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

Optimal. Leaf size=131 \[ \frac {2^{m+\frac {5}{2}} (a \sec (c+d x)+a)^{3/2} \left (\frac {1}{\sec (c+d x)+1}\right )^{m+\frac {5}{2}} (e \tan (c+d x))^{m+1} F_1\left (\frac {m+1}{2};m+\frac {3}{2},1;\frac {m+3}{2};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d e (m+1)} \]

[Out]

2^(5/2+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)))^(5/2+m)*(a+a*sec(d*x+c))^(3/2)*(e*tan(d*x+c))^(1+m)/d/e/(1+m)

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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} \[ \frac {2^{m+\frac {5}{2}} (a \sec (c+d x)+a)^{3/2} \left (\frac {1}{\sec (c+d x)+1}\right )^{m+\frac {5}{2}} (e \tan (c+d x))^{m+1} F_1\left (\frac {m+1}{2};m+\frac {3}{2},1;\frac {m+3}{2};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(5/2 + 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*S
ec[c + d*x])/(a + a*Sec[c + d*x])]*((1 + Sec[c + d*x])^(-1))^(5/2 + m)*(a + a*Sec[c + d*x])^(3/2)*(e*Tan[c + d
*x])^(1 + m))/(d*e*(1 + m))

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 (a+a \sec (c+d x))^{3/2} (e \tan (c+d x))^m \, dx &=\frac {2^{\frac {5}{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 {5}{2}+m} (a+a \sec (c+d x))^{3/2} (e \tan (c+d x))^{1+m}}{d e (1+m)}\\ \end {align*}

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Mathematica [F]  time = 3.76, size = 0, normalized size = 0.00 \[ \int (a+a \sec (c+d x))^{3/2} (e \tan (c+d x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \tan \left (d x + c\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.61, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}} \left (e \tan \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \left (e \tan {\left (c + d x \right )}\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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