3.493 \(\int \sec ^6(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=330 \[ -\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (-13 a^2 b^2+8 a^4+5 b^4\right )-a \left (-61 a^2 b^2+32 a^4+29 b^4\right ) \sin (c+d x)\right )}{60 d \left (a^2-b^2\right )^2}+\frac{\left (32 a^2-5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \sqrt{a+b \sin (c+d x)}}-\frac{a \left (32 a^2-29 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{5 d}-\frac{\sec ^3(c+d x) (b-8 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{30 d} \]

[Out]

-(Sec[c + d*x]^3*(b - 8*a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(30*d) + (Sec[c + d*x]^5*(b + a*Sin[c + d*x]
)*Sqrt[a + b*Sin[c + d*x]])/(5*d) - (a*(32*a^2 - 29*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a +
b*Sin[c + d*x]])/(60*(a^2 - b^2)*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((32*a^2 - 5*b^2)*EllipticF[(c - Pi/
2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(60*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]
*Sqrt[a + b*Sin[c + d*x]]*(b*(8*a^4 - 13*a^2*b^2 + 5*b^4) - a*(32*a^4 - 61*a^2*b^2 + 29*b^4)*Sin[c + d*x]))/(6
0*(a^2 - b^2)^2*d)

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Rubi [A]  time = 0.697539, antiderivative size = 330, 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 = {2691, 2866, 2752, 2663, 2661, 2655, 2653} \[ -\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (-13 a^2 b^2+8 a^4+5 b^4\right )-a \left (-61 a^2 b^2+32 a^4+29 b^4\right ) \sin (c+d x)\right )}{60 d \left (a^2-b^2\right )^2}+\frac{\left (32 a^2-5 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \sqrt{a+b \sin (c+d x)}}-\frac{a \left (32 a^2-29 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+b) \sqrt{a+b \sin (c+d x)}}{5 d}-\frac{\sec ^3(c+d x) (b-8 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{30 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

-(Sec[c + d*x]^3*(b - 8*a*Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(30*d) + (Sec[c + d*x]^5*(b + a*Sin[c + d*x]
)*Sqrt[a + b*Sin[c + d*x]])/(5*d) - (a*(32*a^2 - 29*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a +
b*Sin[c + d*x]])/(60*(a^2 - b^2)*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + ((32*a^2 - 5*b^2)*EllipticF[(c - Pi/
2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(60*d*Sqrt[a + b*Sin[c + d*x]]) - (Sec[c + d*x]
*Sqrt[a + b*Sin[c + d*x]]*(b*(8*a^4 - 13*a^2*b^2 + 5*b^4) - a*(32*a^4 - 61*a^2*b^2 + 29*b^4)*Sin[c + d*x]))/(6
0*(a^2 - b^2)^2*d)

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 ^6(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{5 d}-\frac{1}{5} \int \frac{\sec ^4(c+d x) \left (-4 a^2+\frac{b^2}{2}-\frac{7}{2} a b \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{\sec ^3(c+d x) (b-8 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{5 d}+\frac{\int \frac{\sec ^2(c+d x) \left (\frac{1}{4} \left (32 a^4-37 a^2 b^2+5 b^4\right )+6 a b \left (a^2-b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )}\\ &=-\frac{\sec ^3(c+d x) (b-8 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{5 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (8 a^4-13 a^2 b^2+5 b^4\right )-a \left (32 a^4-61 a^2 b^2+29 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right )^2 d}-\frac{\int \frac{\frac{1}{8} b^2 \left (8 a^4-13 a^2 b^2+5 b^4\right )+\frac{1}{8} a b \left (32 a^4-61 a^2 b^2+29 b^4\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )^2}\\ &=-\frac{\sec ^3(c+d x) (b-8 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{5 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (8 a^4-13 a^2 b^2+5 b^4\right )-a \left (32 a^4-61 a^2 b^2+29 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right )^2 d}-\frac{\left (a \left (32 a^2-29 b^2\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{120 \left (a^2-b^2\right )}-\frac{1}{120} \left (-32 a^2+5 b^2\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{\sec ^3(c+d x) (b-8 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{5 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (8 a^4-13 a^2 b^2+5 b^4\right )-a \left (32 a^4-61 a^2 b^2+29 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right )^2 d}-\frac{\left (a \left (32 a^2-29 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}{120 \left (a^2-b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{\left (\left (-32 a^2+5 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}{120 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{\sec ^3(c+d x) (b-8 a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{30 d}+\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) \sqrt{a+b \sin (c+d x)}}{5 d}-\frac{a \left (32 a^2-29 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)}}{60 \left (a^2-b^2\right ) d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (32 a^2-5 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}}}{60 d \sqrt{a+b \sin (c+d x)}}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (b \left (8 a^4-13 a^2 b^2+5 b^4\right )-a \left (32 a^4-61 a^2 b^2+29 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 6.24963, size = 364, normalized size = 1.1 \[ \frac{\sqrt{a+b \sin (c+d x)} \left (\frac{\sec (c+d x) \left (-8 a^2 b+32 a^3 \sin (c+d x)-29 a b^2 \sin (c+d x)+5 b^3\right )}{60 \left (a^2-b^2\right )}+\frac{1}{5} \sec ^5(c+d x) (a \sin (c+d x)+b)+\frac{1}{30} \sec ^3(c+d x) (8 a \sin (c+d x)-b)\right )}{d}-\frac{b \left (-\frac{2 \left (8 a^2 b-5 b^3\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (-c-d x+\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}-\frac{\left (32 a^3-29 a b^2\right ) \left (\frac{2 (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{2} \left (-c-d x+\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}-\frac{2 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (-c-d x+\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}\right )}{b}\right )}{120 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(Sqrt[a + b*Sin[c + d*x]]*((Sec[c + d*x]^5*(b + a*Sin[c + d*x]))/5 + (Sec[c + d*x]^3*(-b + 8*a*Sin[c + d*x]))/
30 + (Sec[c + d*x]*(-8*a^2*b + 5*b^3 + 32*a^3*Sin[c + d*x] - 29*a*b^2*Sin[c + d*x]))/(60*(a^2 - b^2))))/d - (b
*((-2*(8*a^2*b - 5*b^3)*EllipticF[(-c + Pi/2 - d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/Sqrt
[a + b*Sin[c + d*x]] - ((32*a^3 - 29*a*b^2)*((2*(a + b)*EllipticE[(-c + Pi/2 - d*x)/2, (2*b)/(a + b)]*Sqrt[(a
+ b*Sin[c + d*x])/(a + b)])/Sqrt[a + b*Sin[c + d*x]] - (2*a*EllipticF[(-c + Pi/2 - d*x)/2, (2*b)/(a + b)]*Sqrt
[(a + b*Sin[c + d*x])/(a + b)])/Sqrt[a + b*Sin[c + d*x]]))/b))/(120*(a - b)*(a + b)*d)

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Maple [B]  time = 0.843, size = 1519, 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)^6*(a+b*sin(d*x+c))^(3/2),x)

[Out]

1/120*(2*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b*(32*a^4-37*a^2*b^2+5*b^4)*sin(d*x+c)*cos(d*x+c)^4+
4*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b*(8*a^4-9*a^2*b^2+b^4)*cos(d*x+c)^2*sin(d*x+c)+24*(cos(d*x
+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b*(a^4-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^2*(32*a^2-29*b^2)*cos(d*x+c)^6-2*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*(32*(-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)+1
/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^4*b-24*(-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)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+
c)+b/(a+b))^(1/2)*a^3*b^2-37*(-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)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a^2*b^3+24*(-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)+1/(a-b)*a)^(1/2)*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(1/2)*a*b^4+5*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*El
lipticF((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)+1/(a-b)*a)^(1/2)*(-b/(a+
b)*sin(d*x+c)+b/(a+b))^(1/2)*b^5-32*(-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)*E
llipticE((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+61*
(-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^3*b^2-29*(-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*b^4-8*a^3*b^2+8*a*b^4)*cos(d*x+c)^4+4*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*
x+c)^2)^(1/2)*a*b^2*(a^2-b^2)*cos(d*x+c)^2+48*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a^3*b^2-48*(cos
(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b^4)/(1+sin(d*x+c))^2/(a-b)/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(
1+sin(d*x+c)))^(1/2)/(a+b)/(sin(d*x+c)-1)^2/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 )^{6}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(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)^6, 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 )^{6} \sin \left (d x + c\right ) + a \sec \left (d x + c\right )^{6}\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)^6*(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*sec(d*x + c)^6*sin(d*x + c) + a*sec(d*x + c)^6)*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)**6*(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)^6*(a+b*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out