3.572 \(\int \frac{(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{9/2}} \, dx\)

Optimal. Leaf size=241 \[ \frac{10 a b \left (a^2-2 b^2\right ) \sqrt{e \cos (c+d x)}}{21 d e^5}-\frac{2 \left (a b-\left (5 a^2-6 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{21 d e^3 (e \cos (c+d x))^{3/2}}+\frac{2 b \left (5 a^2-6 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{21 d e^5}+\frac{2 \left (-12 a^2 b^2+5 a^4+12 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt{e \cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}} \]

[Out]

(10*a*b*(a^2 - 2*b^2)*Sqrt[e*Cos[c + d*x]])/(21*d*e^5) + (2*(5*a^4 - 12*a^2*b^2 + 12*b^4)*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2])/(21*d*e^4*Sqrt[e*Cos[c + d*x]]) + (2*b*(5*a^2 - 6*b^2)*Sqrt[e*Cos[c + d*x]]*(a + b*S
in[c + d*x]))/(21*d*e^5) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^3)/(7*d*e*(e*Cos[c + d*x])^(7/2)) - (2
*(a + b*Sin[c + d*x])^2*(a*b - (5*a^2 - 6*b^2)*Sin[c + d*x]))/(21*d*e^3*(e*Cos[c + d*x])^(3/2))

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Rubi [A]  time = 0.462852, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2691, 2861, 2862, 2669, 2642, 2641} \[ \frac{10 a b \left (a^2-2 b^2\right ) \sqrt{e \cos (c+d x)}}{21 d e^5}-\frac{2 \left (a b-\left (5 a^2-6 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{21 d e^3 (e \cos (c+d x))^{3/2}}+\frac{2 b \left (5 a^2-6 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{21 d e^5}+\frac{2 \left (-12 a^2 b^2+5 a^4+12 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt{e \cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(9/2),x]

[Out]

(10*a*b*(a^2 - 2*b^2)*Sqrt[e*Cos[c + d*x]])/(21*d*e^5) + (2*(5*a^4 - 12*a^2*b^2 + 12*b^4)*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2])/(21*d*e^4*Sqrt[e*Cos[c + d*x]]) + (2*b*(5*a^2 - 6*b^2)*Sqrt[e*Cos[c + d*x]]*(a + b*S
in[c + d*x]))/(21*d*e^5) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^3)/(7*d*e*(e*Cos[c + d*x])^(7/2)) - (2
*(a + b*Sin[c + d*x])^2*(a*b - (5*a^2 - 6*b^2)*Sin[c + d*x]))/(21*d*e^3*(e*Cos[c + d*x])^(3/2))

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 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(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))^{9/2}} \, dx &=\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}}-\frac{2 \int \frac{(a+b \sin (c+d x))^2 \left (-\frac{5 a^2}{2}+3 b^2+\frac{1}{2} a b \sin (c+d x)\right )}{(e \cos (c+d x))^{5/2}} \, dx}{7 e^2}\\ &=\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}}-\frac{2 (a+b \sin (c+d x))^2 \left (a b-\left (5 a^2-6 b^2\right ) \sin (c+d x)\right )}{21 d e^3 (e \cos (c+d x))^{3/2}}+\frac{4 \int \frac{(a+b \sin (c+d x)) \left (\frac{1}{4} a \left (5 a^2-2 b^2\right )-\frac{3}{4} b \left (5 a^2-6 b^2\right ) \sin (c+d x)\right )}{\sqrt{e \cos (c+d x)}} \, dx}{21 e^4}\\ &=\frac{2 b \left (5 a^2-6 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{21 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}}-\frac{2 (a+b \sin (c+d x))^2 \left (a b-\left (5 a^2-6 b^2\right ) \sin (c+d x)\right )}{21 d e^3 (e \cos (c+d x))^{3/2}}+\frac{8 \int \frac{\frac{3}{8} \left (5 a^4-12 a^2 b^2+12 b^4\right )-\frac{15}{8} a b \left (a^2-2 b^2\right ) \sin (c+d x)}{\sqrt{e \cos (c+d x)}} \, dx}{63 e^4}\\ &=\frac{10 a b \left (a^2-2 b^2\right ) \sqrt{e \cos (c+d x)}}{21 d e^5}+\frac{2 b \left (5 a^2-6 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{21 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}}-\frac{2 (a+b \sin (c+d x))^2 \left (a b-\left (5 a^2-6 b^2\right ) \sin (c+d x)\right )}{21 d e^3 (e \cos (c+d x))^{3/2}}+\frac{\left (5 a^4-12 a^2 b^2+12 b^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{21 e^4}\\ &=\frac{10 a b \left (a^2-2 b^2\right ) \sqrt{e \cos (c+d x)}}{21 d e^5}+\frac{2 b \left (5 a^2-6 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{21 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}}-\frac{2 (a+b \sin (c+d x))^2 \left (a b-\left (5 a^2-6 b^2\right ) \sin (c+d x)\right )}{21 d e^3 (e \cos (c+d x))^{3/2}}+\frac{\left (\left (5 a^4-12 a^2 b^2+12 b^4\right ) \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 e^4 \sqrt{e \cos (c+d x)}}\\ &=\frac{10 a b \left (a^2-2 b^2\right ) \sqrt{e \cos (c+d x)}}{21 d e^5}+\frac{2 \left (5 a^4-12 a^2 b^2+12 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d e^4 \sqrt{e \cos (c+d x)}}+\frac{2 b \left (5 a^2-6 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{21 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{7 d e (e \cos (c+d x))^{7/2}}-\frac{2 (a+b \sin (c+d x))^2 \left (a b-\left (5 a^2-6 b^2\right ) \sin (c+d x)\right )}{21 d e^3 (e \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.879653, size = 177, normalized size = 0.73 \[ \frac{\sec ^4(c+d x) \sqrt{e \cos (c+d x)} \left (60 a^2 b^2 \sin (c+d x)-12 a^2 b^2 \sin (3 (c+d x))+4 \left (-12 a^2 b^2+5 a^4+12 b^4\right ) \cos ^{\frac{7}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+48 a^3 b+17 a^4 \sin (c+d x)+5 a^4 \sin (3 (c+d x))-56 a b^3 \cos (2 (c+d x))-8 a b^3+3 b^4 \sin (c+d x)-9 b^4 \sin (3 (c+d x))\right )}{42 d e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(9/2),x]

[Out]

(Sqrt[e*Cos[c + d*x]]*Sec[c + d*x]^4*(48*a^3*b - 8*a*b^3 - 56*a*b^3*Cos[2*(c + d*x)] + 4*(5*a^4 - 12*a^2*b^2 +
12*b^4)*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 17*a^4*Sin[c + d*x] + 60*a^2*b^2*Sin[c + d*x] + 3*b^4*
Sin[c + d*x] + 5*a^4*Sin[3*(c + d*x)] - 12*a^2*b^2*Sin[3*(c + d*x)] - 9*b^4*Sin[3*(c + d*x)]))/(42*d*e^5)

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Maple [B]  time = 5.161, size = 1067, normalized size = 4.4 \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))^(9/2),x)

[Out]

-2/21/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*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^4*(72*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-96*a^2*b^2*cos(1/2*d*x+1/2*c)*sin
(1/2*d*x+1/2*c)^6+96*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^2-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b
^4+40*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^4*
sin(1/2*d*x+1/2*c)^6+96*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/
2*c),2^(1/2))*b^4*sin(1/2*d*x+1/2*c)^6-60*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt
icF(cos(1/2*d*x+1/2*c),2^(1/2))*a^4*sin(1/2*d*x+1/2*c)^4-144*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2
*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^4*sin(1/2*d*x+1/2*c)^4+144*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/
2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2*sin(1/2*d*x+1/2*c)^4-96*(2*sin(1
/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2*sin(1/2*d*
x+1/2*c)^6+30*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/
2))*a^4*sin(1/2*d*x+1/2*c)^2+72*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/
2*d*x+1/2*c),2^(1/2))*b^4*sin(1/2*d*x+1/2*c)^2+40*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-72*b^4*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-40*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+16*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)^2-112*a*b^3*sin(1/2*d*x+1/2*c)^5+112*a*b^3*sin(1/2
*d*x+1/2*c)^3+12*a^3*b*sin(1/2*d*x+1/2*c)-16*a*b^3*sin(1/2*d*x+1/2*c)-72*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2*sin(1/2*d*x+1/2*c)^2+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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^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{9}{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))^(9/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(9/2), 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^{5} \cos \left (d x + c\right )^{5}}, 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))^(9/2),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^5*cos(d*x + c)^5), 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))**(9/2),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{9}{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))^(9/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(9/2), x)