3.237 \(\int \frac{1}{(a \cos (c+d x)+b \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac{2 \sqrt{a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{d \left (a^2+b^2\right ) \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}}-\frac{2 (b \cos (c+d x)-a \sin (c+d x))}{d \left (a^2+b^2\right ) \sqrt{a \cos (c+d x)+b \sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0576028, antiderivative size = 138, normalized size of antiderivative = 1., 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, 2639} \[ -\frac{2 \sqrt{a \cos (c+d x)+b \sin (c+d x)} E\left (\left .\frac{1}{2} \left (c+d x-\tan ^{-1}(a,b)\right )\right |2\right )}{d \left (a^2+b^2\right ) \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}}-\frac{2 (b \cos (c+d x)-a \sin (c+d x))}{d \left (a^2+b^2\right ) \sqrt{a \cos (c+d x)+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-3/2),x]

[Out]

(-2*(b*Cos[c + d*x] - a*Sin[c + d*x]))/((a^2 + b^2)*d*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]]) - (2*EllipticE[(c
+ d*x - ArcTan[a, b])/2, 2]*Sqrt[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)*d*Sqrt[(a*Cos[c + d*x] + b*Si
n[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))^{3/2}} \, dx &=-\frac{2 (b \cos (c+d x)-a \sin (c+d x))}{\left (a^2+b^2\right ) d \sqrt{a \cos (c+d x)+b \sin (c+d x)}}-\frac{\int \sqrt{a \cos (c+d x)+b \sin (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac{2 (b \cos (c+d x)-a \sin (c+d x))}{\left (a^2+b^2\right ) d \sqrt{a \cos (c+d x)+b \sin (c+d x)}}-\frac{\sqrt{a \cos (c+d x)+b \sin (c+d x)} \int \sqrt{\cos \left (c+d x-\tan ^{-1}(a,b)\right )} \, dx}{\left (a^2+b^2\right ) \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))}{\left (a^2+b^2\right ) d \sqrt{a \cos (c+d x)+b \sin (c+d x)}}-\frac{2 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)}}{\left (a^2+b^2\right ) d \sqrt{\frac{a \cos (c+d x)+b \sin (c+d x)}{\sqrt{a^2+b^2}}}}\\ \end{align*}

Mathematica [C]  time = 3.11977, size = 219, normalized size = 1.59 \[ \frac{\frac{\tan \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \sqrt{a \sqrt{\frac{b^2}{a^2}+1} \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )} \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )\right )}{\sqrt{\sin ^2\left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )}}-\tan \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right ) \sqrt{a \sqrt{\frac{b^2}{a^2}+1} \cos \left (-\tan ^{-1}\left (\frac{b}{a}\right )+c+d x\right )}-\frac{2 b \cos (c+d x)}{\sqrt{a \cos (c+d x)+b \sin (c+d x)}}+\frac{2 a \sin (c+d x)}{\sqrt{a \cos (c+d x)+b \sin (c+d x)}}}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cos[c + d*x] + b*Sin[c + d*x])^(-3/2),x]

[Out]

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

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Maple [A]  time = 1.742, size = 228, normalized size = 1.7 \begin{align*}{\frac{1}{\cos \left ( dx+c-\arctan \left ( -a,b \right ) \right ) d} \left ( 2\,\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 ) }{\it EllipticE} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },1/2\,\sqrt{2} \right ) -\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 ) }{\it EllipticF} \left ( \sqrt{1+\sin \left ( dx+c-\arctan \left ( -a,b \right ) \right ) },{\frac{\sqrt{2}}{2}} \right ) -2\, \left ( \cos \left ( dx+c-\arctan \left ( -a,b \right ) \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}}{\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))^(3/2),x)

[Out]

(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)*Ell
ipticE((1+sin(d*x+c-arctan(-a,b)))^(1/2),1/2*2^(1/2))-(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))-2*cos
(d*x+c-arctan(-a,b))^2)/(a^2+b^2)^(1/2)/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., 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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + b*sin(d*x + c))^(-3/2), 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 )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}, 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))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \cos{\left (c + d x \right )} + b \sin{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \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))**(3/2),x)

[Out]

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

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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{3}{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))^(3/2),x, algorithm="giac")

[Out]

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