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

Optimal. Leaf size=359 \[ -\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac{2 b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (-15 a^2 b^2+4 a^4-21 b^4\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^3}+\frac{2 a \left (a^2-3 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}-\frac{\left (-15 a^2 b^2+4 a^4-21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{6 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

[Out]

(2*b*Sec[c + d*x]^3)/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - ((4*a^4 - 15*a^2*b^2 - 21*b^4)*EllipticE[(c -
Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(6*(a^2 - b^2)^3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)])
+ (2*a*(a^2 - 3*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(3*(a^2
- b^2)^2*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(8*a*b - (a^2 + 7*b^2)*Sin[c
+ d*x]))/(3*(a^2 - b^2)^2*d) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*b*(a^2 - 33*b^2) - (4*a^4 - 15*a^2*b^
2 - 21*b^4)*Sin[c + d*x]))/(6*(a^2 - b^2)^3*d)

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Rubi [A]  time = 0.61347, antiderivative size = 359, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 d \left (a^2-b^2\right )^2}+\frac{2 b \sec ^3(c+d x)}{d \left (a^2-b^2\right ) \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (-15 a^2 b^2+4 a^4-21 b^4\right ) \sin (c+d x)\right )}{6 d \left (a^2-b^2\right )^3}+\frac{2 a \left (a^2-3 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}-\frac{\left (-15 a^2 b^2+4 a^4-21 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{6 d \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*Sec[c + d*x]^3)/((a^2 - b^2)*d*Sqrt[a + b*Sin[c + d*x]]) - ((4*a^4 - 15*a^2*b^2 - 21*b^4)*EllipticE[(c -
Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(6*(a^2 - b^2)^3*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)])
+ (2*a*(a^2 - 3*b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(3*(a^2
- b^2)^2*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(8*a*b - (a^2 + 7*b^2)*Sin[c
+ d*x]))/(3*(a^2 - b^2)^2*d) - (Sec[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(a*b*(a^2 - 33*b^2) - (4*a^4 - 15*a^2*b^
2 - 21*b^4)*Sin[c + d*x]))/(6*(a^2 - b^2)^3*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 ^4(c+d x)}{(a+b \sin (c+d x))^{3/2}} \, dx &=\frac{2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{2 \int \frac{\sec ^4(c+d x) \left (-\frac{a}{2}+\frac{7}{2} b \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{a^2-b^2}\\ &=\frac{2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac{2 \int \frac{\sec ^2(c+d x) \left (a \left (a^2-3 b^2\right )+\frac{3}{4} b \left (a^2+7 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=\frac{2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac{2 \int \frac{\frac{1}{8} a b^2 \left (a^2-33 b^2\right )+\frac{1}{8} b \left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^3}\\ &=\frac{2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}+\frac{\left (a \left (a^2-3 b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}-\frac{\left (4 a^4-15 a^2 b^2-21 b^4\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{12 \left (a^2-b^2\right )^3}\\ &=\frac{2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac{\left (\left (4 a^4-15 a^2 b^2-21 b^4\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 \left (a^2-b^2\right )^3 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (a \left (a^2-3 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 \left (a^2-b^2\right )^2 \sqrt{a+b \sin (c+d x)}}\\ &=\frac{2 b \sec ^3(c+d x)}{\left (a^2-b^2\right ) d \sqrt{a+b \sin (c+d x)}}-\frac{\left (4 a^4-15 a^2 b^2-21 b^4\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 \left (a^2-b^2\right )^3 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{2 a \left (a^2-3 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 \left (a^2-b^2\right )^2 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b-\left (a^2+7 b^2\right ) \sin (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (a b \left (a^2-33 b^2\right )-\left (4 a^4-15 a^2 b^2-21 b^4\right ) \sin (c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 2.9746, size = 348, normalized size = 0.97 \[ \frac{-4 a \left (-4 a^2 b^2+a^4+3 b^4\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\left (-15 a^3 b^2-15 a^2 b^3+4 a^4 b+4 a^5-21 a b^4-21 b^5\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+\frac{1}{8} \sec ^3(c+d x) \left (-64 a^3 b^2 \sin (c+d x)-32 a^3 b^2 \sin (3 (c+d x))+\left (84 a^2 b^3-12 a^4 b+56 b^5\right ) \cos (2 (c+d x))+\left (15 a^2 b^3-4 a^4 b+21 b^5\right ) \cos (4 (c+d x))+101 a^2 b^3-24 a^4 b+24 a^5 \sin (c+d x)+8 a^5 \sin (3 (c+d x))+40 a b^4 \sin (c+d x)+24 a b^4 \sin (3 (c+d x))+19 b^5\right )}{6 d (a-b)^3 (a+b)^3 \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^5 + 4*a^4*b - 15*a^3*b^2 - 15*a^2*b^3 - 21*a*b^4 - 21*b^5)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b
)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - 4*a*(a^4 - 4*a^2*b^2 + 3*b^4)*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*(-24*a^4*b + 101*a^2*b^3 + 19*b^5 + (-12*a^4*b +
84*a^2*b^3 + 56*b^5)*Cos[2*(c + d*x)] + (-4*a^4*b + 15*a^2*b^3 + 21*b^5)*Cos[4*(c + d*x)] + 24*a^5*Sin[c + d*x
] - 64*a^3*b^2*Sin[c + d*x] + 40*a*b^4*Sin[c + d*x] + 8*a^5*Sin[3*(c + d*x)] - 32*a^3*b^2*Sin[3*(c + d*x)] + 2
4*a*b^4*Sin[3*(c + d*x)]))/8)/(6*(a - b)^3*(a + b)^3*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 3.27, size = 1646, normalized size = 4.6 \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*(-(-b*sin(d*x+c)-a)*cos(d*x+c)^2)^(1/2)/cos(d*x+c)^5/(a+b*sin(d*x+c))^(3/2)/b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6
)*(-2*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b^2*(a^4-2*a^2*b^2+b^4)-cos(d*x+c)^4*(cos(d*x+c)^2*sin(
d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b^2*(4*a^4-15*a^2*b^2-21*b^4)+4*cos(d*x+c)^2*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d
*x+c)^2)^(1/2)*a*b*(a^4-4*a^2*b^2+3*b^4)*sin(d*x+c)+2*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b*(a^
4-2*a^2*b^2+b^4)*sin(d*x+c)+cos(d*x+c)^2*(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^6-19*(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^4*b^2-6*(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^2*b^4+21*(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))*b^6-4*(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)*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^5*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)*Ellipti
cF((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*b^2+16*(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)*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+18*(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)*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^4-12*(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)*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^5-21*(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)*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^6+a^4*b^2+6*a^2*b^4-7*b^6))/d

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Maxima [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="maxima")

[Out]

Timed out

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

[Out]

integral(-sqrt(b*sin(d*x + c) + a)*sec(d*x + c)^4/(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(-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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{{\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)^4/(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^4/(b*sin(d*x + c) + a)^(3/2), x)