3.336 \(\int \frac{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=120 \[ \frac{(A+2 C) \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 b^2 d \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x)}{2 b^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

[Out]

((A + 2*C)*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(2*b^2*d*Sqrt[b*Cos[c + d*x]]) + (A*Sin[c + d*x])/(2*b^2*
d*Cos[c + d*x]^(3/2)*Sqrt[b*Cos[c + d*x]]) + (B*Sin[c + d*x])/(b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*x]])

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Rubi [A]  time = 0.0919308, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {18, 3021, 2748, 3767, 8, 3770} \[ \frac{(A+2 C) \sqrt{\cos (c+d x)} \tanh ^{-1}(\sin (c+d x))}{2 b^2 d \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x)}{2 b^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(5/2)),x]

[Out]

((A + 2*C)*ArcTanh[Sin[c + d*x]]*Sqrt[Cos[c + d*x]])/(2*b^2*d*Sqrt[b*Cos[c + d*x]]) + (A*Sin[c + d*x])/(2*b^2*
d*Cos[c + d*x]^(3/2)*Sqrt[b*Cos[c + d*x]]) + (B*Sin[c + d*x])/(b^2*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rule 3770

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{A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (b \cos (c+d x))^{5/2}} \, dx &=\frac{\sqrt{\cos (c+d x)} \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{b^2 \sqrt{b \cos (c+d x)}}\\ &=\frac{A \sin (c+d x)}{2 b^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{\sqrt{\cos (c+d x)} \int (2 B+(A+2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx}{2 b^2 \sqrt{b \cos (c+d x)}}\\ &=\frac{A \sin (c+d x)}{2 b^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{\left (B \sqrt{\cos (c+d x)}\right ) \int \sec ^2(c+d x) \, dx}{b^2 \sqrt{b \cos (c+d x)}}+\frac{\left ((A+2 C) \sqrt{\cos (c+d x)}\right ) \int \sec (c+d x) \, dx}{2 b^2 \sqrt{b \cos (c+d x)}}\\ &=\frac{(A+2 C) \tanh ^{-1}(\sin (c+d x)) \sqrt{\cos (c+d x)}}{2 b^2 d \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x)}{2 b^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}-\frac{\left (B \sqrt{\cos (c+d x)}\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{b^2 d \sqrt{b \cos (c+d x)}}\\ &=\frac{(A+2 C) \tanh ^{-1}(\sin (c+d x)) \sqrt{\cos (c+d x)}}{2 b^2 d \sqrt{b \cos (c+d x)}}+\frac{A \sin (c+d x)}{2 b^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{b \cos (c+d x)}}+\frac{B \sin (c+d x)}{b^2 d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.0942971, size = 69, normalized size = 0.57 \[ \frac{\sqrt{\cos (c+d x)} \left (\sin (c+d x) (A+2 B \cos (c+d x))+(A+2 C) \cos ^2(c+d x) \tanh ^{-1}(\sin (c+d x))\right )}{2 d (b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^(5/2)),x]

[Out]

(Sqrt[Cos[c + d*x]]*((A + 2*C)*ArcTanh[Sin[c + d*x]]*Cos[c + d*x]^2 + (A + 2*B*Cos[c + d*x])*Sin[c + d*x]))/(2
*d*(b*Cos[c + d*x])^(5/2))

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Maple [A]  time = 0.404, size = 151, normalized size = 1.3 \begin{align*} -{\frac{1}{2\,d} \left ( A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\ln \left ( -{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) +4\,C \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it Artanh} \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) -2\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -A\sin \left ( dx+c \right ) \right ) \sqrt{\cos \left ( dx+c \right ) } \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x)

[Out]

-1/2/d*(A*cos(d*x+c)^2*ln(-(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))-A*cos(d*x+c)^2*ln(-(-1+cos(d*x+c)-sin(d*x+c)
)/sin(d*x+c))+4*C*cos(d*x+c)^2*arctanh((-1+cos(d*x+c))/sin(d*x+c))-2*B*sin(d*x+c)*cos(d*x+c)-A*sin(d*x+c))*cos
(d*x+c)^(1/2)/(b*cos(d*x+c))^(5/2)

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Maxima [B]  time = 2.34179, size = 1107, normalized size = 9.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/4*(8*B*sqrt(b)*sin(2*d*x + 2*c)/(b^3*cos(2*d*x + 2*c)^2 + b^3*sin(2*d*x + 2*c)^2 + 2*b^3*cos(2*d*x + 2*c) +
b^3) - (4*(sin(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 4*(si
n(4*d*x + 4*c) + 2*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (2*(2*cos(2*d*x +
2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*
c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c))) + 1) + (2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x +
2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c)
+ 1)*log(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))^2 - 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x
+ 2*c) + 1)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*(cos(4*d*x + 4*c) + 2*cos(2*d*x + 2*c) +
1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*A/((b^2*cos(4*d*x + 4*c)^2 + 4*b^2*cos(2*d*x + 2*c)^2
+ b^2*sin(4*d*x + 4*c)^2 + 4*b^2*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*b^2*sin(2*d*x + 2*c)^2 + 4*b^2*cos(2*d
*x + 2*c) + b^2 + 2*(2*b^2*cos(2*d*x + 2*c) + b^2)*cos(4*d*x + 4*c))*sqrt(b)) + 2*C*(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))/b^(5/2))/d

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Fricas [A]  time = 1.71653, size = 659, normalized size = 5.49 \begin{align*} \left [\frac{{\left (A + 2 \, C\right )} \sqrt{b} \cos \left (d x + c\right )^{3} \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 \,{\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{4 \, b^{3} d \cos \left (d x + c\right )^{3}}, -\frac{{\left (A + 2 \, C\right )} \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 )^{3} -{\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, b^{3} d \cos \left (d x + c\right )^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/4*((A + 2*C)*sqrt(b)*cos(d*x + c)^3*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*(2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(
d*x + c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^3), -1/2*((A + 2*C)*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)^3 - (2*B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*sqrt(cos(d*x
+ c))*sin(d*x + c))/(b^3*d*cos(d*x + c)^3)]

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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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(5/2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{\left (b \cos \left (d x + c\right )\right )^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c))^(5/2)*sqrt(cos(d*x + c))), x)