Integrand size = 25, antiderivative size = 131 \[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=\frac {2^{\frac {3}{2}+m} \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2}+m,1,\frac {3+m}{2},-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{\frac {3}{2}+m} \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^{1+m}}{d e (1+m)} \] Output:
2^(3/2+m)*AppellF1(1/2+1/2*m,1,1/2+m,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)))^(3/2+m)*(a+a *sec(d*x+c))^(1/2)*(e*tan(d*x+c))^(1+m)/d/e/(1+m)
\[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=\int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx \] Input:
Integrate[Sqrt[a + a*Sec[c + d*x]]*(e*Tan[c + d*x])^m,x]
Output:
Integrate[Sqrt[a + a*Sec[c + d*x]]*(e*Tan[c + d*x])^m, x]
Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {3042, 4377}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {a \sec (c+d x)+a} (e \tan (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^mdx\) |
\(\Big \downarrow \) 4377 |
\(\displaystyle \frac {2^{m+\frac {3}{2}} \sqrt {a \sec (c+d x)+a} \left (\frac {1}{\sec (c+d x)+1}\right )^{m+\frac {3}{2}} (e \tan (c+d x))^{m+1} \operatorname {AppellF1}\left (\frac {m+1}{2},m+\frac {1}{2},1,\frac {m+3}{2},-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{d e (m+1)}\) |
Input:
Int[Sqrt[a + a*Sec[c + d*x]]*(e*Tan[c + d*x])^m,x]
Output:
(2^(3/2 + m)*AppellF1[(1 + m)/2, 1/2 + 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))^(3/2 + m)*Sqrt[a + a*Sec[c + d*x]]*(e*Tan[c + d*x]) ^(1 + m))/(d*e*(1 + m))
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)*App ellF1[(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]
\[\int \sqrt {a +a \sec \left (d x +c \right )}\, \left (e \tan \left (d x +c \right )\right )^{m}d x\]
Input:
int((a+a*sec(d*x+c))^(1/2)*(e*tan(d*x+c))^m,x)
Output:
int((a+a*sec(d*x+c))^(1/2)*(e*tan(d*x+c))^m,x)
\[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(e*tan(d*x+c))^m,x, algorithm="fricas")
Output:
integral(sqrt(a*sec(d*x + c) + a)*(e*tan(d*x + c))^m, x)
\[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (e \tan {\left (c + d x \right )}\right )^{m}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(1/2)*(e*tan(d*x+c))**m,x)
Output:
Integral(sqrt(a*(sec(c + d*x) + 1))*(e*tan(c + d*x))**m, x)
\[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(e*tan(d*x+c))^m,x, algorithm="maxima")
Output:
integrate(sqrt(a*sec(d*x + c) + a)*(e*tan(d*x + c))^m, x)
\[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \tan \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(1/2)*(e*tan(d*x+c))^m,x, algorithm="giac")
Output:
integrate(sqrt(a*sec(d*x + c) + a)*(e*tan(d*x + c))^m, x)
Timed out. \[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=\int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \] Input:
int((e*tan(c + d*x))^m*(a + a/cos(c + d*x))^(1/2),x)
Output:
int((e*tan(c + d*x))^m*(a + a/cos(c + d*x))^(1/2), x)
\[ \int \sqrt {a+a \sec (c+d x)} (e \tan (c+d x))^m \, dx=e^{m} \sqrt {a}\, \left (\int \tan \left (d x +c \right )^{m} \sqrt {\sec \left (d x +c \right )+1}d x \right ) \] Input:
int((a+a*sec(d*x+c))^(1/2)*(e*tan(d*x+c))^m,x)
Output:
e**m*sqrt(a)*int(tan(c + d*x)**m*sqrt(sec(c + d*x) + 1),x)