3.3.30 \(\int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx\) [230]

Optimal. Leaf size=114 \[ \frac {2^{\frac {5}{2}+n} F_1\left (\frac {3}{4};\frac {1}{2}+n,1;\frac {7}{4};-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {3}{2}+n} (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]

[Out]

1/3*2^(5/2+n)*AppellF1(3/4,1/2+n,1,7/4,(-a+a*sec(d*x+c))/(a+a*sec(d*x+c)),(a-a*sec(d*x+c))/(a+a*sec(d*x+c)))*(
1/(1+sec(d*x+c)))^(3/2+n)*(a+a*sec(d*x+c))^n*tan(d*x+c)^(3/2)/d

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Rubi [A]
time = 0.04, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {3974} \begin {gather*} \frac {2^{n+\frac {5}{2}} \tan ^{\frac {3}{2}}(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+\frac {3}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac {3}{4};n+\frac {1}{2},1;\frac {7}{4};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n*Sqrt[Tan[c + d*x]],x]

[Out]

(2^(5/2 + n)*AppellF1[3/4, 1/2 + n, 1, 7/4, -((a - a*Sec[c + d*x])/(a + a*Sec[c + d*x])), (a - a*Sec[c + d*x])
/(a + a*Sec[c + d*x])]*((1 + Sec[c + d*x])^(-1))^(3/2 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^(3/2))/(3*d)

Rule 3974

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-2^(m
+ n + 1))*(e*Cot[c + d*x])^(m + 1)*((a + b*Csc[c + d*x])^n/(d*e*(m + 1)))*(a/(a + b*Csc[c + d*x]))^(m + n + 1
)*AppellF1[(m + 1)/2, m + n, 1, (m + 3)/2, -(a - b*Csc[c + d*x])/(a + b*Csc[c + d*x]), (a - b*Csc[c + d*x])/(a
+ b*Csc[c + d*x])], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^n \sqrt {\tan (c+d x)} \, dx &=\frac {2^{\frac {5}{2}+n} F_1\left (\frac {3}{4};\frac {1}{2}+n,1;\frac {7}{4};-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {3}{2}+n} (a+a \sec (c+d x))^n \tan ^{\frac {3}{2}}(c+d x)}{3 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(238\) vs. \(2(114)=228\).
time = 3.09, size = 238, normalized size = 2.09 \begin {gather*} \frac {56 F_1\left (\frac {3}{4};\frac {1}{2}+n,1;\frac {7}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) (a (1+\sec (c+d x)))^n \sin \left (\frac {1}{2} (c+d x)\right ) \sqrt {\tan (c+d x)}}{d \left (6 \left (2 F_1\left (\frac {7}{4};\frac {1}{2}+n,2;\frac {11}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-(1+2 n) F_1\left (\frac {7}{4};\frac {3}{2}+n,1;\frac {11}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+21 F_1\left (\frac {3}{4};\frac {1}{2}+n,1;\frac {7}{4};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^n*Sqrt[Tan[c + d*x]],x]

[Out]

(56*AppellF1[3/4, 1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^3*(a*(1 + Sec[c +
d*x]))^n*Sin[(c + d*x)/2]*Sqrt[Tan[c + d*x]])/(d*(6*(2*AppellF1[7/4, 1/2 + n, 2, 11/4, Tan[(c + d*x)/2]^2, -T
an[(c + d*x)/2]^2] - (1 + 2*n)*AppellF1[7/4, 3/2 + n, 1, 11/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*(-1 +
Cos[c + d*x]) + 21*AppellF1[3/4, 1/2 + n, 1, 7/4, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])
))

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \left (\sqrt {\tan }\left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^(1/2),x)

[Out]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^(1/2),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((a+a*sec(d*x+c))^n*tan(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*sqrt(tan(d*x + c)), 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((a+a*sec(d*x+c))^n*tan(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*sqrt(tan(d*x + c)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \sqrt {\tan {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**(1/2),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*sqrt(tan(c + d*x)), 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((a+a*sec(d*x+c))^n*tan(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*sqrt(tan(d*x + c)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^(1/2)*(a + a/cos(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^(1/2)*(a + a/cos(c + d*x))^n, x)

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