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

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Rubi [A]  time = 0.27, antiderivative size = 158, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 6.23, 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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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right )} \sec \left (f x + e\right )^{2}}{d \sin \left (f x + e\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sin \left (f x + e\right ) + a} \sec \left (f x + e\right )^{2}}{\sqrt {d \sin \left (f x + e\right )}}\,{d x} \]

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)

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maple [B]  time = 0.72, size = 666, normalized size = 4.22 \[ -\frac {\left (\sqrt {-\frac {-\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )-b \sin \left (f x +e \right )+\cos \left (f x +e \right ) a -a}{\left (b +\sqrt {-a^{2}+b^{2}}\right ) \sin \left (f x +e \right )}}\, \sqrt {\frac {\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )-b \sin \left (f x +e \right )+\cos \left (f x +e \right ) a -a}{\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )}}\, \sqrt {\frac {a \left (-1+\cos \left (f x +e \right )\right )}{\left (b +\sqrt {-a^{2}+b^{2}}\right ) \sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {-\frac {-\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )-b \sin \left (f x +e \right )+\cos \left (f x +e \right ) a -a}{\left (b +\sqrt {-a^{2}+b^{2}}\right ) \sin \left (f x +e \right )}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}}}}{2}\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {-a^{2}+b^{2}}+\sqrt {-\frac {-\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )-b \sin \left (f x +e \right )+\cos \left (f x +e \right ) a -a}{\left (b +\sqrt {-a^{2}+b^{2}}\right ) \sin \left (f x +e \right )}}\, \sqrt {\frac {\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )-b \sin \left (f x +e \right )+\cos \left (f x +e \right ) a -a}{\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )}}\, \sqrt {\frac {a \left (-1+\cos \left (f x +e \right )\right )}{\left (b +\sqrt {-a^{2}+b^{2}}\right ) \sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {-\frac {-\sqrt {-a^{2}+b^{2}}\, \sin \left (f x +e \right )-b \sin \left (f x +e \right )+\cos \left (f x +e \right ) a -a}{\left (b +\sqrt {-a^{2}+b^{2}}\right ) \sin \left (f x +e \right )}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-a^{2}+b^{2}}}{\sqrt {-a^{2}+b^{2}}}}}{2}\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -\sqrt {2}\, \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -\sqrt {2}\, \cos \left (f x +e \right ) a +\sqrt {2}\, b \sin \left (f x +e \right )+\sqrt {2}\, a \right ) \sin \left (f x +e \right ) \sqrt {2}}{2 f \left (-1+\cos \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {d \sin \left (f x +e \right )}\, \sqrt {a +b \sin \left (f x +e \right )}} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b \sin \left (f x + e\right ) + a} \sec \left (f x + e\right )^{2}}{\sqrt {d \sin \left (f x + e\right )}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + b \sin {\left (e + f x \right )}} \sec ^{2}{\left (e + f x \right )}}{\sqrt {d \sin {\left (e + f x \right )}}}\, dx \]

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]

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

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