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

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

[Out]

-1/30*sec(d*x+c)^3*(b-8*a*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/d+1/5*sec(d*x+c)^5*(b+a*sin(d*x+c))*(a+b*sin(d*x+
c))^(1/2)/d-1/60*sec(d*x+c)*(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(d*x+c))*(a+b*sin(d*x+
c))^(1/2)/(a^2-b^2)^2/d+1/60*a*(32*a^2-29*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)*E
llipticE(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)/(a^2-b^2)/d/((a+b*sin(d*x+c
))/(a+b))^(1/2)-1/60*(32*a^2-5*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(co
s(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 = 0.70, antiderivative size = 330, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 6.28, size = 364, normalized size = 1.10 \[ \frac {\sqrt {a+b \sin (c+d x)} \left (\frac {\sec (c+d x) \left (32 a^3 \sin (c+d x)-8 a^2 b-29 a b^2 \sin (c+d x)+5 b^3\right )}{60 \left (a^2-b^2\right )}+\frac {1}{5} \sec ^5(c+d x) (a \sin (c+d x)+b)+\frac {1}{30} \sec ^3(c+d x) (8 a \sin (c+d x)-b)\right )}{d}-\frac {b \left (-\frac {\left (32 a^3-29 a b^2\right ) \left (\frac {2 (a+b) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{2} \left (-c-d x+\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}-\frac {2 a \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (-c-d x+\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}\right )}{b}-\frac {2 \left (8 a^2 b-5 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (-c-d x+\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{\sqrt {a+b \sin (c+d x)}}\right )}{120 d (a-b) (a+b)} \]

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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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right )^{6} \sin \left (d x + c\right ) + a \sec \left (d x + c\right )^{6}\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]

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

[Out]

Timed out

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maple [B]  time = 1.04, size = 1519, normalized size = 4.60 \[ \text {result too large to display} \]

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+cos(d*x+c)^2*a)^(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+cos(d*x+c)^2*a)^(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+cos(d*x+c)^2*a)^(1/2)*b*(a^4-b^4)*sin(d*x+c)+2*(cos(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)
^(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+cos(d*x+c)^2*a)^(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)*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)*a^4*b-24*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(1/2)*(-b/(a+b)*s
in(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)*a^3*b^2-37*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(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)+1/(a-b)*a)^(1/2)*a^2*b^3+24*(
-b/(a-b)*sin(d*x+c)-b/(a-b))^(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)+1/(a-b)*a)^(1/2)*a*b^4+5*(-b/(a-b)*sin(d*x+c)-b/(a-b))^(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)+1/(a-b)*a)^(1/2)*b^5-32*(-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^5+6
1*(-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^3*b^2-29*(-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*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+cos(d*
x+c)^2*a)^(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+cos(d*x+c)^2*a)^(1/2)*a^3*b^2+48*(c
os(d*x+c)^2*sin(d*x+c)*b+cos(d*x+c)^2*a)^(1/2)*a*b^4)/(-(a+b*sin(d*x+c))*(sin(d*x+c)-1)*(1+sin(d*x+c)))^(1/2)/
(a+b)/(a-b)/(1+sin(d*x+c))^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 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{6}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^(3/2)/cos(c + d*x)^6,x)

[Out]

int((a + b*sin(c + d*x))^(3/2)/cos(c + d*x)^6, 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)**6*(a+b*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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