3.218 \(\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=131 \[ \frac {2^{m-\frac {1}{2}} \left (\frac {1}{\sec (c+d x)+1}\right )^{m-\frac {1}{2}} (e \tan (c+d x))^{m+1} F_1\left (\frac {m+1}{2};m-\frac {3}{2},1;\frac {m+3}{2};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d e (m+1) (a \sec (c+d x)+a)^{3/2}} \]

[Out]

2^(-1/2+m)*AppellF1(1/2+1/2*m,-3/2+m,1,3/2+1/2*m,(-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)))^(-1/2+m)*(e*tan(d*x+c))^(1+m)/d/e/(1+m)/(a+a*sec(d*x+c))^(3/2)

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Rubi [A]  time = 0.10, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {3889} \[ \frac {2^{m-\frac {1}{2}} \left (\frac {1}{\sec (c+d x)+1}\right )^{m-\frac {1}{2}} (e \tan (c+d x))^{m+1} F_1\left (\frac {m+1}{2};m-\frac {3}{2},1;\frac {m+3}{2};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d e (m+1) (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1/2 + m)*AppellF1[(1 + m)/2, -3/2 + m, 1, (3 + m)/2, -((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))^(-1/2 + m)*(e*Tan[c + d*x])^(1 + m))/(d*e*(1 +
m)*(a + a*Sec[c + d*x])^(3/2))

Rule 3889

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*(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])])/(d*e*(m + 1)), x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx &=\frac {2^{-\frac {1}{2}+m} F_1\left (\frac {1+m}{2};-\frac {3}{2}+m,1;\frac {3+m}{2};-\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 {1}{2}+m} (e \tan (c+d x))^{1+m}}{d e (1+m) (a+a \sec (c+d x))^{3/2}}\\ \end {align*}

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Mathematica [F]  time = 10.46, size = 0, normalized size = 0.00 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sec \left (d x + c\right ) + a} \left (e \tan \left (d x + c\right )\right )^{m}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sec(d*x + c) + a)*(e*tan(d*x + c))^m/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.55, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^m/(a + a/cos(c + d*x))^(3/2),x)

[Out]

int((e*tan(c + d*x))^m/(a + a/cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**m/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((e*tan(c + d*x))**m/(a*(sec(c + d*x) + 1))**(3/2), x)

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