3.6.4 \(\int \sec ^6(c+d x) (a+b \sin (c+d x))^{5/2} \, dx\) [504]

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

[Out]

1/5*sec(d*x+c)^5*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^(3/2)/d+1/30*sec(d*x+c)^3*(5*a*b+(8*a^2-3*b^2)*sin(d*x+c))*
(a+b*sin(d*x+c))^(1/2)/d-1/60*sec(d*x+c)*(8*a*b*(a^2-b^2)-(32*a^4-41*a^2*b^2+9*b^4)*sin(d*x+c))*(a+b*sin(d*x+c
))^(1/2)/(a^2-b^2)/d+1/60*(32*a^2-9*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)*Ellipti
cE(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)-
1/60*a*(32*a^2-17*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*P
i+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 = 0.42, antiderivative size = 322, normalized size of antiderivative = 1.00, 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 = {2770, 2940, 2945, 2831, 2742, 2740, 2734, 2732} \begin {gather*} \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 {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 (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 d \left (a^2-b^2\right )}+\frac {\sec ^5(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^{3/2}}{5 d} \end {gather*}

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 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2770

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)*S
in[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && (Integers
Q[2*m, 2*p] || IntegerQ[m])

Rule 2831

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 2940

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 2945

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]

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*}

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Mathematica [A]
time = 6.26, size = 351, normalized size = 1.09 \begin {gather*} \frac {\sqrt {a+b \sin (c+d x)} \left (\frac {1}{60} \sec (c+d x) \left (-8 a b+32 a^2 \sin (c+d x)-9 b^2 \sin (c+d x)\right )+\frac {1}{30} \sec ^3(c+d x) \left (-a b+8 a^2 \sin (c+d x)-3 b^2 \sin (c+d x)\right )+\frac {1}{5} \sec ^5(c+d x) \left (2 a b+a^2 \sin (c+d x)+b^2 \sin (c+d x)\right )\right )}{d}-\frac {b \left (-\frac {16 a b 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}}}{\sqrt {a+b \sin (c+d x)}}-\frac {\left (32 a^2-9 b^2\right ) \left (\frac {2 (a+b) E\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}}}{\sqrt {a+b \sin (c+d x)}}-\frac {2 a 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}}}{\sqrt {a+b \sin (c+d x)}}\right )}{b}\right )}{120 d} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(1359\) vs. \(2(364)=728\).
time = 2.75, size = 1360, normalized size = 4.22

method	result	size
default	\(\text {Expression too large to display}\)	\(1360\)

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,method=_RETURNVERBOSE)

[Out]

1/60*((b*cos(d*x+c)^2*sin(d*x+c)+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*(b*cos(d*
x+c)^2*sin(d*x+c)+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*(b*cos(d*x+c)^2*sin(d*x+c)+
a*cos(d*x+c)^2)^(1/2)*a*b*(a^2+3*b^2)*sin(d*x+c)-(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*b^2*(32*a^2-
9*b^2)*cos(d*x+c)^6+(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*(32*(-b/(a+b)*sin(d*x+c)+b/(a+b))^(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)+1/(a-b)*a)^(1/2)*(-b/(a
-b)*sin(d*x+c)-b/(a-b))^(1/2)*a^4-41*(-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/(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^
2+9*(-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
/(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^4-32*(-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)*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+24*(-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+17*(-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^3-9*(-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+8*a^2*b^2-3*b^4)*cos(d*x+c)^4+2*(b*cos(d*x+c)^2*sin(d*x+c)+a*c
os(d*x+c)^2)^(1/2)*b^2*(a^2-9*b^2)*cos(d*x+c)^2+36*(b*cos(d*x+c)^2*sin(d*x+c)+a*cos(d*x+c)^2)^(1/2)*a^2*b^2+12
*(b*cos(d*x+c)^2*sin(d*x+c)+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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 561, normalized size = 1.74 \begin {gather*} \frac {2 \, \sqrt {2} {\left (32 \, a^{3} - 21 \, a b^{2}\right )} \sqrt {i \, b} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right ) + 2 \, \sqrt {2} {\left (32 \, a^{3} - 21 \, a b^{2}\right )} \sqrt {-i \, b} \cos \left (d x + c\right )^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right ) - 3 \, \sqrt {2} {\left (-32 i \, a^{2} b + 9 i \, b^{3}\right )} \sqrt {i \, b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right )\right ) - 3 \, \sqrt {2} {\left (32 i \, a^{2} b - 9 i \, b^{3}\right )} \sqrt {-i \, b} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right ) - 6 \, {\left (8 \, a b^{2} \cos \left (d x + c\right )^{4} + 2 \, a b^{2} \cos \left (d x + c\right )^{2} - 24 \, a b^{2} - {\left ({\left (32 \, a^{2} b - 9 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 12 \, a^{2} b + 12 \, b^{3} + 2 \, {\left (8 \, a^{2} b - 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}}{360 \, b d \cos \left (d x + c\right )^{5}} \end {gather*}

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]

1/360*(2*sqrt(2)*(32*a^3 - 21*a*b^2)*sqrt(I*b)*cos(d*x + c)^5*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8
/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b) + 2*sqrt(2)*(32*a^3 - 21
*a*b^2)*sqrt(-I*b)*cos(d*x + c)^5*weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b
^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b) - 3*sqrt(2)*(-32*I*a^2*b + 9*I*b^3)*sqrt(I*b)*cos(d
*x + c)^5*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*
(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b))
- 3*sqrt(2)*(32*I*a^2*b - 9*I*b^3)*sqrt(-I*b)*cos(d*x + c)^5*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(
-8*I*a^3 + 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3
*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) - 6*(8*a*b^2*cos(d*x + c)^4 + 2*a*b^2*cos(d*x + c)^2 - 24
*a*b^2 - ((32*a^2*b - 9*b^3)*cos(d*x + c)^4 + 12*a^2*b + 12*b^3 + 2*(8*a^2*b - 3*b^3)*cos(d*x + c)^2)*sin(d*x
+ c))*sqrt(b*sin(d*x + c) + a))/(b*d*cos(d*x + c)^5)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^(5/2)/cos(c + d*x)^6,x)

[Out]

\text{Hanged}

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