3.1.93 \(\int \frac {\sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [93]

Optimal. Leaf size=68 \[ \frac {A \tanh ^{-1}(\sin (c+d x)) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+\frac {C \sqrt {b \cos (c+d x)} \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 68, 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 = {17, 3093, 3855} \begin {gather*} \frac {A \sqrt {b \cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt {\cos (c+d x)}}+\frac {C \sin (c+d x) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 17

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

Rule 3093

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 44, normalized size = 0.65 \begin {gather*} \frac {\sqrt {b \cos (c+d x)} \left (A \tanh ^{-1}(\sin (c+d x))+C \sin (c+d x)\right )}{d \sqrt {\cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.30, size = 55, normalized size = 0.81

method	result	size
default	\(-\frac {\left (2 A \arctanh \left (\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )-C \sin \left (d x +c \right )\right ) \sqrt {b \cos \left (d x +c \right )}}{d \sqrt {\cos \left (d x +c \right )}}\)	\(55\)
risch	\(-\frac {i \sqrt {b \cos \left (d x +c \right )}\, C \,{\mathrm e}^{i \left (d x +c \right )}}{2 \sqrt {\cos \left (d x +c \right )}\, d}+\frac {i \sqrt {b \cos \left (d x +c \right )}\, C \,{\mathrm e}^{-i \left (d x +c \right )}}{2 \sqrt {\cos \left (d x +c \right )}\, d}-\frac {\sqrt {b \cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}+\frac {\sqrt {b \cos \left (d x +c \right )}\, A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\cos \left (d x +c \right )}\, d}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.64, size = 80, normalized size = 1.18 \begin {gather*} \frac {A \sqrt {b} {\left (\log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} + 2 \, C \sqrt {b} \sin \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 201, normalized size = 2.96 \begin {gather*} \left [\frac {A \sqrt {b} \cos \left (d x + c\right ) \log \left (-\frac {b \cos \left (d x + c\right )^{3} - 2 \, \sqrt {b \cos \left (d x + c\right )} \sqrt {b} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{3}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )}, -\frac {A \sqrt {-b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )} \sqrt {-b} \sin \left (d x + c\right )}{b \sqrt {\cos \left (d x + c\right )}}\right ) \cos \left (d x + c\right ) - \sqrt {b \cos \left (d x + c\right )} C \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {b \cos {\left (c + d x \right )}} \left (A + C \cos ^{2}{\left (c + d x \right )}\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)*(b*cos(d*x+c))**(1/2)/cos(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(b*cos(c + d*x))*(A + C*cos(c + d*x)**2)/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)*(b*cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),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 [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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