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

Optimal. Leaf size=187 \[ \frac{52 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{21 d}+\frac{2 a^3 \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d}+\frac{6 a^3 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{52 a^3 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{28 a^3 \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}-\frac{28 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

[Out]

(-28*a^3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (52*a^3*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (28*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (52*a^3
*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (6*a^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*a^3*Sec[c + d*x]
^(7/2)*Sin[c + d*x])/(7*d)

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Rubi [A]  time = 0.189881, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3791, 3768, 3771, 2639, 2641} \[ \frac{2 a^3 \sin (c+d x) \sec ^{\frac{7}{2}}(c+d x)}{7 d}+\frac{6 a^3 \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{52 a^3 \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{21 d}+\frac{28 a^3 \sin (c+d x) \sqrt{\sec (c+d x)}}{5 d}+\frac{52 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 d}-\frac{28 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-28*a^3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (52*a^3*Sqrt[Cos[c + d*x]]*E
llipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(21*d) + (28*a^3*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (52*a^3
*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(21*d) + (6*a^3*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(5*d) + (2*a^3*Sec[c + d*x]
^(7/2)*Sin[c + d*x])/(7*d)

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 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]

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]

Rubi steps

\begin{align*} \int \sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^3 \, dx &=\int \left (a^3 \sec ^{\frac{3}{2}}(c+d x)+3 a^3 \sec ^{\frac{5}{2}}(c+d x)+3 a^3 \sec ^{\frac{7}{2}}(c+d x)+a^3 \sec ^{\frac{9}{2}}(c+d x)\right ) \, dx\\ &=a^3 \int \sec ^{\frac{3}{2}}(c+d x) \, dx+a^3 \int \sec ^{\frac{9}{2}}(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^{\frac{5}{2}}(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^{\frac{7}{2}}(c+d x) \, dx\\ &=\frac{2 a^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{d}+\frac{2 a^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{d}+\frac{6 a^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a^3 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{7} \left (5 a^3\right ) \int \sec ^{\frac{5}{2}}(c+d x) \, dx-a^3 \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+a^3 \int \sqrt{\sec (c+d x)} \, dx+\frac{1}{5} \left (9 a^3\right ) \int \sec ^{\frac{3}{2}}(c+d x) \, dx\\ &=\frac{28 a^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{52 a^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{6 a^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a^3 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{21} \left (5 a^3\right ) \int \sqrt{\sec (c+d x)} \, dx-\frac{1}{5} \left (9 a^3\right ) \int \frac{1}{\sqrt{\sec (c+d x)}} \, dx+\left (a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\left (a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{2 a^3 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{2 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{d}+\frac{28 a^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{52 a^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{6 a^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a^3 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}+\frac{1}{21} \left (5 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx-\frac{1}{5} \left (9 a^3 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{28 a^3 \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{52 a^3 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{\sec (c+d x)}}{21 d}+\frac{28 a^3 \sqrt{\sec (c+d x)} \sin (c+d x)}{5 d}+\frac{52 a^3 \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac{6 a^3 \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac{2 a^3 \sec ^{\frac{7}{2}}(c+d x) \sin (c+d x)}{7 d}\\ \end{align*}

Mathematica [C]  time = 2.47255, size = 287, normalized size = 1.53 \[ \frac{a^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^3 \left (\frac{294 \csc (c) \cos (d x)+(63 \cos (c+d x)+65 \cos (2 (c+d x))+80) \tan (c+d x) \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x)}-\frac{2 i \sqrt{2} e^{-i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^3(c+d x) \left (147 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+65 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+147 \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{420 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(((-2*I)*Sqrt[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*
Cos[c + d*x]^3*(147*(1 + E^((2*I)*(c + d*x))) + 147*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeom
etric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 65*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c +
d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (294*Co
s[d*x]*Csc[c] + (80 + 63*Cos[c + d*x] + 65*Cos[2*(c + d*x)])*Sec[c + d*x]^2*Tan[c + d*x])/Sec[c + d*x]^(5/2)))
/(420*d)

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Maple [B]  time = 2.418, size = 439, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-3/10*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)^3-56/5*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-
2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+848/105*(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))-28/5
*(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-1/28*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)^4-26/21*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)/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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a^3*sec(d*x + c)^4 + 3*a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + a^3*sec(d*x + c))*sqrt(sec(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(sec(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} \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(sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

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