3.503 \(\int \sec ^4(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=238 \[ \frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (\left (4 a^2-3 b^2\right ) \sin (c+d x)+a b\right )}{6 d}+\frac {2 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2-3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{3 d} \]

[Out]

1/3*sec(d*x+c)^3*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^(3/2)/d+1/6*sec(d*x+c)*(a*b+(4*a^2-3*b^2)*sin(d*x+c))*(a+b*
sin(d*x+c))^(1/2)/d+1/6*(4*a^2-3*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(
cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/d/((a+b*sin(d*x+c))/(a+b))^(1/2)-2/3
*a*(a^2-b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x)
,2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 238, normalized size of antiderivative = 1.00, 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, 2861, 2752, 2663, 2661, 2655, 2653} \[ \frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (\left (4 a^2-3 b^2\right ) \sin (c+d x)+a b\right )}{6 d}+\frac {2 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \sin (c+d x)}}-\frac {\left (4 a^2-3 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

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

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]

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 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 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 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 2861

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*(d + c*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)*Simp[a*c*(p +
2) + b*d*m + b*c*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x
])

Rubi steps

\begin {align*} \int \sec ^4(c+d x) (a+b \sin (c+d x))^{5/2} \, dx &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{3 d}-\frac {1}{3} \int \sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (-2 a^2+\frac {3 b^2}{2}-\frac {1}{2} a b \sin (c+d x)\right ) \, dx\\ &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{3 d}+\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b+\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 d}+\frac {1}{3} \int \frac {-\frac {a b^2}{4}-\frac {1}{4} b \left (4 a^2-3 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{3 d}+\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b+\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 d}+\frac {1}{3} \left (a \left (a^2-b^2\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx+\frac {1}{12} \left (-4 a^2+3 b^2\right ) \int \sqrt {a+b \sin (c+d x)} \, dx\\ &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{3 d}+\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b+\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 d}+\frac {\left (\left (-4 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}{12 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {\left (a \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}{3 \sqrt {a+b \sin (c+d x)}}\\ &=\frac {\sec ^3(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{3 d}-\frac {\left (4 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)}}{6 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}+\frac {2 a \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}}}{3 d \sqrt {a+b \sin (c+d x)}}+\frac {\sec (c+d x) \sqrt {a+b \sin (c+d x)} \left (a b+\left (4 a^2-3 b^2\right ) \sin (c+d x)\right )}{6 d}\\ \end {align*}

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Mathematica [A]  time = 3.52, size = 259, normalized size = 1.09 \[ \frac {-4 a \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\left (4 a^3+4 a^2 b-3 a b^2-3 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+\frac {1}{8} \sec ^3(c+d x) \left (24 a^3 \sin (c+d x)+8 a^3 \sin (3 (c+d x))-4 \left (3 a^2 b+2 b^3\right ) \cos (2 (c+d x))+\left (3 b^3-4 a^2 b\right ) \cos (4 (c+d x))+40 a^2 b+40 a b^2 \sin (c+d x)-8 a b^2 \sin (3 (c+d x))+5 b^3\right )}{6 d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

((4*a^3 + 4*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)] - 4*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)]
+ (Sec[c + d*x]^3*(40*a^2*b + 5*b^3 - 4*(3*a^2*b + 2*b^3)*Cos[2*(c + d*x)] + (-4*a^2*b + 3*b^3)*Cos[4*(c + d*
x)] + 24*a^3*Sin[c + d*x] + 40*a*b^2*Sin[c + d*x] + 8*a^3*Sin[3*(c + d*x)] - 8*a*b^2*Sin[3*(c + d*x)]))/8)/(6*
d*Sqrt[a + b*Sin[c + d*x]])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (2 \, a b \sec \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (b^{2} \cos \left (d x + c\right )^{2} - a^{2} - b^{2}\right )} \sec \left (d x + c\right )^{4}\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((2*a*b*sec(d*x + c)^4*sin(d*x + c) - (b^2*cos(d*x + c)^2 - a^2 - b^2)*sec(d*x + c)^4)*sqrt(b*sin(d*x
+ c) + a), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.91, size = 1249, normalized size = 5.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*sin(d*x+c))^(5/2),x)

[Out]

1/6*(-4*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*a*b*(a^2-b^2)*sin(d*x+c)*cos(d*x+c)^2-2*(cos(d*x+c)^2
*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*a*b*(a^2+3*b^2)*sin(d*x+c)+(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2
)*b^2*(4*a^2-3*b^2)*cos(d*x+c)^4-(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*(4*(-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/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^4-7*(-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/(a+b)*sin(d*x+c)+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)+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/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*b^4-4*(-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)*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)*a^3*b+3*(-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)*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)*a^2*b^2+4*(-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)*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)*a*b^3-3*(-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)*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^4+a^2*b^2-5*b^4)*cos(d*x+c)^2-6*(cos(d*x+c)^2*sin(d*x+c)
*b+cos(d*x+c)^2*a)^(1/2)*a^2*b^2-2*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*b^4)/(-(a+b*sin(d*x+c))*(s
in(d*x+c)-1)*(1+sin(d*x+c)))^(1/2)/(sin(d*x+c)-1)/(1+sin(d*x+c))/b/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^(5/2)*sec(d*x + c)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^(5/2)/cos(c + d*x)^4,x)

[Out]

int((a + b*sin(c + d*x))^(5/2)/cos(c + d*x)^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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