(f*x+e)^2+2*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(1/
2)*sin(f*x+e)+a*(ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a+ln(2/(-1+sin(
f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos
(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a-ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a
))*b)*cos(f*x+e)^2)/(a+b)^(3/2)/cos(f*x+e)^2/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.74399, size = 852, normalized size = 10.39 \begin{align*} \left [\frac{\sqrt{a + b} a \cos \left (f x + e\right )^{2} \log \left (\frac{{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \,{\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}{\left (a + b\right )} \sin \left (f x + e\right )}{8 \,{\left (a + b\right )} f \cos \left (f x + e\right )^{2}}, -\frac{a \sqrt{-a - b} \arctan \left (\frac{{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{2 \,{\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}{\left (a + b\right )} \sin \left (f x + e\right )}{4 \,{\left (a + b\right )} f \cos \left (f x + e\right )^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(a + b)*a*cos(f*x + e)^2*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b + 2*b^2)*cos(f*x
+ e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*sin(f*x + e) +
8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*sin(f*x + e))/((a + b)*f*c
os(f*x + e)^2), -1/4*(a*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2
+ a + b)*sqrt(-a - b)/(((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e)))*cos(f*x + e)^2 - 2*sqrt
(-b*cos(f*x + e)^2 + a + b)*(a + b)*sin(f*x + e))/((a + b)*f*cos(f*x + e)^2)]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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