3.1478 \(\int \frac{\sec ^2(e+f x) \sqrt{a+b \sin (e+f x)}}{\sqrt{d \sin (e+f x)}} \, dx\)

Optimal. Leaf size=158 \[ \frac{\sec (e+f x) \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}{d f}-\frac{\sqrt{a+b} \tan (e+f x) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right )}{\sqrt{d} f} \]

[Out]

(Sec[e + f*x]*Sqrt[d*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(d*f) - (Sqrt[a + b]*Sqrt[(a*(1 - Csc[e + f*x]))/
(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticF[ArcSin[(Sqrt[d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]
*Sqrt[d*Sin[e + f*x]])], -((a + b)/(a - b))]*Tan[e + f*x])/(Sqrt[d]*f)

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Rubi [A]  time = 0.265491, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {2888, 2816} \[ \frac{\sec (e+f x) \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}{d f}-\frac{\sqrt{a+b} \tan (e+f x) \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (\csc (e+f x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right )}{\sqrt{d} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*Sqrt[d*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(d*f) - (Sqrt[a + b]*Sqrt[(a*(1 - Csc[e + f*x]))/
(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticF[ArcSin[(Sqrt[d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]
*Sqrt[d*Sin[e + f*x]])], -((a + b)/(a - b))]*Tan[e + f*x])/(Sqrt[d]*f)

Rule 2888

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) +
(f_.)*(x_)]], x_Symbol] :> Simp[(2*(g*Cos[e + f*x])^(p + 1)*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^m)/(d*f*
g*(2*m + 1)), x] + Dist[(2*a*m)/(g^2*(2*m + 1)), Int[((g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1))/S
qrt[d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && EqQ[m + p + 3/2,
0]

Rule 2816

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(-2*
Tan[e + f*x]*Rt[(a + b)/d, 2]*Sqrt[(a*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*Ellipt
icF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[d*Sin[e + f*x]]*Rt[(a + b)/d, 2])], -((a + b)/(a - b))])/(a*f), x] /
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && PosQ[(a + b)/d]

Rubi steps

\begin{align*} \int \frac{\sec ^2(e+f x) \sqrt{a+b \sin (e+f x)}}{\sqrt{d \sin (e+f x)}} \, dx &=\frac{\sec (e+f x) \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}{d f}+\frac{1}{2} a \int \frac{1}{\sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}} \, dx\\ &=\frac{\sec (e+f x) \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}}{d f}-\frac{\sqrt{a+b} \sqrt{\frac{a (1-\csc (e+f x))}{a+b}} \sqrt{\frac{a (1+\csc (e+f x))}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b \sin (e+f x)}}{\sqrt{a+b} \sqrt{d \sin (e+f x)}}\right )|-\frac{a+b}{a-b}\right ) \tan (e+f x)}{\sqrt{d} f}\\ \end{align*}

Mathematica [A]  time = 6.18182, size = 198, normalized size = 1.25 \[ \frac{4 a^2 \sin ^4\left (\frac{1}{4} (2 e+2 f x-\pi )\right ) \sec (e+f x) \sqrt{-\frac{(a+b) \sin (e+f x) (a+b \sin (e+f x))}{a^2 (\sin (e+f x)-1)^2}} \sqrt{-\frac{(a+b) \cot ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{a-b}} F\left (\sin ^{-1}\left (\sqrt{-\frac{a+b \sin (e+f x)}{a (\sin (e+f x)-1)}}\right )|\frac{2 a}{a-b}\right )+(a+b) \tan (e+f x) (a+b \sin (e+f x))}{f (a+b) \sqrt{d \sin (e+f x)} \sqrt{a+b \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*Sqrt[-(((a + b)*Cot[(2*e - Pi + 2*f*x)/4]^2)/(a - b))]*EllipticF[ArcSin[Sqrt[-((a + b*Sin[e + f*x])/(a*
(-1 + Sin[e + f*x])))]], (2*a)/(a - b)]*Sec[e + f*x]*Sqrt[-(((a + b)*Sin[e + f*x]*(a + b*Sin[e + f*x]))/(a^2*(
-1 + Sin[e + f*x])^2))]*Sin[(2*e - Pi + 2*f*x)/4]^4 + (a + b)*(a + b*Sin[e + f*x])*Tan[e + f*x])/((a + b)*f*Sq
rt[d*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])

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Maple [B]  time = 0.481, size = 650, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*2^(1/2)*((((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1
/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)*a-a)/(-a^2+b^2)^(1/2)/sin(f*x+e))^(1/2)*((-1+cos(f*x
+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticF((((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*
a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*sin(f*x
+e)*cos(f*x+e)*(-a^2+b^2)^(1/2)+(((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2)
)/sin(f*x+e))^(1/2)*(((-a^2+b^2)^(1/2)*sin(f*x+e)-b*sin(f*x+e)+cos(f*x+e)*a-a)/(-a^2+b^2)^(1/2)/sin(f*x+e))^(1
/2)*((-1+cos(f*x+e))*a/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2)*EllipticF((((-a^2+b^2)^(1/2)*sin(f*x+e)+b*sin(f*
x+e)-cos(f*x+e)*a+a)/(b+(-a^2+b^2)^(1/2))/sin(f*x+e))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2)
)^(1/2))*sin(f*x+e)*cos(f*x+e)*b-sin(f*x+e)*cos(f*x+e)*2^(1/2)*b+sin(f*x+e)*2^(1/2)*b-cos(f*x+e)*2^(1/2)*a+2^(
1/2)*a)*sin(f*x+e)/(-1+cos(f*x+e))/cos(f*x+e)/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e))*sec(f*x + e)^2/(d*sin(f*x + e)), x)

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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)**2*(a+b*sin(f*x+e))**(1/2)/(d*sin(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{\sqrt{b \sin \left (f x + e\right ) + a} \sec \left (f x + e\right )^{2}}{\sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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