3.1.39 \(\int \frac {(b \sec (c+d x))^n (A+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [39]

Optimal. Leaf size=142 \[ -\frac {2 C (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (A (3-2 n)+C (5-2 n)) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (7-2 n);\frac {1}{4} (11-2 n);\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (7-2 n) \sec ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-2*C*(b*sec(d*x+c))^n*sin(d*x+c)/d/(3-2*n)/sec(d*x+c)^(3/2)-2*(A*(3-2*n)+C*(5-2*n))*hypergeom([1/2, 7/4-1/2*n]
,[11/4-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^n*sin(d*x+c)/d/(4*n^2-20*n+21)/sec(d*x+c)^(7/2)/(sin(d*x+c)^2)^(1/2
)

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Rubi [A]
time = 0.09, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {20, 4131, 3857, 2722} \begin {gather*} -\frac {2 (A (3-2 n)+C (5-2 n)) \sin (c+d x) (b \sec (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (7-2 n);\frac {1}{4} (11-2 n);\cos ^2(c+d x)\right )}{d (3-2 n) (7-2 n) \sqrt {\sin ^2(c+d x)} \sec ^{\frac {7}{2}}(c+d x)}-\frac {2 C \sin (c+d x) (b \sec (c+d x))^n}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(-2*C*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 2*n)*Sec[c + d*x]^(3/2)) - (2*(A*(3 - 2*n) + C*(5 - 2*n))*Hyper
geometric2F1[1/2, (7 - 2*n)/4, (11 - 2*n)/4, Cos[c + d*x]^2]*(b*Sec[c + d*x])^n*Sin[c + d*x])/(d*(3 - 2*n)*(7
- 2*n)*Sec[c + d*x]^(7/2)*Sqrt[Sin[c + d*x]^2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[b^IntPart[n]*((b*v)^FracPart[n]/(a^IntPart[n]
*(a*v)^FracPart[n])), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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

Rule 4131

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]

Rubi steps

\begin {align*} \int \frac {(b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {5}{2}+n}(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=-\frac {2 C (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (\left (C \left (-\frac {5}{2}+n\right )+A \left (-\frac {3}{2}+n\right )\right ) \sec ^{-n}(c+d x) (b \sec (c+d x))^n\right ) \int \sec ^{-\frac {5}{2}+n}(c+d x) \, dx}{-\frac {3}{2}+n}\\ &=-\frac {2 C (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (\left (C \left (-\frac {5}{2}+n\right )+A \left (-\frac {3}{2}+n\right )\right ) \cos ^{\frac {1}{2}+n}(c+d x) \sqrt {\sec (c+d x)} (b \sec (c+d x))^n\right ) \int \cos ^{\frac {5}{2}-n}(c+d x) \, dx}{-\frac {3}{2}+n}\\ &=-\frac {2 C (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (A (3-2 n)+C (5-2 n)) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (7-2 n);\frac {1}{4} (11-2 n);\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (3-2 n) (7-2 n) \sec ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.45, size = 338, normalized size = 2.38 \begin {gather*} -\frac {i 2^{-\frac {1}{2}+n} e^{-\frac {1}{2} i (4 c+d (-1+2 n) x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-\frac {1}{2}+n} \left (1+e^{2 i (c+d x)}\right )^{-\frac {1}{2}+n} \left (\frac {A e^{\frac {1}{2} i d (-5+2 n) x} \, _2F_1\left (-\frac {1}{2}+n,\frac {1}{4} (-5+2 n);\frac {1}{4} (-1+2 n);-e^{2 i (c+d x)}\right )}{-5+2 n}+\frac {e^{\frac {1}{2} i (4 c+d (-1+2 n) x)} \left (2 (A+2 C) (3+2 n) \, _2F_1\left (-\frac {1}{2}+n,\frac {1}{4} (-1+2 n);\frac {1}{4} (3+2 n);-e^{2 i (c+d x)}\right )+A e^{2 i (c+d x)} (-1+2 n) \, _2F_1\left (-\frac {1}{2}+n,\frac {1}{4} (3+2 n);\frac {1}{4} (7+2 n);-e^{2 i (c+d x)}\right )\right )}{-3+4 n+4 n^2}\right ) \sec ^{-2-n}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{d (A+2 C+A \cos (2 c+2 d x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

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

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Maple [F]
time = 0.60, size = 0, normalized size = 0.00 \[\int \frac {\left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )}{\sec \left (d x +c \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)

[Out]

int((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n/sec(d*x + c)^(5/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n/sec(d*x + c)^(5/2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))**n*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2),x)

[Out]

Integral((b*sec(c + d*x))**n*(A + C*sec(c + d*x)**2)/sec(c + d*x)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n/sec(d*x + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^n}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n)/(1/cos(c + d*x))^(5/2),x)

[Out]

int(((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n)/(1/cos(c + d*x))^(5/2), x)

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