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

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

[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]])

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Rubi [A]  time = 0.0859234, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3076, 3078, 2639} \[ -\frac{6 \sqrt{a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{5 d \left (a^2+b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}}-\frac{6 (b \cos (c+d x)-a \sin (c+d x))}{5 d \left (a^2+b^2\right )^2 \sqrt{a \cos (c+d x)+b \sin (c+d x)}}-\frac{2 (b \cos (c+d x)-a \sin (c+d x))}{5 d \left (a^2+b^2\right ) (a \cos (c+d x)+b \sin (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[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 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])

Rule 2639

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

Rubi steps

\begin{align*} \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{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]  time = 2.42064, size = 277, normalized size = 1.41 \[ \frac{\frac{\cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \left (3 b \sin \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )-3 \sqrt{\sin ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )} \left (b \sin \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )-2 a \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )\right )}{\sqrt{\sin ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )} \left (a \sqrt{\frac{b^2}{a^2}+1} \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )^{3/2}}-\frac{2 \left (3 a^2 \cos ^3(c+d x)-a b \sin (c+d x)+6 a b \sin (c+d x) \cos ^2(c+d x)+b^2 \left (3 \sin ^2(c+d x)+1\right ) \cos (c+d x)\right )}{(a \cos (c+d x)+b \sin (c+d x))^{5/2}}}{5 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 2.812, size = 309, normalized size = 1.6 \begin{align*}{\frac{1}{5\, \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{2} \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \cos \left ( dx+c-\arctan \left ( -a,b \right ) \right ) d}\sqrt{{a}^{2}+{b}^{2}} \left ( 6\,\sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) +2}\sqrt{-\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) } \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{2}{\it EllipticE} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },1/2\,\sqrt{2} \right ) -3\,\sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) }\sqrt{-2\,\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) +2}\sqrt{-\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) } \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{2}{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },1/2\,\sqrt{2} \right ) +6\, \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{4}-4\, \left ( \sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{2}-2 \right ){\frac{1}{\sqrt{\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \sqrt{{a}^{2}+{b}^{2}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/sin(d*x+c-arctan(-a,b))^2*(a^2+b^2)^(1/2)*(6*(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)*sin(d*x+c-arctan(-a,b))^2*EllipticE((1+sin(d*x+c-arctan(-a,b)))^(1/
2),1/2*2^(1/2))-3*(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)*sin(d*x+c-arctan(-a,b))^2*EllipticF((1+sin(d*x+c-arctan(-a,b)))^(1/2),1/2*2^(1/2))+6*sin(d*x+c-arc
tan(-a,b))^4-4*sin(d*x+c-arctan(-a,b))^2-2)/(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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )}}{{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 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 ) +{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*cos(d*x + c) + b*sin(d*x + c))/((a^4 - 6*a^2*b^2 + b^4)*cos(d*x + c)^4 + b^4 + 2*(3*a^2*b^2 -
b^4)*cos(d*x + c)^2 + 4*(a*b^3*cos(d*x + c) + (a^3*b - a*b^3)*cos(d*x + c)^3)*sin(d*x + c)), 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(1/(a*cos(d*x+c)+b*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \cos \left (d x + c\right ) + b \sin \left (d x + c\right )\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)