3.94 \(\int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=142 \[ \frac{2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (5 A+8 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 d}+\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}+\frac{2 a B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

[Out]

(2*a^(5/2)*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^3*(35*A + 32*B)*Sin[c + d*x])/
(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(5*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*B*(a
+ a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.414053, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2976, 2981, 2773, 206} \[ \frac{2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}}+\frac{2 a^2 (5 A+8 B) \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 d}+\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{d}+\frac{2 a B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*A*ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (2*a^3*(35*A + 32*B)*Sin[c + d*x])/
(15*d*Sqrt[a + a*Cos[c + d*x]]) + (2*a^2*(5*A + 8*B)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a*B*(a
+ a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*d)

Rule 2976

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

Rule 2981

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2*b*B*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(2*n + 3)*Sqr
t[a + b*Sin[e + f*x]]), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[n, -1]

Rule 2773

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec (c+d x) \, dx &=\frac{2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{2}{5} \int (a+a \cos (c+d x))^{3/2} \left (\frac{5 a A}{2}+\frac{1}{2} a (5 A+8 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{2 a^2 (5 A+8 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\frac{4}{15} \int \sqrt{a+a \cos (c+d x)} \left (\frac{15 a^2 A}{4}+\frac{1}{4} a^2 (35 A+32 B) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac{2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (5 A+8 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}+\left (a^2 A\right ) \int \sqrt{a+a \cos (c+d x)} \sec (c+d x) \, dx\\ &=\frac{2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (5 A+8 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}-\frac{\left (2 a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,-\frac{a \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}\\ &=\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a+a \cos (c+d x)}}\right )}{d}+\frac{2 a^3 (35 A+32 B) \sin (c+d x)}{15 d \sqrt{a+a \cos (c+d x)}}+\frac{2 a^2 (5 A+8 B) \sqrt{a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a B (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.380174, size = 104, normalized size = 0.73 \[ \frac{a^2 \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)} \left (\sin \left (\frac{1}{2} (c+d x)\right ) (2 (5 A+14 B) \cos (c+d x)+80 A+3 B \cos (2 (c+d x))+89 B)+15 \sqrt{2} A \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(15*Sqrt[2]*A*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + (80*A + 89*
B + 2*(5*A + 14*B)*Cos[c + d*x] + 3*B*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(15*d)

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Maple [B]  time = 4.168, size = 311, normalized size = 2.2 \begin{align*}{\frac{1}{15\,d}{a}^{{\frac{3}{2}}}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 24\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-20\,\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a}\sqrt{2} \left ( A+4\,B \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+90\,A\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+15\,A\ln \left ( -4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,a}{-2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) a+15\,A\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a\sqrt{2}\cos \left ( 1/2\,dx+c/2 \right ) +2\,a}{2\,\cos \left ( 1/2\,dx+c/2 \right ) +\sqrt{2}}} \right ) a+120\,B\sqrt{2}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{a} \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^(3/2)*cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*B*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^
(1/2)*sin(1/2*d*x+1/2*c)^4-20*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*2^(1/2)*(A+4*B)*sin(1/2*d*x+1/2*c)^2+90*A
*a^(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+15*A*ln(-4/(-2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^(1/2)*2^(1/2)*(a
*sin(1/2*d*x+1/2*c)^2)^(1/2)-a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+15*A*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a^
(1/2)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+a*2^(1/2)*cos(1/2*d*x+1/2*c)+2*a))*a+120*B*2^(1/2)*(a*sin(1/2*d*x
+1/2*c)^2)^(1/2)*a^(1/2))/sin(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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Maxima [A]  time = 2.64634, size = 82, normalized size = 0.58 \begin{align*} \frac{{\left (3 \, \sqrt{2} a^{2} \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 25 \, \sqrt{2} a^{2} \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 150 \, \sqrt{2} a^{2} \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} B \sqrt{a}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(3*sqrt(2)*a^2*sin(5/2*d*x + 5/2*c) + 25*sqrt(2)*a^2*sin(3/2*d*x + 3/2*c) + 150*sqrt(2)*a^2*sin(1/2*d*x +
1/2*c))*B*sqrt(a)/d

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Fricas [A]  time = 1.7643, size = 462, normalized size = 3.25 \begin{align*} \frac{15 \,{\left (A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + 4 \,{\left (3 \, B a^{2} \cos \left (d x + c\right )^{2} +{\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (d x + c\right ) +{\left (40 \, A + 43 \, B\right )} a^{2}\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{30 \,{\left (d \cos \left (d x + c\right ) + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*(A*a^2*cos(d*x + c) + A*a^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x +
c) + a)*sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3*B*a^2*cos(d*x
+ c)^2 + (5*A + 14*B)*a^2*cos(d*x + c) + (40*A + 43*B)*a^2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c))/(d*cos(d*x
+ c) + d)

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

[Out]

Timed out

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Giac [A]  time = 3.8328, size = 306, normalized size = 2.15 \begin{align*} \frac{\frac{15 \, A a^{\frac{7}{2}} \log \left (\frac{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}{{\left | 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt{2}{\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} + \frac{2 \,{\left (45 \, \sqrt{2} A a^{5} + 60 \, \sqrt{2} B a^{5} +{\left (80 \, \sqrt{2} A a^{5} + 80 \, \sqrt{2} B a^{5} +{\left (35 \, \sqrt{2} A a^{5} + 32 \, \sqrt{2} B a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{5}{2}}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*A*a^(7/2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)
*abs(a) - 6*a)/abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs(a)
- 6*a))/abs(a) + 2*(45*sqrt(2)*A*a^5 + 60*sqrt(2)*B*a^5 + (80*sqrt(2)*A*a^5 + 80*sqrt(2)*B*a^5 + (35*sqrt(2)*A
*a^5 + 32*sqrt(2)*B*a^5)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x +
1/2*c)^2 + a)^(5/2))/d