3.1078 \(\int \cos ^{\frac{3}{2}}(c+d x) (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=48 \[ \frac{2 (A+3 C) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

[Out]

(2*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.0600691, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4066, 3014, 2641} \[ \frac{2 (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(2*(A + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*A*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 4066

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[b^2, Int
[(b*Cos[e + f*x])^(m - 2)*(C + A*Cos[e + f*x]^2), x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !IntegerQ[m]

Rule 3014

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[
e + f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e + f*
x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

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]

Rubi steps

\begin{align*} \int \cos ^{\frac{3}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\int \frac{C+A \cos ^2(c+d x)}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} (A+3 C) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 (A+3 C) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 A \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 0.940077, size = 124, normalized size = 2.58 \[ -\frac{4 \sin (c) \sqrt{\cos (c+d x)} \left (A \cos ^2(c+d x)+C\right ) \left ((A+3 C) \sqrt{\csc ^2(c)} \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )-A \csc (c) \sin (c+d x)\right )}{3 d (A \cos (2 (c+d x))+A+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2),x]

[Out]

(-4*Sqrt[Cos[c + d*x]]*(C + A*Cos[c + d*x]^2)*Sin[c]*((A + 3*C)*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^
2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]] - A*Csc[c]*Sin[
c + d*x]))/(3*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [B]  time = 1.782, size = 228, normalized size = 4.8 \begin{align*} -{\frac{2}{3\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 4\,A\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+A\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{2} \right ) -2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) A+3\,C\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x)

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+A*(s
in(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))-2*sin(1/2*d*
x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A+3*C*(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(co
s(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*
d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)*sec(d*x + c)^2 + A*cos(d*x + c))*sqrt(cos(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(cos(d*x+c)**(3/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*cos(d*x + c)^(3/2), x)