3.491 \(\int \sec ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=168 \[ \frac{\left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}+\frac{\sec (c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{d}-\frac{a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

[Out]

(Sec[c + d*x]*(b + a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/d - (a*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^2 - b^2)*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]])

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Rubi [A]  time = 0.203109, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2691, 2752, 2663, 2661, 2655, 2653} \[ \frac{\left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{a+b \sin (c+d x)}}+\frac{\sec (c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{d}-\frac{a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(Sec[c + d*x]*(b + a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/d - (a*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^2 - b^2)*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]])

Rule 2691

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 - 1)*(b + a*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 - 2)*(b^2*(m - 1) + a^2*(p + 2) + a*b*(m + p + 1)*Sin
[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (IntegersQ[
2*m, 2*p] || IntegerQ[m])

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) (a+b \sin (c+d x))^{3/2} \, dx &=\frac{\sec (c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{d}-\int \frac{\frac{b^2}{2}+\frac{1}{2} a b \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=\frac{\sec (c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{d}-\frac{1}{2} a \int \sqrt{a+b \sin (c+d x)} \, dx-\frac{1}{2} \left (-a^2+b^2\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=\frac{\sec (c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{d}-\frac{\left (a \sqrt{a+b \sin (c+d x)}\right ) \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 (\left (-a^2+b^2\right ) \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{\sec (c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{d}-\frac{a 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{\left (a^2-b^2\right ) 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)}}\\ \end{align*}

Mathematica [A]  time = 0.634687, size = 163, normalized size = 0.97 \[ \frac{-\left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+a^2 \tan (c+d x)+a b \sec (c+d x)+a b \sin (c+d x) \tan (c+d x)+a (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+b^2 \tan (c+d x)}{d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

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

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Maple [B]  time = 0.614, size = 633, normalized size = 3.8 \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 ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{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{b}{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}^{3}-\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{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{b}{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{b}^{2}-\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{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{b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){a}^{2}b+\sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{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{b}{a-b}}}{\it EllipticF} \left ( \sqrt{{\frac{b\sin \left ( dx+c \right ) }{a-b}}+{\frac{a}{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ){b}^{3}-a{b}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+{a}^{2}b\sin \left ( dx+c \right ) +{b}^{3}\sin \left ( dx+c \right ) +2\,a{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))^(3/2),x)

[Out]

1/b*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(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)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+
b))^(1/2))*a^3-(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(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)*EllipticE((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a*b^2-(b/(a-b)*sin(d*x+c)+1/
(a-b)*a)^(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)*EllipticF((b/(a-b)*sin(
d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b+(b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(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)*EllipticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))
^(1/2))*b^3-a*b^2*cos(d*x+c)^2+a^2*b*sin(d*x+c)+b^3*sin(d*x+c)+2*a*b^2)/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1+s
in(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{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \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))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^(3/2)*sec(d*x + c)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a \sec \left (d x + c\right )^{2}\right )} \sqrt{b \sin \left (d x + c\right ) + a}, 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))^(3/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^2*sin(d*x + c) + a*sec(d*x + c)^2)*sqrt(b*sin(d*x + c) + a), 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(d*x+c)**2*(a+b*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [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(d*x+c)^2*(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out