3.314 \(\int (a+a \cos (c+d x))^4 \sec ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=159 \[ \frac{2 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{8 a^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{32 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{56 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

[Out]

(56*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (32*a^4*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^4*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a^4*Sin
[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

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Rubi [A]  time = 0.211547, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3238, 3791, 3769, 3771, 2639, 2641, 3768} \[ \frac{2 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a^4 \sin (c+d x) \sqrt{\sec (c+d x)}}{d}+\frac{8 a^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{32 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{56 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^(3/2),x]

[Out]

(56*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (32*a^4*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^4*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a^4*Sin
[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^4 \sec ^{\frac{3}{2}}(c+d x) \, dx &=\int \frac{(a+a \sec (c+d x))^4}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\int \left (\frac{a^4}{\sec ^{\frac{5}{2}}(c+d x)}+\frac{4 a^4}{\sec ^{\frac{3}{2}}(c+d x)}+\frac{6 a^4}{\sqrt{\sec (c+d x)}}+4 a^4 \sqrt{\sec (c+d x)}+a^4 \sec ^{\frac{3}{2}}(c+d x)\right ) \, dx\\ &=a^4 \int \frac{1}{\sec ^{\frac{5}{2}}(c+d x)} \, dx+a^4 \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \sqrt{\sec (c+d x)} \, dx+\left (6 a^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{2 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{5} \left (3 a^4\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx-a^4 \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\frac{1}{3} \left (4 a^4\right ) \int \sqrt{\sec (c+d x)} \, dx+\left (4 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\left (6 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{12 a^4 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{8 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{2 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{1}{5} \left (3 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx-\left (a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (4 a^4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{56 a^4 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{5 d}+\frac{32 a^4 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{3 d}+\frac{2 a^4 \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{8 a^4 \sin (c+d x)}{3 d \sqrt{\sec (c+d x)}}+\frac{2 a^4 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}\\ \end{align*}

Mathematica [C]  time = 1.45534, size = 150, normalized size = 0.94 \[ \frac{a^4 \left (\frac{672 i \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-e^{2 i (c+d x)}\right )}{\sqrt{1+e^{2 i (c+d x)}}}-320 i \sqrt{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-e^{2 i (c+d x)}\right ) \sec (c+d x)+80 \sin (c+d x)+63 \tan (c+d x)+3 \sin (3 (c+d x)) \sec (c+d x)-336 i\right )}{30 d \sqrt{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*Sec[c + d*x]^(3/2),x]

[Out]

(a^4*(-336*I + ((672*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))]
- (320*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]*Sec[c + d*x] +
80*Sin[c + d*x] + 3*Sec[c + d*x]*Sin[3*(c + d*x)] + 63*Tan[c + d*x]))/(30*d*Sqrt[Sec[c + d*x]])

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Maple [A]  time = 2.5, size = 194, normalized size = 1.2 \begin{align*} -{\frac{8\,{a}^{4}}{15\,d} \left ( -6\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) +26\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +20\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -21\,\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}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -19\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^4*sec(d*x+c)^(3/2),x)

[Out]

-8/15*a^4*(-6*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+26*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+20*(2*sin(1/2
*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-21*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-19*sin(1/2*d*x+1/2*c)^2*cos
(1/2*d*x+1/2*c))/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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((a*cos(d*x + c) + a)^4*sec(d*x + c)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{4} \cos \left (d x + c\right )^{4} + 4 \, a^{4} \cos \left (d x + c\right )^{3} + 6 \, a^{4} \cos \left (d x + c\right )^{2} + 4 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sec \left (d x + c\right )^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

integral((a^4*cos(d*x + c)^4 + 4*a^4*cos(d*x + c)^3 + 6*a^4*cos(d*x + c)^2 + 4*a^4*cos(d*x + c) + a^4)*sec(d*x
+ c)^(3/2), x)

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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))**4*sec(d*x+c)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \cos \left (d x + c\right ) + a\right )}^{4} \sec \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*sec(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((a*cos(d*x + c) + a)^4*sec(d*x + c)^(3/2), x)