Integrand size = 25, antiderivative size = 106 \[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\frac {2 a e \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-m}{2},\frac {1}{2} (-2-m),\frac {1}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} (1+\cos (c+d x))^{-m/2} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}}{d} \] Output:
2*a*e*AppellF1(-1/2,-1-1/2*m,1/2-1/2*m,1/2,-cos(d*x+c),cos(d*x+c))*(1-cos( d*x+c))^(1/2-1/2*m)*(a+a*sec(d*x+c))^(1/2)*(e*sin(d*x+c))^(-1+m)/d/((1+cos (d*x+c))^(1/2*m))
Leaf count is larger than twice the leaf count of optimal. \(1243\) vs. \(2(106)=212\).
Time = 9.30 (sec) , antiderivative size = 1243, normalized size of antiderivative = 11.73 \[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx =\text {Too large to display} \] Input:
Integrate[(a + a*Sec[c + d*x])^(3/2)*(e*Sin[c + d*x])^m,x]
Output:
(4*(3 + m)*(AppellF1[(1 + m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2 , -Tan[(c + d*x)/2]^2] + AppellF1[(1 + m)/2, 1/2, m, (3 + m)/2, Tan[(c + d *x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[(c + d*x)/2]^3*(a*(1 + Sec[c + d*x]))^(3/2)*Sin[(c + d*x)/2]*(e*Sin[c + d*x])^m)/(d*(1 + m)*(6*AppellF 1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*m*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2 ]^2, -Tan[(c + d*x)/2]^2] - 2*m*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2 , Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*m*AppellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + A ppellF1[(3 + m)/2, 3/2, m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2 ]^2] - 4*m*AppellF1[(3 + m)/2, 3/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 6*AppellF1[(3 + m)/2, 5/2, m, (5 + m)/2, Tan[(c + d *x)/2]^2, -Tan[(c + d*x)/2]^2] + 6*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] + 2*m*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] + 2*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, - Tan[(c + d*x)/2]^2]*Cos[c + d*x] + 2*m*AppellF1[(3 + m)/2, -1/2, 2 + m,...
Time = 0.71 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.58, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4364, 3042, 3365, 152, 152, 150}
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.
\(\displaystyle \int (a \sec (c+d x)+a)^{3/2} (e \sin (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2} \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^mdx\) |
\(\Big \downarrow \) 4364 |
\(\displaystyle \frac {\sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} \int \frac {(-\cos (c+d x) a-a)^{3/2} (e \sin (c+d x))^m}{(-\cos (c+d x))^{3/2}}dx}{\sqrt {a (-\cos (c+d x))-a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} \int \frac {\left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^m \left (-\sin \left (c+d x+\frac {\pi }{2}\right ) a-a\right )^{3/2}}{\left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{\sqrt {a (-\cos (c+d x))-a}}\) |
\(\Big \downarrow \) 3365 |
\(\displaystyle -\frac {e \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}} (e \sin (c+d x))^{m-1} \int \frac {(-\cos (c+d x) a-a)^{\frac {m+2}{2}} (a \cos (c+d x)-a)^{\frac {m-1}{2}}}{(-\cos (c+d x))^{3/2}}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle \frac {a e \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}} (e \sin (c+d x))^{m-1} \int \frac {(\cos (c+d x)+1)^{\frac {m+2}{2}} (a \cos (c+d x)-a)^{\frac {m-1}{2}}}{(-\cos (c+d x))^{3/2}}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 152 |
\(\displaystyle \frac {a e \sqrt {-\cos (c+d x)} \sqrt {a \sec (c+d x)+a} (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} (e \sin (c+d x))^{m-1} \int \frac {(1-\cos (c+d x))^{\frac {m-1}{2}} (\cos (c+d x)+1)^{\frac {m+2}{2}}}{(-\cos (c+d x))^{3/2}}d\cos (c+d x)}{d}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 a e \sqrt {a \sec (c+d x)+a} (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{-m/2} (a (-\cos (c+d x))-a)^{\frac {1-m}{2}+\frac {m}{2}-\frac {1}{2}} (a \cos (c+d x)-a)^{\frac {1-m}{2}+\frac {m-1}{2}} \operatorname {AppellF1}\left (-\frac {1}{2},\frac {1-m}{2},\frac {1}{2} (-m-2),\frac {1}{2},\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{d}\) |
Input:
Int[(a + a*Sec[c + d*x])^(3/2)*(e*Sin[c + d*x])^m,x]
Output:
(2*a*e*AppellF1[-1/2, (1 - m)/2, (-2 - m)/2, 1/2, Cos[c + d*x], -Cos[c + d *x]]*(1 - Cos[c + d*x])^((1 - m)/2)*(-a - a*Cos[c + d*x])^(-1/2 + (1 - m)/ 2 + m/2)*(-a + a*Cos[c + d*x])^((1 - m)/2 + (-1 + m)/2)*Sqrt[a + a*Sec[c + d*x]]*(e*Sin[c + d*x])^(-1 + m))/(d*(1 + Cos[c + d*x])^(m/2))
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*((g*Cos [e + f*x])^(p - 1)/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e + f*x]) ^((p - 1)/2))) Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[Sin[e + f*x]^FracPart[m]*((a + b*Csc[e + f*x] )^FracPart[m]/(b + a*Sin[e + f*x])^FracPart[m]) Int[(g*Cos[e + f*x])^p*(( b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x], x] /; FreeQ[{a, b, e, f, g, m, p }, x] && (EqQ[a^2 - b^2, 0] || IntegersQ[2*m, p])
\[\int \left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}} \left (e \sin \left (d x +c \right )\right )^{m}d x\]
Input:
int((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x)
Output:
int((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x)
\[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x, algorithm="fricas")
Output:
integral((a*sec(d*x + c) + a)^(3/2)*(e*sin(d*x + c))^m, x)
Timed out. \[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))**(3/2)*(e*sin(d*x+c))**m,x)
Output:
Timed out
\[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)^(3/2)*(e*sin(d*x + c))^m, x)
\[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)^(3/2)*(e*sin(d*x + c))^m, x)
Timed out. \[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^m\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int((e*sin(c + d*x))^m*(a + a/cos(c + d*x))^(3/2),x)
Output:
int((e*sin(c + d*x))^m*(a + a/cos(c + d*x))^(3/2), x)
\[ \int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx=e^{m} \sqrt {a}\, a \left (\int \sin \left (d x +c \right )^{m} \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )d x +\int \sin \left (d x +c \right )^{m} \sqrt {\sec \left (d x +c \right )+1}d x \right ) \] Input:
int((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x)
Output:
e**m*sqrt(a)*a*(int(sin(c + d*x)**m*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x) + int(sin(c + d*x)**m*sqrt(sec(c + d*x) + 1),x))