3.6.73 \(\int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{11/2}} \, dx\) [573]

Optimal. Leaf size=264 \[ \frac {2 a b \left (21 a^2-22 b^2\right ) (e \cos (c+d x))^{3/2}}{45 d e^7}-\frac {2 \left (7 a^4-12 a^2 b^2+4 b^4\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d e^6 \sqrt {\cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}}-\frac {2 (a+b \sin (c+d x)) \left (b \left (7 a^2-6 b^2\right )-a \left (21 a^2-22 b^2\right ) \sin (c+d x)\right )}{45 d e^5 \sqrt {e \cos (c+d x)}}+\frac {2 (a+b \sin (c+d x))^2 \left (a b+\left (7 a^2-6 b^2\right ) \sin (c+d x)\right )}{45 d e^3 (e \cos (c+d x))^{5/2}} \]

[Out]

2/45*a*b*(21*a^2-22*b^2)*(e*cos(d*x+c))^(3/2)/d/e^7+2/9*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^3/d/e/(e*cos(d*x+c))
^(9/2)+2/45*(a+b*sin(d*x+c))^2*(a*b+(7*a^2-6*b^2)*sin(d*x+c))/d/e^3/(e*cos(d*x+c))^(5/2)-2/45*(a+b*sin(d*x+c))
*(b*(7*a^2-6*b^2)-a*(21*a^2-22*b^2)*sin(d*x+c))/d/e^5/(e*cos(d*x+c))^(1/2)-2/15*(7*a^4-12*a^2*b^2+4*b^4)*(cos(
1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*(e*cos(d*x+c))^(1/2)/d/e^6/co
s(d*x+c)^(1/2)

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Rubi [A]
time = 0.32, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2770, 2940, 2748, 2721, 2719} \begin {gather*} \frac {2 a b \left (21 a^2-22 b^2\right ) (e \cos (c+d x))^{3/2}}{45 d e^7}-\frac {2 \left (b \left (7 a^2-6 b^2\right )-a \left (21 a^2-22 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))}{45 d e^5 \sqrt {e \cos (c+d x)}}+\frac {2 \left (\left (7 a^2-6 b^2\right ) \sin (c+d x)+a b\right ) (a+b \sin (c+d x))^2}{45 d e^3 (e \cos (c+d x))^{5/2}}-\frac {2 \left (7 a^4-12 a^2 b^2+4 b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {e \cos (c+d x)}}{15 d e^6 \sqrt {\cos (c+d x)}}+\frac {2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(11/2),x]

[Out]

(2*a*b*(21*a^2 - 22*b^2)*(e*Cos[c + d*x])^(3/2))/(45*d*e^7) - (2*(7*a^4 - 12*a^2*b^2 + 4*b^4)*Sqrt[e*Cos[c + d
*x]]*EllipticE[(c + d*x)/2, 2])/(15*d*e^6*Sqrt[Cos[c + d*x]]) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^3
)/(9*d*e*(e*Cos[c + d*x])^(9/2)) - (2*(a + b*Sin[c + d*x])*(b*(7*a^2 - 6*b^2) - a*(21*a^2 - 22*b^2)*Sin[c + d*
x]))/(45*d*e^5*Sqrt[e*Cos[c + d*x]]) + (2*(a + b*Sin[c + d*x])^2*(a*b + (7*a^2 - 6*b^2)*Sin[c + d*x]))/(45*d*e
^3*(e*Cos[c + d*x])^(5/2))

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 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 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])

Rubi steps

\begin {align*} \int \frac {(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{11/2}} \, dx &=\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}}-\frac {2 \int \frac {(a+b \sin (c+d x))^2 \left (-\frac {7 a^2}{2}+3 b^2-\frac {1}{2} a b \sin (c+d x)\right )}{(e \cos (c+d x))^{7/2}} \, dx}{9 e^2}\\ &=\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}}+\frac {2 (a+b \sin (c+d x))^2 \left (a b+\left (7 a^2-6 b^2\right ) \sin (c+d x)\right )}{45 d e^3 (e \cos (c+d x))^{5/2}}+\frac {4 \int \frac {(a+b \sin (c+d x)) \left (\frac {1}{4} a \left (21 a^2-22 b^2\right )-\frac {1}{4} b \left (7 a^2-6 b^2\right ) \sin (c+d x)\right )}{(e \cos (c+d x))^{3/2}} \, dx}{45 e^4}\\ &=\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}}-\frac {2 (a+b \sin (c+d x)) \left (b \left (7 a^2-6 b^2\right )-a \left (21 a^2-22 b^2\right ) \sin (c+d x)\right )}{45 d e^5 \sqrt {e \cos (c+d x)}}+\frac {2 (a+b \sin (c+d x))^2 \left (a b+\left (7 a^2-6 b^2\right ) \sin (c+d x)\right )}{45 d e^3 (e \cos (c+d x))^{5/2}}-\frac {8 \int \sqrt {e \cos (c+d x)} \left (\frac {3}{8} \left (7 a^4-12 a^2 b^2+4 b^4\right )+\frac {3}{8} a b \left (21 a^2-22 b^2\right ) \sin (c+d x)\right ) \, dx}{45 e^6}\\ &=\frac {2 a b \left (21 a^2-22 b^2\right ) (e \cos (c+d x))^{3/2}}{45 d e^7}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}}-\frac {2 (a+b \sin (c+d x)) \left (b \left (7 a^2-6 b^2\right )-a \left (21 a^2-22 b^2\right ) \sin (c+d x)\right )}{45 d e^5 \sqrt {e \cos (c+d x)}}+\frac {2 (a+b \sin (c+d x))^2 \left (a b+\left (7 a^2-6 b^2\right ) \sin (c+d x)\right )}{45 d e^3 (e \cos (c+d x))^{5/2}}-\frac {\left (7 a^4-12 a^2 b^2+4 b^4\right ) \int \sqrt {e \cos (c+d x)} \, dx}{15 e^6}\\ &=\frac {2 a b \left (21 a^2-22 b^2\right ) (e \cos (c+d x))^{3/2}}{45 d e^7}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}}-\frac {2 (a+b \sin (c+d x)) \left (b \left (7 a^2-6 b^2\right )-a \left (21 a^2-22 b^2\right ) \sin (c+d x)\right )}{45 d e^5 \sqrt {e \cos (c+d x)}}+\frac {2 (a+b \sin (c+d x))^2 \left (a b+\left (7 a^2-6 b^2\right ) \sin (c+d x)\right )}{45 d e^3 (e \cos (c+d x))^{5/2}}-\frac {\left (\left (7 a^4-12 a^2 b^2+4 b^4\right ) \sqrt {e \cos (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 e^6 \sqrt {\cos (c+d x)}}\\ &=\frac {2 a b \left (21 a^2-22 b^2\right ) (e \cos (c+d x))^{3/2}}{45 d e^7}-\frac {2 \left (7 a^4-12 a^2 b^2+4 b^4\right ) \sqrt {e \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d e^6 \sqrt {\cos (c+d x)}}+\frac {2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}}-\frac {2 (a+b \sin (c+d x)) \left (b \left (7 a^2-6 b^2\right )-a \left (21 a^2-22 b^2\right ) \sin (c+d x)\right )}{45 d e^5 \sqrt {e \cos (c+d x)}}+\frac {2 (a+b \sin (c+d x))^2 \left (a b+\left (7 a^2-6 b^2\right ) \sin (c+d x)\right )}{45 d e^3 (e \cos (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 1.61, size = 219, normalized size = 0.83 \begin {gather*} \frac {\sqrt {e \cos (c+d x)} \sec ^5(c+d x) \left (320 a^3 b+32 a b^3-288 a b^3 \cos (2 (c+d x))-48 \left (7 a^4-12 a^2 b^2+4 b^4\right ) \cos ^{\frac {9}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+150 a^4 \sin (c+d x)+360 a^2 b^2 \sin (c+d x)+60 b^4 \sin (c+d x)+91 a^4 \sin (3 (c+d x))-156 a^2 b^2 \sin (3 (c+d x))-8 b^4 \sin (3 (c+d x))+21 a^4 \sin (5 (c+d x))-36 a^2 b^2 \sin (5 (c+d x))+12 b^4 \sin (5 (c+d x))\right )}{360 d e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(11/2),x]

[Out]

(Sqrt[e*Cos[c + d*x]]*Sec[c + d*x]^5*(320*a^3*b + 32*a*b^3 - 288*a*b^3*Cos[2*(c + d*x)] - 48*(7*a^4 - 12*a^2*b
^2 + 4*b^4)*Cos[c + d*x]^(9/2)*EllipticE[(c + d*x)/2, 2] + 150*a^4*Sin[c + d*x] + 360*a^2*b^2*Sin[c + d*x] + 6
0*b^4*Sin[c + d*x] + 91*a^4*Sin[3*(c + d*x)] - 156*a^2*b^2*Sin[3*(c + d*x)] - 8*b^4*Sin[3*(c + d*x)] + 21*a^4*
Sin[5*(c + d*x)] - 36*a^2*b^2*Sin[5*(c + d*x)] + 12*b^4*Sin[5*(c + d*x)]))/(360*d*e^6)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1415\) vs. \(2(268)=536\).
time = 34.28, size = 1416, normalized size = 5.36

method	result	size
default	\(\text {Expression too large to display}\)	\(1416\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x,method=_RETURNVERBOSE)

[Out]

-2/45/(16*sin(1/2*d*x+1/2*c)^8-32*sin(1/2*d*x+1/2*c)^6+24*sin(1/2*d*x+1/2*c)^4-8*sin(1/2*d*x+1/2*c)^2+1)/sin(1
/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/e^5*(-384*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10+1344*
a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+768*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-1064*a^4*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(
1/2*d*x+1/2*c),2^(1/2))*a^4+12*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2
*d*x+1/2*c),2^(1/2))*b^4+1152*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*
d*x+1/2*c),2^(1/2))*a^2*b^2*sin(1/2*d*x+1/2*c)^6-576*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2*sin(1/2*d*x+1/2*c)^8+504*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^4*sin(1/2*d*x+1/2*c)^4+288*(sin(1/2*d*x+1/
2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b^4*sin(1/2*d*x+1/2*c)^4-
168*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^4*si
n(1/2*d*x+1/2*c)^2-96*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)
^2-1)^(1/2)*b^4*sin(1/2*d*x+1/2*c)^2+1152*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10-2304*a^2*b^2*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+1824*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-672*a^2*b^2*cos(1/2*d*x
+1/2*c)*sin(1/2*d*x+1/2*c)^4+36*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+104*b^4*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^4-488*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+392*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^4-66*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-12*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+144*a*b^3*si
n(1/2*d*x+1/2*c)^5-144*a*b^3*sin(1/2*d*x+1/2*c)^3-20*a^3*b*sin(1/2*d*x+1/2*c)+16*a*b^3*sin(1/2*d*x+1/2*c)-36*(
sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+336
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4*sin(1
/2*d*x+1/2*c)^8-384*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c)
,2^(1/2))*b^4*sin(1/2*d*x+1/2*c)^6-672*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^10-864*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^2*b^2*sin(1/2*d*x+1/2*c)^4+288
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^2*b^2*s
in(1/2*d*x+1/2*c)^2+192*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/
2*c),2^(1/2))*b^4*sin(1/2*d*x+1/2*c)^8-672*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip
ticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4*sin(1/2*d*x+1/2*c)^6)/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((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x, algorithm="maxima")

[Out]

e^(-11/2)*integrate((b*sin(d*x + c) + a)^4/cos(d*x + c)^(11/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 243, normalized size = 0.92 \begin {gather*} -\frac {{\left (3 \, \sqrt {2} {\left (7 i \, a^{4} - 12 i \, a^{2} b^{2} + 4 i \, b^{4}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-7 i \, a^{4} + 12 i \, a^{2} b^{2} - 4 i \, b^{4}\right )} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (36 \, a b^{3} \cos \left (d x + c\right )^{2} - 20 \, a^{3} b - 20 \, a b^{3} - {\left (3 \, {\left (7 \, a^{4} - 12 \, a^{2} b^{2} + 4 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 5 \, a^{4} + 30 \, a^{2} b^{2} + 5 \, b^{4} + {\left (7 \, a^{4} - 12 \, a^{2} b^{2} - 11 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}\right )} e^{\left (-\frac {11}{2}\right )}}{45 \, d \cos \left (d x + c\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x, algorithm="fricas")

[Out]

-1/45*(3*sqrt(2)*(7*I*a^4 - 12*I*a^2*b^2 + 4*I*b^4)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(
-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt(2)*(-7*I*a^4 + 12*I*a^2*b^2 - 4*I*b^4)*cos(d*x + c)^5*weierstr
assZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(36*a*b^3*cos(d*x + c)^2 - 20*a^
3*b - 20*a*b^3 - (3*(7*a^4 - 12*a^2*b^2 + 4*b^4)*cos(d*x + c)^4 + 5*a^4 + 30*a^2*b^2 + 5*b^4 + (7*a^4 - 12*a^2
*b^2 - 11*b^4)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(cos(d*x + c)))*e^(-11/2)/(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((a+b*sin(d*x+c))**4/(e*cos(d*x+c))**(11/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^4/(e*cos(d*x+c))^(11/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^4*e^(-11/2)/cos(d*x + c)^(11/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(c + d*x))^4/(e*cos(c + d*x))^(11/2),x)

[Out]

int((a + b*sin(c + d*x))^4/(e*cos(c + d*x))^(11/2), x)

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