Integrand size = 26, antiderivative size = 141 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{8 \sqrt {2} d}-\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))^2}+\frac {i a^5 \sqrt {a+i a \tan (c+d x)}}{8 d \left (a^2-i a^2 \tan (c+d x)\right )} \] Output:
1/16*I*a^(7/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/ 2)/d-1/2*I*a^5*(a+I*a*tan(d*x+c))^(1/2)/d/(a-I*a*tan(d*x+c))^2+1/8*I*a^5*( a+I*a*tan(d*x+c))^(1/2)/d/(a^2-I*a^2*tan(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.10 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.38 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i a^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3,\frac {5}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^{3/2}}{12 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
((-1/12*I)*a^2*Hypergeometric2F1[3/2, 3, 5/2, (1 + I*Tan[c + d*x])/2]*(a + I*a*Tan[c + d*x])^(3/2))/d
Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3968, 51, 52, 73, 219}
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.
\(\displaystyle \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{7/2}}{\sec (c+d x)^4}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i a^5 \int \frac {\sqrt {i \tan (c+d x) a+a}}{(a-i a \tan (c+d x))^3}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {i a^5 \left (\frac {\sqrt {a+i a \tan (c+d x)}}{2 (a-i a \tan (c+d x))^2}-\frac {1}{4} \int \frac {1}{(a-i a \tan (c+d x))^2 \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))\right )}{d}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {i a^5 \left (\frac {1}{4} \left (-\frac {\int \frac {1}{(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{4 a}-\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )+\frac {\sqrt {a+i a \tan (c+d x)}}{2 (a-i a \tan (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {i a^5 \left (\frac {1}{4} \left (-\frac {\int \frac {1}{a^2 \tan ^2(c+d x)+2 a}d\sqrt {i \tan (c+d x) a+a}}{2 a}-\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )+\frac {\sqrt {a+i a \tan (c+d x)}}{2 (a-i a \tan (c+d x))^2}\right )}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {i a^5 \left (\frac {1}{4} \left (-\frac {i \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2}}\right )}{2 \sqrt {2} a^{3/2}}-\frac {\sqrt {a+i a \tan (c+d x)}}{2 a (a-i a \tan (c+d x))}\right )+\frac {\sqrt {a+i a \tan (c+d x)}}{2 (a-i a \tan (c+d x))^2}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^4*(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
((-I)*a^5*(Sqrt[a + I*a*Tan[c + d*x]]/(2*(a - I*a*Tan[c + d*x])^2) + (((-1 /2*I)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[2]])/(Sqrt[2]*a^(3/2)) - Sqrt[a + I*a*Tan[c + d*x]]/(2*a*(a - I*a*Tan[c + d*x])))/4))/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 484 vs. \(2 (114 ) = 228\).
Time = 66.06 (sec) , antiderivative size = 485, normalized size of antiderivative = 3.44
method | result | size |
default | \(-\frac {\left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-2 \cos \left (d x +c \right )-2\right ) \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \left (-2 \cos \left (d x +c \right )^{3}-2 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+1\right ) \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (2 \cos \left (d x +c \right )+2\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (-2 \cos \left (d x +c \right )^{3}-2 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )+1\right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-1-4 \cos \left (d x +c \right )\right )+\cos \left (d x +c \right ) \left (-4 \cos \left (d x +c \right )^{2}-3 \cos \left (d x +c \right )+1\right )\right ) \cos \left (d x +c \right )^{3} \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{3}}{8 d \left (i \left (2 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right )+2 \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )-1\right )}\) | \(485\) |
Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
-1/8/d*(sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)-2)*arctanh(2^(1/2)*(cot(d*x+c )-csc(d*x+c))/(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2)) *(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*(-2*cos(d*x+c)^3-2*cos(d*x+c)^2+cos( d*x+c)+1)*arctanh(2^(1/2)*(cot(d*x+c)-csc(d*x+c))/(cot(d*x+c)^2-2*cot(d*x+ c)*csc(d*x+c)+csc(d*x+c)^2-1)^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I* sin(d*x+c)*cos(d*x+c)*(2*cos(d*x+c)+2)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+ c)+1))^(1/2)*2^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+(-2*cos(d*x+c)^3- 2*cos(d*x+c)^2+cos(d*x+c)+1)*arctan(1/2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/ 2)*2^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+I*sin(d*x+c)*cos(d*x+c)*(-1 -4*cos(d*x+c))+cos(d*x+c)*(-4*cos(d*x+c)^2-3*cos(d*x+c)+1))*cos(d*x+c)^3*( -tan(d*x+c)+I)^3*(a*(1+I*tan(d*x+c)))^(1/2)*a^3/(I*(2*cos(d*x+c)+1)*sin(d* x+c)+2*cos(d*x+c)^2+cos(d*x+c)-1)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (106) = 212\).
Time = 0.08 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.85 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {\sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) - \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} d \log \left (\frac {4 \, {\left (a^{4} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{7}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{3}}\right ) + \sqrt {2} {\left (-2 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 3 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a^{3} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{16 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
Output:
1/16*(sqrt(1/2)*sqrt(-a^7/d^2)*d*log(4*(a^4*e^(I*d*x + I*c) - sqrt(2)*sqrt (1/2)*sqrt(-a^7/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) - sqrt(1/2)*sqrt(-a^7/d^2)*d*log(4*(a^ 4*e^(I*d*x + I*c) - sqrt(2)*sqrt(1/2)*sqrt(-a^7/d^2)*(-I*d*e^(2*I*d*x + 2* I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^3) + sqr t(2)*(-2*I*a^3*e^(5*I*d*x + 5*I*c) - 3*I*a^3*e^(3*I*d*x + 3*I*c) - I*a^3*e ^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/d
Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(a+I*a*tan(d*x+c))**(7/2),x)
Output:
Timed out
Time = 0.14 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.98 \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {i \, {\left (\sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5} + 2 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{6}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 4 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 4 \, a^{2}}\right )}}{32 \, a d} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
Output:
-1/32*I*(sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a ))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 4*((I*a*tan(d*x + c) + a)^(3/2)*a^5 + 2*sqrt(I*a*tan(d*x + c) + a)*a^6)/((I*a*tan(d*x + c) + a) ^2 - 4*(I*a*tan(d*x + c) + a)*a + 4*a^2))/(a*d)
Exception generated. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Timed out. \[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int {\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2} \,d x \] Input:
int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(7/2),x)
Output:
int(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(7/2), x)
\[ \int \cos ^4(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\sqrt {a}\, a^{3} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )^{3}d x \right ) i -3 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )^{2}d x \right )+3 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4} \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cos \left (d x +c \right )^{4}d x \right ) \] Input:
int(cos(d*x+c)^4*(a+I*a*tan(d*x+c))^(7/2),x)
Output:
sqrt(a)*a**3*( - int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4*tan(c + d*x) **3,x)*i - 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4*tan(c + d*x)**2, x) + 3*int(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4*tan(c + d*x),x)*i + in t(sqrt(tan(c + d*x)*i + 1)*cos(c + d*x)**4,x))