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

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

[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))

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Rubi [A]  time = 0.479028, antiderivative size = 264, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2691, 2861, 2669, 2640, 2639} \[ \frac{2 a b \left (21 a^2-22 b^2\right ) (e \cos (c+d x))^{3/2}}{45 d e^7}+\frac{2 \left (\left (7 a^2-6 b^2\right ) \sin (c+d x)+a b\right ) (a+b \sin (c+d x))^2}{45 d e^3 (e \cos (c+d x))^{5/2}}-\frac{2 \left (b \left (7 a^2-6 b^2\right )-a \left (21 a^2-22 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))}{45 d e^5 \sqrt{e \cos (c+d x)}}-\frac{2 \left (-12 a^2 b^2+7 a^4+4 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{15 d e^6 \sqrt{\cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{9 d e (e \cos (c+d x))^{9/2}} \]

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 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 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
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 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

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

Mathematica [A]  time = 1.59691, size = 219, normalized size = 0.83 \[ \frac{\sec ^5(c+d x) \sqrt{e \cos (c+d x)} \left (360 a^2 b^2 \sin (c+d x)-156 a^2 b^2 \sin (3 (c+d x))-36 a^2 b^2 \sin (5 (c+d x))-48 \left (-12 a^2 b^2+7 a^4+4 b^4\right ) \cos ^{\frac{9}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+320 a^3 b+150 a^4 \sin (c+d x)+91 a^4 \sin (3 (c+d x))+21 a^4 \sin (5 (c+d x))-288 a b^3 \cos (2 (c+d x))+32 a b^3+60 b^4 \sin (c+d x)-8 b^4 \sin (3 (c+d x))+12 b^4 \sin (5 (c+d x))\right )}{360 d e^6} \]

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]  time = 7.477, size = 1416, normalized size = 5.4 \begin{align*} \text{result too large to display} \end{align*}

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)

[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*(-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+104*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+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*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+504*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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*EllipticE(cos(1/2*d*x+1/2*c),2
^(1/2))*(sin(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)^2-96*EllipticE(co
s(1/2*d*x+1/2*c),2^(1/2))*(sin(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+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^4-384
*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)^6-488*b^4*cos(1/2*d*x+1/2*c)*s
in(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*sin(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)-576*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b^2*sin(1/2*d*x+1/2*c)^8+1152*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*b^2*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+336*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*sin(1/2*d*x+1/2*c)^8+192*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^4*sin(1/2*d*x+1/2*c)^8-672*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)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^4*sin(1/2*d*x+1/2*c)^6-384*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*b^4*sin(1/2*d
*x+1/2*c)^6-864*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*a^2*b^2*sin(1/2*d*x+1/2*c)^2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac{11}{2}}}\,{d x} \end{align*}

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]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{e \cos \left (d x + c\right )}}{e^{6} \cos \left (d x + c\right )^{6}}, x\right ) \end{align*}

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]

integral((b^4*cos(d*x + c)^4 + a^4 + 6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)*cos(d*x + c)^2 - 4*(a*b^3*cos(d*x +
c)^2 - a^3*b - a*b^3)*sin(d*x + c))*sqrt(e*cos(d*x + c))/(e^6*cos(d*x + c)^6), 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((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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac{11}{2}}}\,{d x} \end{align*}

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*cos(d*x + c))^(11/2), x)