3.17 \(\int (b \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=110 \[ -\frac{2 b^2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 b (5 A+3 C) \sin (c+d x) \sqrt{b \sec (c+d x)}}{5 d}+\frac{2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \]

[Out]

(-2*b^2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*(5*A + 3*C
)*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*C*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

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Rubi [A]  time = 0.0825354, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {4046, 3768, 3771, 2639} \[ -\frac{2 b^2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 b (5 A+3 C) \sin (c+d x) \sqrt{b \sec (c+d x)}}{5 d}+\frac{2 C \tan (c+d x) (b \sec (c+d x))^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (2*b*(5*A + 3*C
)*Sqrt[b*Sec[c + d*x]]*Sin[c + d*x])/(5*d) + (2*C*(b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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]

Rubi steps

\begin{align*} \int (b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{2 C (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{5} (5 A+3 C) \int (b \sec (c+d x))^{3/2} \, dx\\ &=\frac{2 b (5 A+3 C) \sqrt{b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 C (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac{1}{5} \left (b^2 (5 A+3 C)\right ) \int \frac{1}{\sqrt{b \sec (c+d x)}} \, dx\\ &=\frac{2 b (5 A+3 C) \sqrt{b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 C (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}-\frac{\left (b^2 (5 A+3 C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}\\ &=-\frac{2 b^2 (5 A+3 C) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d \sqrt{\cos (c+d x)} \sqrt{b \sec (c+d x)}}+\frac{2 b (5 A+3 C) \sqrt{b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{2 C (b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [C]  time = 1.24154, size = 184, normalized size = 1.67 \[ \frac{4 i e^{i (c+d x)} \cos ^3(c+d x) (b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left ((5 A+3 C) \left (1+e^{2 i (c+d x)}\right )^{5/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-e^{2 i (c+d x)}\right )-3 \left (5 A \left (1+e^{2 i (c+d x)}\right )^2+C \left (8 e^{2 i (c+d x)}+3 e^{4 i (c+d x)}+1\right )\right )\right )}{15 d \left (1+e^{2 i (c+d x)}\right )^2 (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*I)/15)*E^(I*(c + d*x))*Cos[c + d*x]^3*(-3*(5*A*(1 + E^((2*I)*(c + d*x)))^2 + C*(1 + 8*E^((2*I)*(c + d*x))
+ 3*E^((4*I)*(c + d*x)))) + (5*A + 3*C)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^(
(2*I)*(c + d*x))])*(b*Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2))/(d*(1 + E^((2*I)*(c + d*x)))^2*(A + 2*C + A*
Cos[2*(c + d*x)]))

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Maple [C]  time = 0.273, size = 670, normalized size = 6.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x)

[Out]

-2/5/d*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^2*(5*I*A*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-5*I*A*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+3*I*C*cos(d*x+c)^3*(1/(co
s(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-3*I*
C*cos(d*x+c)^3*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+
c),I)*sin(d*x+c)+5*I*A*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1
+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-5*I*A*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^
(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+3*I*C*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-3*I*C*cos(d*x+c)^2*(1/(cos(d*x
+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)+5*A*cos(d
*x+c)^3+3*C*cos(d*x+c)^3-5*A*cos(d*x+c)^2-2*C*cos(d*x+c)^2-C)*(b/cos(d*x+c))^(3/2)/sin(d*x+c)^5/cos(d*x+c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*b*sec(d*x + c)^3 + A*b*sec(d*x + c))*sqrt(b*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((b*sec(d*x+c))**(3/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(3/2)*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^(3/2), x)