3.4.27 \(\int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) [327]

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

[Out]

1/2*a*arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/f/(a+b)^(1/2)+1/2*sec(f*x+e)*(a+b*sin(f*x+e)^2)
^(1/2)*tan(f*x+e)/f

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Rubi [A]
time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3269, 386, 385, 212} \begin {gather*} \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} \end {gather*}

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 212

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

Rule 385

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 386

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 3269

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]

Rubi steps

\begin {align*} \int \sec ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=\frac {\text {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 \text {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 \text {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*}

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Mathematica [A]
time = 1.52, size = 164, normalized size = 2.00 \begin {gather*} \frac {\sin (e+f x) \left (\sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {\frac {(a+b) \sin ^2(e+f x)}{a}}}{\sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}}+(2 a+b-b \cos (2 (e+f x))) \sec ^2(e+f x) \sqrt {\frac {(a+b) \sin ^2(e+f x)}{a}}\right )}{4 f \sqrt {\frac {(a+b) \sin ^2(e+f x)}{a}} \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(289\) vs. \(2(70)=140\).
time = 24.38, size = 290, normalized size = 3.54

method	result	size
default	\(\frac {2 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {a +b}\, b \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {a +b}\, \sin \left (f x +e \right )+a \left (\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a +\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a -\ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{4 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{2} f}\)	\(290\)

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,method=_RETURNVERBOSE)

[Out]

1/4*(2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(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/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a+ln(2/(sin(f*x+
e)-1)*((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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (70) = 140\).
time = 0.49, size = 337, normalized size = 4.11 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sec ^{3}{\left (e + f x \right )}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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