Integrand size = 23, antiderivative size = 154 \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\frac {a e^{3/2} \arctan \left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a e^{3/2} \text {arctanh}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a e^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{3 d \sqrt {e \sin (c+d x)}}-\frac {2 a e \sqrt {e \sin (c+d x)}}{d}-\frac {2 a e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d} \] Output:
a*e^(3/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))/d+a*e^(3/2)*arctanh((e*sin( d*x+c))^(1/2)/e^(1/2))/d+2/3*a*e^2*InverseJacobiAM(1/2*c-1/4*Pi+1/2*d*x,2^ (1/2))*sin(d*x+c)^(1/2)/d/(e*sin(d*x+c))^(1/2)-2*a*e*(e*sin(d*x+c))^(1/2)/ d-2/3*a*e*cos(d*x+c)*(e*sin(d*x+c))^(1/2)/d
Time = 0.55 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.10 \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\frac {a (e \sin (c+d x))^{3/2} \left (12 \arctan \left (\sqrt {\sin (c+d x)}\right )+6 \text {arctanh}\left (\sqrt {\sin (c+d x)}\right )-8 \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )-3 \log \left (1-\sqrt {\sin (c+d x)}\right )+3 \log \left (1+\sqrt {\sin (c+d x)}\right )-24 \sqrt {\sin (c+d x)}-8 \cos (c+d x) \sec (2 (c+d x)) \sqrt {\sin (c+d x)}+16 \cos (c+d x) \sec (2 (c+d x)) \sin ^{\frac {5}{2}}(c+d x)\right )}{12 d \sin ^{\frac {3}{2}}(c+d x)} \] Input:
Integrate[(a + a*Sec[c + d*x])*(e*Sin[c + d*x])^(3/2),x]
Output:
(a*(e*Sin[c + d*x])^(3/2)*(12*ArcTan[Sqrt[Sin[c + d*x]]] + 6*ArcTanh[Sqrt[ Sin[c + d*x]]] - 8*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] - 3*Log[1 - Sqrt[Si n[c + d*x]]] + 3*Log[1 + Sqrt[Sin[c + d*x]]] - 24*Sqrt[Sin[c + d*x]] - 8*C os[c + d*x]*Sec[2*(c + d*x)]*Sqrt[Sin[c + d*x]] + 16*Cos[c + d*x]*Sec[2*(c + d*x)]*Sin[c + d*x]^(5/2)))/(12*d*Sin[c + d*x]^(3/2))
Time = 0.73 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.97, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.826, Rules used = {3042, 4360, 25, 25, 3042, 3317, 3042, 3044, 27, 262, 266, 756, 216, 219, 3115, 3042, 3121, 3042, 3120}
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.
\(\displaystyle \int (a \sec (c+d x)+a) (e \sin (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right ) \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{3/2}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\left (\sec (c+d x) (a (-\cos (c+d x))-a) (e \sin (c+d x))^{3/2}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\left ((\cos (c+d x) a+a) \sec (c+d x) (e \sin (c+d x))^{3/2}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a) (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (-e \cos \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int (e \sin (c+d x))^{3/2}dx+a \int \sec (c+d x) (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int (e \sin (c+d x))^{3/2}dx+a \int \frac {(e \sin (c+d x))^{3/2}}{\cos (c+d x)}dx\) |
\(\Big \downarrow \) 3044 |
\(\displaystyle \frac {a \int \frac {e^2 (e \sin (c+d x))^{3/2}}{e^2-e^2 \sin ^2(c+d x)}d(e \sin (c+d x))}{d e}+a \int (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a e \int \frac {(e \sin (c+d x))^{3/2}}{e^2-e^2 \sin ^2(c+d x)}d(e \sin (c+d x))}{d}+a \int (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {a e \left (e^2 \int \frac {1}{\sqrt {e \sin (c+d x)} \left (e^2-e^2 \sin ^2(c+d x)\right )}d(e \sin (c+d x))-2 \sqrt {e \sin (c+d x)}\right )}{d}+a \int (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {a e \left (2 e^2 \int \frac {1}{e^2-e^4 \sin ^4(c+d x)}d\sqrt {e \sin (c+d x)}-2 \sqrt {e \sin (c+d x)}\right )}{d}+a \int (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {a e \left (2 e^2 \left (\frac {\int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e}+\frac {\int \frac {1}{e^2 \sin ^2(c+d x)+e}d\sqrt {e \sin (c+d x)}}{2 e}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}+a \int (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {a e \left (2 e^2 \left (\frac {\int \frac {1}{e-e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e}+\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}+a \int (e \sin (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a \int (e \sin (c+d x))^{3/2}dx+\frac {a e \left (2 e^2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {1}{3} e^2 \int \frac {1}{\sqrt {e \sin (c+d x)}}dx-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle a \left (\frac {e^2 \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {e^2 \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)}}dx}{3 \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )+\frac {a e \left (2 e^2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {a e \left (2 e^2 \left (\frac {\arctan \left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}+\frac {\text {arctanh}\left (\sqrt {e} \sin (c+d x)\right )}{2 e^{3/2}}\right )-2 \sqrt {e \sin (c+d x)}\right )}{d}+a \left (\frac {2 e^2 \sqrt {\sin (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d \sqrt {e \sin (c+d x)}}-\frac {2 e \cos (c+d x) \sqrt {e \sin (c+d x)}}{3 d}\right )\) |
Input:
Int[(a + a*Sec[c + d*x])*(e*Sin[c + d*x])^(3/2),x]
Output:
(a*e*(2*e^2*(ArcTan[Sqrt[e]*Sin[c + d*x]]/(2*e^(3/2)) + ArcTanh[Sqrt[e]*Si n[c + d*x]]/(2*e^(3/2))) - 2*Sqrt[e*Sin[c + d*x]]))/d + a*((2*e^2*Elliptic F[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) - (2*e*Cos[c + d*x]*Sqrt[e*Sin[c + d*x]])/(3*d))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ Symbol] :> Simp[1/(a*f) Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a *Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && !(I ntegerQ[(m - 1)/2] && LtQ[0, m, n])
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {a \,e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+a \,e^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-2 a e \sqrt {e \sin \left (d x +c \right )}-\frac {a \,e^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(154\) |
parts | \(-\frac {a \,e^{2} \left (\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {a \left (e^{\frac {3}{2}} \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )+e^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-2 e \sqrt {e \sin \left (d x +c \right )}\right )}{d}\) | \(156\) |
Input:
int((a+a*sec(d*x+c))*(e*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
(a*e^(3/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))+a*e^(3/2)*arctanh((e*sin(d *x+c))^(1/2)/e^(1/2))-2*a*e*(e*sin(d*x+c))^(1/2)-1/3*a*e^2*((-sin(d*x+c)+1 )^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*EllipticF((-sin(d*x+c)+1)^ (1/2),1/2*2^(1/2))-2*sin(d*x+c)^3+2*sin(d*x+c))/cos(d*x+c)/(e*sin(d*x+c))^ (1/2))/d
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 591, normalized size of antiderivative = 3.84 \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))^(3/2),x, algorithm="fricas")
Output:
[-1/24*(6*a*sqrt(-e)*e*arctan(1/4*(cos(d*x + c)^2 - 6*sin(d*x + c) - 2)*sq rt(e*sin(d*x + c))*sqrt(-e)/(e*cos(d*x + c)^2 - e*sin(d*x + c) - e)) - 3*a *sqrt(-e)*e*log((e*cos(d*x + c)^4 - 72*e*cos(d*x + c)^2 + 8*(7*cos(d*x + c )^2 - (cos(d*x + c)^2 - 8)*sin(d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(-e) + 28*(e*cos(d*x + c)^2 - 2*e)*sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*co s(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2)*sin(d*x + c) + 8)) - 16*a*sqrt(-1/2* I*e)*e*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) - 16*a*sqr t(1/2*I*e)*e*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 16 *(a*e*cos(d*x + c) + 3*a*e)*sqrt(e*sin(d*x + c)))/d, 1/24*(6*a*e^(3/2)*arc tan(1/4*(cos(d*x + c)^2 + 6*sin(d*x + c) - 2)*sqrt(e*sin(d*x + c))*sqrt(e) /(e*cos(d*x + c)^2 + e*sin(d*x + c) - e)) + 3*a*e^(3/2)*log((e*cos(d*x + c )^4 - 72*e*cos(d*x + c)^2 - 8*(7*cos(d*x + c)^2 + (cos(d*x + c)^2 - 8)*sin (d*x + c) - 8)*sqrt(e*sin(d*x + c))*sqrt(e) - 28*(e*cos(d*x + c)^2 - 2*e)* sin(d*x + c) + 72*e)/(cos(d*x + c)^4 - 8*cos(d*x + c)^2 + 4*(cos(d*x + c)^ 2 - 2)*sin(d*x + c) + 8)) + 16*a*sqrt(-1/2*I*e)*e*weierstrassPInverse(4, 0 , cos(d*x + c) + I*sin(d*x + c)) + 16*a*sqrt(1/2*I*e)*e*weierstrassPInvers e(4, 0, cos(d*x + c) - I*sin(d*x + c)) - 16*(a*e*cos(d*x + c) + 3*a*e)*sqr t(e*sin(d*x + c)))/d]
Timed out. \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))**(3/2),x)
Output:
Timed out
\[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((a*sec(d*x + c) + a)*(e*sin(d*x + c))^(3/2), x)
\[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )} \left (e \sin \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))^(3/2),x, algorithm="giac")
Output:
integrate((a*sec(d*x + c) + a)*(e*sin(d*x + c))^(3/2), x)
Timed out. \[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^{3/2}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \] Input:
int((e*sin(c + d*x))^(3/2)*(a + a/cos(c + d*x)),x)
Output:
int((e*sin(c + d*x))^(3/2)*(a + a/cos(c + d*x)), x)
\[ \int (a+a \sec (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\sqrt {e}\, a e \left (\int \sqrt {\sin \left (d x +c \right )}\, \sec \left (d x +c \right ) \sin \left (d x +c \right )d x +\int \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )d x \right ) \] Input:
int((a+a*sec(d*x+c))*(e*sin(d*x+c))^(3/2),x)
Output:
sqrt(e)*a*e*(int(sqrt(sin(c + d*x))*sec(c + d*x)*sin(c + d*x),x) + int(sqr t(sin(c + d*x))*sin(c + d*x),x))