3.238 \(\int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=142 \[ \frac {2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{3 d \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^{3/2}} \]

[Out]

-2/3*(b*cos(d*x+c)-a*sin(d*x+c))/(a^2+b^2)/d/(a*cos(d*x+c)+b*sin(d*x+c))^(3/2)+2/3*(cos(1/2*c+1/2*d*x-1/2*arct
an(a,b))^2)^(1/2)/cos(1/2*c+1/2*d*x-1/2*arctan(a,b))*EllipticF(sin(1/2*c+1/2*d*x-1/2*arctan(a,b)),2^(1/2))*((a
*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^(1/2))^(1/2)/(a^2+b^2)/d/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3076, 3078, 2641} \[ \frac {2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{3 d \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{3 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(3*(a^2 + b^2)*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^(3/2)) + (2*Elliptic
F[(c + d*x - ArcTan[a, b])/2, 2]*Sqrt[(a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2]])/(3*(a^2 + b^2)*d*Sqr
t[a*Cos[c + d*x] + b*Sin[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]

Rule 3076

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*Cos[c + d*x] -
a*Sin[c + d*x])*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1))/(d*(n + 1)*(a^2 + b^2)), x] + Dist[(n + 2)/((n + 1
)*(a^2 + b^2)), Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && LtQ[n, -1] && NeQ[n, -2]

Rule 3078

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[c + d*x] +
b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2])^n, Int[Cos[c + d*x - ArcTan[a, b]]^n, x]
, x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])

Rubi steps

\begin {align*} \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{5/2}} \, dx &=-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{3 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{3/2}}+\frac {\int \frac {1}{\sqrt {a \cos (c+d x)+b \sin (c+d x)}} \, dx}{3 \left (a^2+b^2\right )}\\ &=-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{3 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{3/2}}+\frac {\sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}} \int \frac {1}{\sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )}} \, dx}{3 \left (a^2+b^2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}\\ &=-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{3 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{3/2}}+\frac {2 F\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}{3 \left (a^2+b^2\right ) d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 1.68, size = 145, normalized size = 1.02 \[ \frac {2 \left (\frac {\tan \left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right ) \sqrt {\cos ^2\left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2\left (c+d x+\tan ^{-1}\left (\frac {a}{b}\right )\right )\right )}{\sqrt {b \sqrt {\frac {a^2}{b^2}+1} \sin \left (\tan ^{-1}\left (\frac {a}{b}\right )+c+d x\right )}}+\frac {a \sin (c+d x)-b \cos (c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}}\right )}{3 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-(b*Cos[c + d*x]) + a*Sin[c + d*x])/(a*Cos[c + d*x] + b*Sin[c + d*x])^(3/2) + (Sqrt[Cos[c + d*x + ArcTan[
a/b]]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[c + d*x + ArcTan[a/b]]^2]*Tan[c + d*x + ArcTan[a/b]])/Sqrt[S
qrt[1 + a^2/b^2]*b*Sin[c + d*x + ArcTan[a/b]]]))/(3*(a^2 + b^2)*d)

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fricas [F]  time = 1.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}{3 \, a b^{2} \cos \left (d x + c\right ) + {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (b^{3} + {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*cos(d*x + c) + b*sin(d*x + c))/(3*a*b^2*cos(d*x + c) + (a^3 - 3*a*b^2)*cos(d*x + c)^3 + (b^3 +
(3*a^2*b - b^3)*cos(d*x + c)^2)*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.34, size = 178, normalized size = 1.25 \[ \frac {\sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \sqrt {-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )}\, \EllipticF \left (\sqrt {1+\sin \left (d x +c -\arctan \left (-a , b\right )\right )}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c -\arctan \left (-a , b\right )\right )-2 \left (\cos ^{2}\left (d x +c -\arctan \left (-a , b\right )\right )\right )}{3 \sin \left (d x +c -\arctan \left (-a , b\right )\right ) \left (a^{2}+b^{2}\right ) \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cos(c + d*x) + b*sin(c + d*x))^(5/2),x)

[Out]

int(1/(a*cos(c + d*x) + b*sin(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(1/(a*cos(d*x+c)+b*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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