3.394 \(\int \frac{a B+b B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=133 \[ \frac{2 b^2 B \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}+\frac{2 b B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{2 b B \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{2 B \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

(2*b*B*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (2*B*EllipticF[(c + d*x)/2, 2])/(3*a*d) + (2*b^2*B*EllipticPi[(2*b
)/(a + b), (c + d*x)/2, 2])/(a^2*(a + b)*d) + (2*B*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) - (2*b*B*Sin[c + d
*x])/(a^2*d*Sqrt[Cos[c + d*x]])

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Rubi [A]  time = 0.559779, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {21, 2802, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{2 b^2 B \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d (a+b)}+\frac{2 b B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{2 b B \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{2 B \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*B*EllipticE[(c + d*x)/2, 2])/(a^2*d) + (2*B*EllipticF[(c + d*x)/2, 2])/(3*a*d) + (2*b^2*B*EllipticPi[(2*b
)/(a + b), (c + d*x)/2, 2])/(a^2*(a + b)*d) + (2*B*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)) - (2*b*B*Sin[c + d
*x])/(a^2*d*Sqrt[Cos[c + d*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(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)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

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]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{a B+b B \cos (c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=B \int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\\ &=\frac{2 B \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}+\frac{(2 B) \int \frac{-\frac{3 b}{2}+\frac{1}{2} a \cos (c+d x)+\frac{1}{2} b \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{3 a}\\ &=\frac{2 B \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b B \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}+\frac{(4 B) \int \frac{\frac{1}{4} \left (a^2+3 b^2\right )+a b \cos (c+d x)+\frac{3}{4} b^2 \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^2}\\ &=\frac{2 B \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b B \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}-\frac{(4 B) \int \frac{-\frac{1}{4} b \left (a^2+3 b^2\right )-\frac{1}{4} a b^2 \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{3 a^2 b}+\frac{(b B) \int \sqrt{\cos (c+d x)} \, dx}{a^2}\\ &=\frac{2 b B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 B \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b B \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}+\frac{B \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{3 a}+\frac{\left (b^2 B\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^2}\\ &=\frac{2 b B E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac{2 B F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a d}+\frac{2 b^2 B \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 (a+b) d}+\frac{2 B \sin (c+d x)}{3 a d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b B \sin (c+d x)}{a^2 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 4.06872, size = 215, normalized size = 1.62 \[ \frac{B \left (\frac{2 \left (2 a^2+9 b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{6 \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt{\sin ^2(c+d x)}}+\frac{4 \sin (c+d x) (a-3 b \cos (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)}+8 a \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )\right )}{6 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*((2*(2*a^2 + 9*b^2)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 8*a*(2*EllipticF[(c + d*x)/2, 2] -
(2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)) + (4*(a - 3*b*Cos[c + d*x])*Sin[c + d*x])/Cos[c + d*
x]^(3/2) + (6*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*
x]]], -1] + (2*a^2 - b^2)*EllipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*Sqrt[Sin[c + d
*x]^2])))/(6*a^2*d)

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Maple [B]  time = 8.419, size = 452, normalized size = 3.4 \begin{align*} -2\,{\frac{\sqrt{- \left ( -2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}B}{\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}d} \left ( -2\,{\frac{{b}^{3}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}}{{a}^{2} \left ( -2\,ab+2\,{b}^{2} \right ) \sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{\it EllipticPi} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,-2\,{\frac{b}{a-b}},\sqrt{2} \right ) }-{\frac{b \left ( -\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}\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) +2\,\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }{{a}^{2} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }}+{\frac{1}{a} \left ( -1/6\,{\frac{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\cos \left ( 1/2\,dx+c/2 \right ) }{ \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{2}}}+1/3\,{\frac{\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*B*(-2/a^2*b^3/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^
2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos
(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-1/a^2*b*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*sin(1/2*d*x+1/2
*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x
+1/2*c)^2-1)+1/a*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/
2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(
1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(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(-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+b*B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [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+b*B*cos(d*x+c))/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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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+b*B*cos(d*x+c))/cos(d*x+c)**(5/2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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