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

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

[Out]

2/3*(b+a*sin(d*x+c))*(a+b*sin(d*x+c))^3/d/e/(e*cos(d*x+c))^(3/2)+2/3*(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)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)/d/e^2/(e*cos(d*x+c))^(1/2
)+2/3*a*b*(a^2+14*b^2)*(e*cos(d*x+c))^(1/2)/d/e^3+2/3*b*(a^2+2*b^2)*(a+b*sin(d*x+c))*(e*cos(d*x+c))^(1/2)/d/e^
3+2/3*a*b*(a+b*sin(d*x+c))^2*(e*cos(d*x+c))^(1/2)/d/e^3

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Rubi [A]  time = 0.45, antiderivative size = 216, 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 = {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 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]

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 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 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])

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*}

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Mathematica [A]  time = 1.19, size = 137, normalized size = 0.63 \[ \frac {4 a^4 \sin (c+d x)+16 a^3 b+24 a^2 b^2 \sin (c+d x)+4 \left (a^4-12 a^2 b^2-4 b^4\right ) \cos ^{\frac {3}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+24 a b^3 \cos (2 (c+d x))+40 a b^3+5 b^4 \sin (c+d x)+b^4 \sin (3 (c+d x))}{6 d e (e \cos (c+d x))^{3/2}} \]

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))

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {e \cos \left (d x + c\right )}}{e^{3} \cos \left (d x + c\right )^{3}}, x\right ) \]

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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [B]  time = 2.66, size = 575, normalized size = 2.66 \[ -\frac {2 \left (8 b^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a^{4} \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, a^{2} b^{2} \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, b^{4} \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+48 a \,b^{3} \left (\sin ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 b^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{4}+12 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b^{2}+4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{4}+2 a^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 a^{2} b^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-48 a \,b^{3} \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 b^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a^{3} b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+16 a \,b^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 \left (2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{2} d} \]

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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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)*EllipticF(cos(1/2*d*x+1/2*c),
2^(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)^2-8*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elliptic
F(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{4}}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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