3.481 \(\int \sec ^2(c+d x) \sqrt{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=149 \[ \frac{\tan (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}-\frac{\sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

[Out]

-((EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(d*Sqrt[(a + b*Sin[c + d*x])/(a + b)
])) + (a*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c
+ d*x]]) + (Sqrt[a + b*Sin[c + d*x]]*Tan[c + d*x])/d

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Rubi [A]  time = 0.170638, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2690, 12, 2752, 2663, 2661, 2655, 2653} \[ \frac{\tan (c+d x) \sqrt{a+b \sin (c+d x)}}{d}+\frac{a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}-\frac{\sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

-((EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(d*Sqrt[(a + b*Sin[c + d*x])/(a + b)
])) + (a*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(d*Sqrt[a + b*Sin[c
+ d*x]]) + (Sqrt[a + b*Sin[c + d*x]]*Tan[c + d*x])/d

Rule 2690

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[((g*C
os[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*Sin[e + f*x])/(f*g*(p + 1)), x] + Dist[1/(g^2*(p + 1)), Int[(g*Cos
[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1)*(a*(p + 2) + b*(m + p + 2)*Sin[e + f*x]), x], x] /; FreeQ[{a,
b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[0, m, 1] && LtQ[p, -1] && (IntegersQ[2*m, 2*p] || IntegerQ[m])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

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

Mathematica [A]  time = 2.78513, size = 127, normalized size = 0.85 \[ \frac{\tan (c+d x) (a+b \sin (c+d x))-a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+(a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]],x]

[Out]

((a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - a*EllipticF[(-2*
c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + (a + b*Sin[c + d*x])*Tan[c + d*x])/(d*S
qrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 0.486, size = 617, normalized size = 4.1 \begin{align*} -{\frac{1}{bd\cos \left ( dx+c \right ) }\sqrt{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) b+a \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ({\it EllipticF} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) \sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}ab-{\it EllipticF} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) \sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}{b}^{2}-\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){a}^{2}+\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a-b}}-{\frac{b}{a-b}}}\sqrt{-{\frac{b\sin \left ( dx+c \right ) }{a+b}}+{\frac{b}{a+b}}}\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}}{\it EllipticE} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){b}^{2}+{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-ab\sin \left ( dx+c \right ) -{b}^{2} \right ){\frac{1}{\sqrt{- \left ( a+b\sin \left ( dx+c \right ) \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{a+b\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(c + d*x))*sec(c + d*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^2, x)