3.504 \(\int \sec ^6(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=322 \[ \frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (8 a^2-3 b^2\right ) \sin (c+d x)+5 a b\right )}{30 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b \left (a^2-b^2\right )-\left (-41 a^2 b^2+32 a^4+9 b^4\right ) \sin (c+d x)\right )}{60 d \left (a^2-b^2\right )}+\frac{a \left (32 a^2-17 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \sqrt{a+b \sin (c+d x)}}-\frac{\left (32 a^2-9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{5 d} \]

[Out]

(Sec[c + d*x]^5*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(3/2))/(5*d) - ((32*a^2 - 9*b^2)*EllipticE[(c - Pi/2
+ d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(60*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (a*(32*a^2 - 1
7*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]^3*Sqrt[a + b*Sin[c + d*x]]*(5*a*b + (8*a^2 - 3*b^2)*Sin[c + d*x]))/(30*d) - (Sec[c +
d*x]*Sqrt[a + b*Sin[c + d*x]]*(8*a*b*(a^2 - b^2) - (32*a^4 - 41*a^2*b^2 + 9*b^4)*Sin[c + d*x]))/(60*(a^2 - b^
2)*d)

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Rubi [A]  time = 0.672409, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {2691, 2861, 2866, 2752, 2663, 2661, 2655, 2653} \[ \frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (\left (8 a^2-3 b^2\right ) \sin (c+d x)+5 a b\right )}{30 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b \left (a^2-b^2\right )-\left (-41 a^2 b^2+32 a^4+9 b^4\right ) \sin (c+d x)\right )}{60 d \left (a^2-b^2\right )}+\frac{a \left (32 a^2-17 b^2\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \sqrt{a+b \sin (c+d x)}}-\frac{\left (32 a^2-9 b^2\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{60 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(Sec[c + d*x]^5*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^(3/2))/(5*d) - ((32*a^2 - 9*b^2)*EllipticE[(c - Pi/2
+ d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(60*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (a*(32*a^2 - 1
7*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]^3*Sqrt[a + b*Sin[c + d*x]]*(5*a*b + (8*a^2 - 3*b^2)*Sin[c + d*x]))/(30*d) - (Sec[c +
d*x]*Sqrt[a + b*Sin[c + d*x]]*(8*a*b*(a^2 - b^2) - (32*a^4 - 41*a^2*b^2 + 9*b^4)*Sin[c + d*x]))/(60*(a^2 - b^
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 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
])

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))^{5/2} \, dx &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac{1}{5} \int \sec ^4(c+d x) \sqrt{a+b \sin (c+d x)} \left (-4 a^2+\frac{3 b^2}{2}-\frac{5}{2} a b \sin (c+d x)\right ) \, dx\\ &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (5 a b+\left (8 a^2-3 b^2\right ) \sin (c+d x)\right )}{30 d}+\frac{1}{15} \int \frac{\sec ^2(c+d x) \left (\frac{1}{4} a \left (32 a^2-17 b^2\right )+\frac{3}{4} b \left (8 a^2-3 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (5 a b+\left (8 a^2-3 b^2\right ) \sin (c+d x)\right )}{30 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b \left (a^2-b^2\right )-\left (32 a^4-41 a^2 b^2+9 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right ) d}-\frac{\int \frac{a b^2 \left (a^2-b^2\right )+\frac{1}{8} b \left (32 a^4-41 a^2 b^2+9 b^4\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{15 \left (a^2-b^2\right )}\\ &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (5 a b+\left (8 a^2-3 b^2\right ) \sin (c+d x)\right )}{30 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b \left (a^2-b^2\right )-\left (32 a^4-41 a^2 b^2+9 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right ) d}+\frac{1}{120} \left (a \left (32 a^2-17 b^2\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx-\frac{1}{120} \left (32 a^2-9 b^2\right ) \int \sqrt{a+b \sin (c+d x)} \, dx\\ &=\frac{\sec ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{5 d}+\frac{\sec ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (5 a b+\left (8 a^2-3 b^2\right ) \sin (c+d x)\right )}{30 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b \left (a^2-b^2\right )-\left (32 a^4-41 a^2 b^2+9 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right ) d}-\frac{\left (\left (32 a^2-9 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 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (a \left (32 a^2-17 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 ^5(c+d x) (b+a \sin (c+d x)) (a+b \sin (c+d x))^{3/2}}{5 d}-\frac{\left (32 a^2-9 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 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{a \left (32 a^2-17 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 ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (5 a b+\left (8 a^2-3 b^2\right ) \sin (c+d x)\right )}{30 d}-\frac{\sec (c+d x) \sqrt{a+b \sin (c+d x)} \left (8 a b \left (a^2-b^2\right )-\left (32 a^4-41 a^2 b^2+9 b^4\right ) \sin (c+d x)\right )}{60 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 6.25912, size = 351, normalized size = 1.09 \[ \frac{\sqrt{a+b \sin (c+d x)} \left (\frac{1}{5} \sec ^5(c+d x) \left (a^2 \sin (c+d x)+2 a b+b^2 \sin (c+d x)\right )+\frac{1}{30} \sec ^3(c+d x) \left (8 a^2 \sin (c+d x)-a b-3 b^2 \sin (c+d x)\right )+\frac{1}{60} \sec (c+d x) \left (32 a^2 \sin (c+d x)-8 a b-9 b^2 \sin (c+d x)\right )\right )}{d}-\frac{b \left (-\frac{\left (32 a^2-9 b^2\right ) \left (\frac{2 (a+b) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} E\left (\frac{1}{2} \left (-c-d x+\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}-\frac{2 a \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (-c-d x+\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}\right )}{b}-\frac{16 a b \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (-c-d x+\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{\sqrt{a+b \sin (c+d x)}}\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + b*Sin[c + d*x])^(5/2),x]

[Out]

(Sqrt[a + b*Sin[c + d*x]]*((Sec[c + d*x]*(-8*a*b + 32*a^2*Sin[c + d*x] - 9*b^2*Sin[c + d*x]))/60 + (Sec[c + d*
x]^3*(-(a*b) + 8*a^2*Sin[c + d*x] - 3*b^2*Sin[c + d*x]))/30 + (Sec[c + d*x]^5*(2*a*b + a^2*Sin[c + d*x] + b^2*
Sin[c + d*x]))/5))/d - (b*((-16*a*b*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^2 - 9*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*d)

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Maple [B]  time = 0.66, size = 1360, 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)^6*(a+b*sin(d*x+c))^(5/2),x)

[Out]

1/60*((cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a*b*(32*a^2-17*b^2)*sin(d*x+c)*cos(d*x+c)^4+8*(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*sin(d*x+c)+12*(cos(d*x+c)^2*sin(d*x+c)*b+
a*cos(d*x+c)^2)^(1/2)*a*b*(a^2+3*b^2)*sin(d*x+c)-(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b^2*(32*a^2-
9*b^2)*cos(d*x+c)^6+(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)*(
-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^4-41*(-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+9*(-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^4-32*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+24*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+17*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^3-9*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+8*a^2*b^2-3*b^4)*cos(d*x+c)^4+2*(cos(d*x+c)^2*sin(d*x+c)*b+a*c
os(d*x+c)^2)^(1/2)*b^2*(a^2-9*b^2)*cos(d*x+c)^2+36*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*a^2*b^2+12
*(cos(d*x+c)^2*sin(d*x+c)*b+a*cos(d*x+c)^2)^(1/2)*b^4)/(1+sin(d*x+c))^2/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1+s
in(d*x+c)))^(1/2)/(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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^(5/2)*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 (2 \, a b \sec \left (d x + c\right )^{6} \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 )^{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))^(5/2),x, algorithm="fricas")

[Out]

integral((2*a*b*sec(d*x + c)^6*sin(d*x + c) - (b^2*cos(d*x + c)^2 - a^2 - b^2)*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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

Timed out