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

Optimal. Leaf size=251 \[ -\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )^2}+\frac{2 b \sec (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{\left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

[Out]

(2*b*Sec[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - ((a^2 + 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*
b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/((a^2 - b^2)^2*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)])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) -
(Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a*b - (a^2 + 3*b^2)*Sin[c + d*x]))/((a^2 - b^2)^2*d)

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Rubi [A]  time = 0.365161, antiderivative size = 251, 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 = {2694, 2866, 2752, 2663, 2661, 2655, 2653} \[ -\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{d \left (a^2-b^2\right )^2}+\frac{2 b \sec (c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}+\frac{a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{\left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sec[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - ((a^2 + 3*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*
b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/((a^2 - b^2)^2*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)])/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) -
(Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a*b - (a^2 + 3*b^2)*Sin[c + d*x]))/((a^2 - b^2)^2*d)

Rule 2694

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

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*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 \frac{\sec ^2(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx &=\frac{2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{2 \int \frac{\sec ^2(c+d x) \left (-\frac{a}{2}+\frac{3}{2} b \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}+\frac{2 \int \frac{-a b^2-\frac{1}{4} b \left (a^2+3 b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac{2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}+\frac{a \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )}-\frac{\left (a^2+3 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac{2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}-\frac{\left (\left (a^2+3 b^2\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{2 \left (a^2-b^2\right )^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 \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}\\ &=\frac{2 b \sec (c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\left (a^2+3 b^2\right ) 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)}}{\left (a^2-b^2\right )^2 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}}}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a b-\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 1.65651, size = 205, normalized size = 0.82 \[ \frac{-a \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\left (a^2 b+a^3+3 a b^2+3 b^3\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-\frac{1}{2} \sec (c+d x) \left (-2 a \left (a^2-b^2\right ) \sin (c+d x)+b \left (a^2+3 b^2\right ) \cos (2 (c+d x))+3 a^2 b+b^3\right )}{d (a-b)^2 (a+b)^2 \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3 + a^2*b + 3*a*b^2 + 3*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a
+ b)] - a*(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)] - (Se
c[c + d*x]*(3*a^2*b + b^3 + b*(a^2 + 3*b^2)*Cos[2*(c + d*x)] - 2*a*(a^2 - b^2)*Sin[c + d*x]))/2)/((a - b)^2*(a
+ b)^2*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 0.788, size = 1062, normalized size = 4.2 \begin{align*} \text{result too large to display} \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)-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^4+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^2-3*(-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^4-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^3*b-3*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^2*b^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^3+3*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^4-a^2*b^2*cos(d*x+c)^2-3*b^4*cos(d*x+c)^2+a^3*b*sin(d*x+c)-a*b^3*sin(d*x+c)-a^2*b^2+b^4)/(-
(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1+sin(d*x+c)))^(1/2)/(a+b)/(a-b)/(a^2-b^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 \frac{\sec \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{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(sec(d*x + c)^2/(b*sin(d*x + c) + a)^(3/2), x)

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{2}{\left (c + d x \right )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, 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))**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{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="giac")

[Out]

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