3.570 \(\int \frac{(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=216 \[ \frac{2 a b \left (a^2+14 b^2\right ) \sqrt{e \cos (c+d x)}}{3 d e^3}+\frac{2 b \left (a^2+2 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 \left (-12 a^2 b^2+a^4-4 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt{e \cos (c+d x)}}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{3 d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}} \]

[Out]

(2*a*b*(a^2 + 14*b^2)*Sqrt[e*Cos[c + d*x]])/(3*d*e^3) + (2*(a^4 - 12*a^2*b^2 - 4*b^4)*Sqrt[Cos[c + d*x]]*Ellip
ticF[(c + d*x)/2, 2])/(3*d*e^2*Sqrt[e*Cos[c + d*x]]) + (2*b*(a^2 + 2*b^2)*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c +
d*x]))/(3*d*e^3) + (2*a*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x])^2)/(3*d*e^3) + (2*(b + a*Sin[c + d*x])*(a
+ b*Sin[c + d*x])^3)/(3*d*e*(e*Cos[c + d*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.445278, antiderivative size = 216, 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, 2862, 2669, 2642, 2641} \[ \frac{2 a b \left (a^2+14 b^2\right ) \sqrt{e \cos (c+d x)}}{3 d e^3}+\frac{2 b \left (a^2+2 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 \left (-12 a^2 b^2+a^4-4 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt{e \cos (c+d x)}}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{3 d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(5/2),x]

[Out]

(2*a*b*(a^2 + 14*b^2)*Sqrt[e*Cos[c + d*x]])/(3*d*e^3) + (2*(a^4 - 12*a^2*b^2 - 4*b^4)*Sqrt[Cos[c + d*x]]*Ellip
ticF[(c + d*x)/2, 2])/(3*d*e^2*Sqrt[e*Cos[c + d*x]]) + (2*b*(a^2 + 2*b^2)*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c +
d*x]))/(3*d*e^3) + (2*a*b*Sqrt[e*Cos[c + d*x]]*(a + b*Sin[c + d*x])^2)/(3*d*e^3) + (2*(b + a*Sin[c + d*x])*(a
+ b*Sin[c + d*x])^3)/(3*d*e*(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 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))^{5/2}} \, dx &=\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}}-\frac{2 \int \frac{(a+b \sin (c+d x))^2 \left (-\frac{a^2}{2}+3 b^2+\frac{5}{2} a b \sin (c+d x)\right )}{\sqrt{e \cos (c+d x)}} \, dx}{3 e^2}\\ &=\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{3 d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}}-\frac{4 \int \frac{(a+b \sin (c+d x)) \left (-\frac{5}{4} a \left (a^2-10 b^2\right )+\frac{15}{4} b \left (a^2+2 b^2\right ) \sin (c+d x)\right )}{\sqrt{e \cos (c+d x)}} \, dx}{15 e^2}\\ &=\frac{2 b \left (a^2+2 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{3 d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}}-\frac{8 \int \frac{-\frac{15}{8} \left (a^4-12 a^2 b^2-4 b^4\right )+\frac{15}{8} a b \left (a^2+14 b^2\right ) \sin (c+d x)}{\sqrt{e \cos (c+d x)}} \, dx}{45 e^2}\\ &=\frac{2 a b \left (a^2+14 b^2\right ) \sqrt{e \cos (c+d x)}}{3 d e^3}+\frac{2 b \left (a^2+2 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{3 d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}}+\frac{\left (a^4-12 a^2 b^2-4 b^4\right ) \int \frac{1}{\sqrt{e \cos (c+d x)}} \, dx}{3 e^2}\\ &=\frac{2 a b \left (a^2+14 b^2\right ) \sqrt{e \cos (c+d x)}}{3 d e^3}+\frac{2 b \left (a^2+2 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{3 d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}}+\frac{\left (\left (a^4-12 a^2 b^2-4 b^4\right ) \sqrt{\cos (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 e^2 \sqrt{e \cos (c+d x)}}\\ &=\frac{2 a b \left (a^2+14 b^2\right ) \sqrt{e \cos (c+d x)}}{3 d e^3}+\frac{2 \left (a^4-12 a^2 b^2-4 b^4\right ) \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d e^2 \sqrt{e \cos (c+d x)}}+\frac{2 b \left (a^2+2 b^2\right ) \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))}{3 d e^3}+\frac{2 a b \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2}{3 d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{3 d e (e \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.15919, size = 137, normalized size = 0.63 \[ \frac{24 a^2 b^2 \sin (c+d x)+4 \left (-12 a^2 b^2+a^4-4 b^4\right ) \cos ^{\frac{3}{2}}(c+d x) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+16 a^3 b+4 a^4 \sin (c+d x)+24 a b^3 \cos (2 (c+d x))+40 a b^3+5 b^4 \sin (c+d x)+b^4 \sin (3 (c+d x))}{6 d e (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(5/2),x]

[Out]

(16*a^3*b + 40*a*b^3 + 24*a*b^3*Cos[2*(c + d*x)] + 4*(a^4 - 12*a^2*b^2 - 4*b^4)*Cos[c + d*x]^(3/2)*EllipticF[(
c + d*x)/2, 2] + 4*a^4*Sin[c + d*x] + 24*a^2*b^2*Sin[c + d*x] + 5*b^4*Sin[c + d*x] + b^4*Sin[3*(c + d*x)])/(6*
d*e*(e*Cos[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [B]  time = 2.206, size = 575, normalized size = 2.7 \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))^(5/2),x)

[Out]

-2/3/(2*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^2*(8*b^4*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^6+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)*EllipticF(cos(1/2*
d*x+1/2*c),2^(1/2))*a^4*sin(1/2*d*x+1/2*c)^2-24*(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)^2-8*(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+48*a*b^3*sin(1/2*d*x+1/2*c
)^5-8*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-(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)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^4+2*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+12*a^2*b^2*cos(1/2*d*x+
1/2*c)*sin(1/2*d*x+1/2*c)^2-48*a*b^3*sin(1/2*d*x+1/2*c)^3+4*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+4*a^3*
b*sin(1/2*d*x+1/2*c)+16*a*b^3*sin(1/2*d*x+1/2*c))/d

________________________________________________________________________________________

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{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(5/2), x)

________________________________________________________________________________________

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^{3} \cos \left (d x + c\right )^{3}}, 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))^(5/2),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^3*cos(d*x + c)^3), x)

________________________________________________________________________________________

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))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^4/(e*cos(d*x + c))^(5/2), x)