3.492 \(\int \sec ^4(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=218 \[ \frac{\left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{6 d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{3 d}-\frac{\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{6 d}-\frac{2 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

[Out]

-(Sec[c + d*x]*(b - 4*a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(6*d) + (Sec[c + d*x]^3*(b + a*Sin[c + d*x])*S
qrt[a + b*Sin[c + d*x]])/(3*d) - (2*a*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(
3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((4*a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a
+ b*Sin[c + d*x])/(a + b)])/(6*d*Sqrt[a + b*Sin[c + d*x]])

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Rubi [A]  time = 0.438024, antiderivative size = 218, 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 = {2691, 2866, 2752, 2663, 2661, 2655, 2653} \[ \frac{\left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{6 d \sqrt{a+b \sin (c+d x)}}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{3 d}-\frac{\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{6 d}-\frac{2 a \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

-(Sec[c + d*x]*(b - 4*a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(6*d) + (Sec[c + d*x]^3*(b + a*Sin[c + d*x])*S
qrt[a + b*Sin[c + d*x]])/(3*d) - (2*a*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(
3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((4*a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a
+ b*Sin[c + d*x])/(a + b)])/(6*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 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 \sec ^4(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{3 d}-\frac{1}{3} \int \frac{\sec ^2(c+d x) \left (-2 a^2+\frac{b^2}{2}-\frac{3}{2} a b \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{6 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{3 d}+\frac{\int \frac{-\frac{1}{4} b^2 \left (a^2-b^2\right )-a b \left (a^2-b^2\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{6 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{3 d}-\frac{1}{3} a \int \sqrt{a+b \sin (c+d x)} \, dx+\frac{1}{12} \left (4 a^2-b^2\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{6 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{3 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}{3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (\left (4 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}{12 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{\sec (c+d x) (b-4 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{6 d}+\frac{\sec ^3(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{3 d}-\frac{2 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)}}{3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (4 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}}}{6 d \sqrt{a+b \sin (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.33819, size = 211, normalized size = 0.97 \[ \frac{-4 \left (4 a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\sec ^3(c+d x) \left (12 a^2 \sin (c+d x)+4 a^2 \sin (3 (c+d x))-6 a b \cos (2 (c+d x))-2 a b \cos (4 (c+d x))+12 a b+7 b^2 \sin (c+d x)-b^2 \sin (3 (c+d x))\right )+16 a (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )}{24 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(16*a*(a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - 4*(4*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)] + Sec[c + d*x]^3*(12*a
*b - 6*a*b*Cos[2*(c + d*x)] - 2*a*b*Cos[4*(c + d*x)] + 12*a^2*Sin[c + d*x] + 7*b^2*Sin[c + d*x] + 4*a^2*Sin[3*
(c + d*x)] - b^2*Sin[3*(c + d*x)]))/(24*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 0.568, size = 937, normalized size = 4.3 \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)^4*(a+b*sin(d*x+c))^(3/2),x)

[Out]

1/6*(-(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b*(4*a^2-b^2)*sin(d*x+c)*cos(d*x+c)^2-2*(cos(d*x+c)^2*s
in(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b*(a^2+b^2)*sin(d*x+c)+4*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*
b^2*cos(d*x+c)^4-(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*(4*(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^3-4*(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^2-4*(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)*Elli
pticF((b/(a-b)*sin(d*x+c)+1/(a-b)*a)^(1/2),((a-b)/(a+b))^(1/2))*a^2*b+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)*EllipticF((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))*b^3+a*b^
2)*cos(d*x+c)^2-4*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b^2)/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1
+sin(d*x+c)))^(1/2)/(1+sin(d*x+c))/(sin(d*x+c)-1)/b/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 )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(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)^4, 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 )^{4} \sin \left (d x + c\right ) + a \sec \left (d x + c\right )^{4}\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)^4*(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^4*sin(d*x + c) + a*sec(d*x + c)^4)*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)**4*(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)^4*(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out