\(\int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx\) [239]

Optimal result

Integrand size = 21, antiderivative size = 197 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{5/2}}-\frac {6 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right )^2 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {6 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{5 \left (a^2+b^2\right )^2 d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \]

[Out]

-2/5*(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))^(5/2)-6/5*(b*cos(d*x+c)-a*sin(d*x+c))
/(a^2+b^2)^2/d/(a*cos(d*x+c)+b*sin(d*x+c))^(1/2)-6/5*(cos(1/2*c+1/2*d*x-1/2*arctan(a,b))^2)^(1/2)/cos(1/2*c+1/
2*d*x-1/2*arctan(a,b))*EllipticE(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))^(1/2)
/(a^2+b^2)^2/d/((a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3155, 3157, 2719} \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=-\frac {6 \sqrt {a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{5 d \left (a^2+b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}}-\frac {6 (b \cos (c+d x)-a \sin (c+d x))}{5 d \left (a^2+b^2\right )^2 \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^{5/2}} \]

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-7/2),x]

[Out]

(-2*(b*Cos[c + d*x] - a*Sin[c + d*x]))/(5*(a^2 + b^2)*d*(a*Cos[c + d*x] + b*Sin[c + d*x])^(5/2)) - (6*(b*Cos[c
+ d*x] - a*Sin[c + d*x]))/(5*(a^2 + b^2)^2*d*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]]) - (6*EllipticE[(c + d*x -
ArcTan[a, b])/2, 2]*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]])/(5*(a^2 + b^2)^2*d*Sqrt[(a*Cos[c + d*x] + b*Sin[c
+ d*x])/Sqrt[a^2 + b^2]])

Rule 2719

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

Rule 3155

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 3157

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*} \text {integral}& = -\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{5/2}}+\frac {3 \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx}{5 \left (a^2+b^2\right )} \\ & = -\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{5/2}}-\frac {6 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right )^2 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {3 \int \sqrt {a \cos (c+d x)+b \sin (c+d x)} \, dx}{5 \left (a^2+b^2\right )^2} \\ & = -\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{5/2}}-\frac {6 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right )^2 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {\left (3 \sqrt {a \cos (c+d x)+b \sin (c+d x)}\right ) \int \sqrt {\cos \left (c+d x-\tan ^{-1}(a,b)\right )} \, dx}{5 \left (a^2+b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \\ & = -\frac {2 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right ) d (a \cos (c+d x)+b \sin (c+d x))^{5/2}}-\frac {6 (b \cos (c+d x)-a \sin (c+d x))}{5 \left (a^2+b^2\right )^2 d \sqrt {a \cos (c+d x)+b \sin (c+d x)}}-\frac {6 E\left (\left .\frac {1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right ) \sqrt {a \cos (c+d x)+b \sin (c+d x)}}{5 \left (a^2+b^2\right )^2 d \sqrt {\frac {a \cos (c+d x)+b \sin (c+d x)}{\sqrt {a^2+b^2}}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.93 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=\frac {-\frac {2 \left (3 a^2 \cos ^3(c+d x)-a b \sin (c+d x)+6 a b \cos ^2(c+d x) \sin (c+d x)+b^2 \cos (c+d x) \left (1+3 \sin ^2(c+d x)\right )\right )}{(a \cos (c+d x)+b \sin (c+d x))^{5/2}}+\frac {\cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right ) \left (3 b \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right ) \sin \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )-3 \sqrt {\sin ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )} \left (-2 a \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )+b \sin \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right )\right )}{\left (a \sqrt {1+\frac {b^2}{a^2}} \cos \left (c+d x-\arctan \left (\frac {b}{a}\right )\right )\right )^{3/2} \sqrt {\sin ^2\left (c+d x-\arctan \left (\frac {b}{a}\right )\right )}}}{5 b \left (a^2+b^2\right ) d} \]

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-7/2),x]

[Out]

((-2*(3*a^2*Cos[c + d*x]^3 - a*b*Sin[c + d*x] + 6*a*b*Cos[c + d*x]^2*Sin[c + d*x] + b^2*Cos[c + d*x]*(1 + 3*Si
n[c + d*x]^2)))/(a*Cos[c + d*x] + b*Sin[c + d*x])^(5/2) + (Cos[c + d*x - ArcTan[b/a]]*(3*b*HypergeometricPFQ[{
-1/2, -1/4}, {3/4}, Cos[c + d*x - ArcTan[b/a]]^2]*Sin[c + d*x - ArcTan[b/a]] - 3*Sqrt[Sin[c + d*x - ArcTan[b/a
]]^2]*(-2*a*Cos[c + d*x - ArcTan[b/a]] + b*Sin[c + d*x - ArcTan[b/a]])))/((a*Sqrt[1 + b^2/a^2]*Cos[c + d*x - A
rcTan[b/a]])^(3/2)*Sqrt[Sin[c + d*x - ArcTan[b/a]]^2]))/(5*b*(a^2 + b^2)*d)

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.51

method	result	size
default	\(\frac {\sqrt {a^{2}+b^{2}}\, \left (6 \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{\frac {7}{2}} \operatorname {EllipticE}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}\, \sqrt {2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )+2}\, \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{\frac {7}{2}} \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c -\arctan \left (-a , b\right )\right )+1}, \frac {\sqrt {2}}{2}\right )+6 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{5}-4 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{3}-2 \sin \left (d x +c -\arctan \left (-a , b\right )\right )\right )}{5 \sin \left (d x +c -\arctan \left (-a , b\right )\right )^{3} \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \cos \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {\sin \left (d x +c -\arctan \left (-a , b\right )\right ) \sqrt {a^{2}+b^{2}}}\, d}\)	\(297\)

[In]

int(1/(cos(d*x+c)*a+b*sin(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/5*(a^2+b^2)^(1/2)*(6*(-sin(d*x+c-arctan(-a,b))+1)^(1/2)*(2*sin(d*x+c-arctan(-a,b))+2)^(1/2)*sin(d*x+c-arctan
(-a,b))^(7/2)*EllipticE((-sin(d*x+c-arctan(-a,b))+1)^(1/2),1/2*2^(1/2))-3*(-sin(d*x+c-arctan(-a,b))+1)^(1/2)*(
2*sin(d*x+c-arctan(-a,b))+2)^(1/2)*sin(d*x+c-arctan(-a,b))^(7/2)*EllipticF((-sin(d*x+c-arctan(-a,b))+1)^(1/2),
1/2*2^(1/2))+6*sin(d*x+c-arctan(-a,b))^5-4*sin(d*x+c-arctan(-a,b))^3-2*sin(d*x+c-arctan(-a,b)))/sin(d*x+c-arct
an(-a,b))^3/(a^4+2*a^2*b^2+b^4)/cos(d*x+c-arctan(-a,b))/(sin(d*x+c-arctan(-a,b))*(a^2+b^2)^(1/2))^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 552, normalized size of antiderivative = 2.80 \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=\frac {3 \, {\left (-3 i \, \sqrt {2} a b^{2} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, a^{3} + 3 i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (-i \, \sqrt {2} b^{3} + \sqrt {2} {\left (-3 i \, a^{2} b + i \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a - i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} + 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (3 i \, \sqrt {2} a b^{2} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, a^{3} - 3 i \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (i \, \sqrt {2} b^{3} + \sqrt {2} {\left (3 i \, a^{2} b - i \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a + i \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a^{2} - 2 i \, a b - b^{2}\right )}}{a^{2} + b^{2}}, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (5 \, a^{2} b - 4 \, b^{3}\right )} \cos \left (d x + c\right ) - {\left (a^{3} + 4 \, a b^{2} + 3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}{5 \, {\left ({\left (a^{7} - a^{5} b^{2} - 5 \, a^{3} b^{4} - 3 \, a b^{6}\right )} d \cos \left (d x + c\right )^{3} + 3 \, {\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left ({\left (3 \, a^{6} b + 5 \, a^{4} b^{3} + a^{2} b^{5} - b^{7}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/5*(3*(-3*I*sqrt(2)*a*b^2*cos(d*x + c) + sqrt(2)*(-I*a^3 + 3*I*a*b^2)*cos(d*x + c)^3 + (-I*sqrt(2)*b^3 + sqrt
(2)*(-3*I*a^2*b + I*b^3)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(a - I*b)*weierstrassZeta(-4*(a^2 + 2*I*a*b - b^2)/
(a^2 + b^2), 0, weierstrassPInverse(-4*(a^2 + 2*I*a*b - b^2)/(a^2 + b^2), 0, cos(d*x + c) + I*sin(d*x + c))) +
3*(3*I*sqrt(2)*a*b^2*cos(d*x + c) + sqrt(2)*(I*a^3 - 3*I*a*b^2)*cos(d*x + c)^3 + (I*sqrt(2)*b^3 + sqrt(2)*(3*
I*a^2*b - I*b^3)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(a + I*b)*weierstrassZeta(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b
^2), 0, weierstrassPInverse(-4*(a^2 - 2*I*a*b - b^2)/(a^2 + b^2), 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(3
*a^2*b - b^3)*cos(d*x + c)^3 - (5*a^2*b - 4*b^3)*cos(d*x + c) - (a^3 + 4*a*b^2 + 3*(a^3 - 3*a*b^2)*cos(d*x + c
)^2)*sin(d*x + c))*sqrt(a*cos(d*x + c) + b*sin(d*x + c)))/((a^7 - a^5*b^2 - 5*a^3*b^4 - 3*a*b^6)*d*cos(d*x + c
)^3 + 3*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + ((3*a^6*b + 5*a^4*b^3 + a^2*b^5 - b^7)*d*cos(d*x + c)^2
+ (a^4*b^3 + 2*a^2*b^5 + b^7)*d)*sin(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + b*sin(d*x + c))^(-7/2), x)

Giac [F]

\[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=\int { \frac {1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate(1/(a*cos(d*x+c)+b*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + b*sin(d*x + c))^(-7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a \cos (c+d x)+b \sin (c+d x))^{7/2}} \, dx=\int \frac {1}{{\left (a\,\cos \left (c+d\,x\right )+b\,\sin \left (c+d\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(1/(a*cos(c + d*x) + b*sin(c + d*x))^(7/2),x)

[Out]

int(1/(a*cos(c + d*x) + b*sin(c + d*x))^(7/2), x)