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

Optimal. Leaf size=91 \[ \frac{20 a^3 \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}+\frac{4 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^3 \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 a^3 \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

[Out]

(4*a^3*EllipticE[(c + d*x)/2, 2])/d + (20*a^3*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^3*Sin[c + d*x])/(d*Sqrt[
Cos[c + d*x]]) + (2*a^3*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.19675, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4264, 3791, 3769, 3771, 2641, 2639, 3768} \[ \frac{20 a^3 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{4 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^3 \sin (c+d x) \sqrt{\cos (c+d x)}}{3 d}+\frac{2 a^3 \sin (c+d x)}{d \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*EllipticE[(c + d*x)/2, 2])/d + (20*a^3*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*a^3*Sin[c + d*x])/(d*Sqrt[
Cos[c + d*x]]) + (2*a^3*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

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 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 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 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 \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^3}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \left (\frac{a^3}{\sec ^{\frac{3}{2}}(c+d x)}+\frac{3 a^3}{\sqrt{\sec (c+d x)}}+3 a^3 \sqrt{\sec (c+d x)}+a^3 \sec ^{\frac{3}{2}}(c+d x)\right ) \, dx\\ &=\left (a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sec ^{\frac{3}{2}}(c+d x)} \, dx+\left (a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx+\left (3 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\left (3 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx\\ &=\frac{2 a^3 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 a^3 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\left (3 a^3\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\left (3 a^3\right ) \int \sqrt{\cos (c+d x)} \, dx+\frac{1}{3} \left (a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \, dx-\left (a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx\\ &=\frac{6 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{6 a^3 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{2 a^3 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 a^3 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} a^3 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-a^3 \int \sqrt{\cos (c+d x)} \, dx\\ &=\frac{4 a^3 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{d}+\frac{20 a^3 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^3 \sin (c+d x)}{d \sqrt{\cos (c+d x)}}+\frac{2 a^3 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [C]  time = 4.84854, size = 240, normalized size = 2.64 \[ \frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (-6 \cos (c) \sqrt{\sec ^2(c)} \sqrt{\sin ^2\left (\tan ^{-1}(\tan (c))+d x\right )} \csc \left (\tan ^{-1}(\tan (c))+d x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cos ^2\left (\tan ^{-1}(\tan (c))+d x\right )\right )-20 \sin (c) \sqrt{\csc ^2(c)} \cos (c+d x) \sqrt{\cos ^2\left (d x-\tan ^{-1}(\cot (c))\right )} \sec \left (d x-\tan ^{-1}(\cot (c))\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},\sin ^2\left (d x-\tan ^{-1}(\cot (c))\right )\right )+\sin (2 (c+d x))-3 \csc (c) \cos (d x)-9 \csc (c) \cos (2 c+d x)+9 \cot (c) \sqrt{\sec ^2(c)} \cos \left (c-\tan ^{-1}(\tan (c))-d x\right )+3 \cot (c) \sqrt{\sec ^2(c)} \cos \left (c+\tan ^{-1}(\tan (c))+d x\right )\right )}{24 d \sqrt{\cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(-3*Cos[d*x]*Csc[c] - 9*Cos[2*c + d*x]*Csc[c] + 9*Cos[c - d*x - A
rcTan[Tan[c]]]*Cot[c]*Sqrt[Sec[c]^2] + 3*Cos[c + d*x + ArcTan[Tan[c]]]*Cot[c]*Sqrt[Sec[c]^2] - 20*Cos[c + d*x]
*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]
]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + Sin[2*(c + d*x)] - 6*Cos[c]*Csc[d*x + ArcTan[Tan[c]]]*HypergeometricP
FQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(24*d*
Sqrt[Cos[c + d*x]])

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Maple [A]  time = 1.664, size = 172, normalized size = 1.9 \begin{align*} -{\frac{4\,{a}^{3}}{3\,d} \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +5\,{\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 ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-3\,{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \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}}-4\, \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(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^3,x)

[Out]

-4/3*a^3*(2*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)
^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)-4*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 \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(d*x + c) + a)^3*cos(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^{3} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )\right )} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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