3.2.20 \(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {b \cos (c+d x)}} \, dx\) [120]

Optimal. Leaf size=59 \[ \frac {C x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}}+\frac {A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

[Out]

A*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)+C*x*cos(d*x+c)^(1/2)/(b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {18, 3091, 8} \begin {gather*} \frac {A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}}+\frac {C x \sqrt {\cos (c+d x)}}{\sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x*Sqrt[Cos[c + d*x]])/Sqrt[b*Cos[c + d*x]] + (A*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 18

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m - 1/2)*b^(n + 1/2)*(Sqrt[a*v]/Sqrt[b*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]

Rule 3091

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 45, normalized size = 0.76 \begin {gather*} \frac {C d x \cos (c+d x)+A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(C*d*x*Cos[c + d*x] + A*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]])

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Maple [A]
time = 0.29, size = 45, normalized size = 0.76

method	result	size
default	\(\frac {C \cos \left (d x +c \right ) \left (d x +c \right )+A \sin \left (d x +c \right )}{d \sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}}\)	\(45\)
risch	\(\frac {C x \left (\sqrt {\cos }\left (d x +c \right )\right )}{\sqrt {b \cos \left (d x +c \right )}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{\sqrt {b \cos \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )}\, d}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(C*cos(d*x+c)*(d*x+c)+A*sin(d*x+c))/(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(1/2)

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Maxima [A]
time = 0.57, size = 85, normalized size = 1.44 \begin {gather*} \frac {2 \, {\left (\frac {C \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{\sqrt {b}} + \frac {A \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, d x + 2 \, c\right ) + b}\right )}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2*(C*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/sqrt(b) + A*sqrt(b)*sin(2*d*x + 2*c)/(b*cos(2*d*x + 2*c)^2 + b*si
n(2*d*x + 2*c)^2 + 2*b*cos(2*d*x + 2*c) + b))/d

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Fricas [A]
time = 0.43, size = 191, normalized size = 3.24 \begin {gather*} \left [-\frac {C \sqrt {-b} \cos \left (d x + c\right )^{2} \log \left (2 \, b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - b\right ) - 2 \, \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b d \cos \left (d x + c\right )^{2}}, \frac {C \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {b} \cos \left (d x + c\right )^{\frac {3}{2}}}\right ) \cos \left (d x + c\right )^{2} + \sqrt {b \cos \left (d x + c\right )} A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b d \cos \left (d x + c\right )^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/2*(C*sqrt(-b)*cos(d*x + c)^2*log(2*b*cos(d*x + c)^2 + 2*sqrt(b*cos(d*x + c))*sqrt(-b)*sqrt(cos(d*x + c))*s
in(d*x + c) - b) - 2*sqrt(b*cos(d*x + c))*A*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^2), (C*sqrt(b)*
arctan(sqrt(b*cos(d*x + c))*sin(d*x + c)/(sqrt(b)*cos(d*x + c)^(3/2)))*cos(d*x + c)^2 + sqrt(b*cos(d*x + c))*A
*sqrt(cos(d*x + c))*sin(d*x + c))/(b*d*cos(d*x + c)^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\sqrt {b \cos {\left (c + d x \right )}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(b*cos(d*x+c))**(1/2),x)

[Out]

Integral((A + C*cos(c + d*x)**2)/(sqrt(b*cos(c + d*x))*cos(c + d*x)**(3/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.25, size = 84, normalized size = 1.42 \begin {gather*} \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (A\,\sin \left (2\,c+2\,d\,x\right )+C\,d\,x+C\,d\,x\,\cos \left (2\,c+2\,d\,x\right )+A\,1{}\mathrm {i}+A\,\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{b\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(3/2)*(b*cos(c + d*x))^(1/2)),x)

[Out]

((b*cos(c + d*x))^(1/2)*(A*1i + A*cos(2*c + 2*d*x)*1i + A*sin(2*c + 2*d*x) + C*d*x + C*d*x*cos(2*c + 2*d*x)))/
(b*d*cos(c + d*x)^(1/2)*(cos(2*c + 2*d*x) + 1))

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