Optimal. Leaf size=258 \[ \frac {363 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}+\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.84, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3556, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {363 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 63
Rule 203
Rule 205
Rule 217
Rule 3544
Rule 3556
Rule 3597
Rule 3599
Rule 3601
Rubi steps
\begin {align*} \int \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {1}{4} a \int \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {15 a}{2}+\frac {17}{2} i a \tan (c+d x)\right ) \, dx\\ &=\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {1}{12} \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {85 i a^2}{4}+\frac {107}{4} a^2 \tan (c+d x)\right ) \, dx\\ &=\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-\frac {321 a^3}{8}-\frac {447}{8} i a^3 \tan (c+d x)\right ) \, dx}{24 a}\\ &=-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {447 i a^4}{16}-\frac {1089}{16} a^4 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{24 a^2}\\ &=-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {1}{128} (363 i a) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+\left (4 i a^2\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (363 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{128 d}+\frac {\left (8 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\\ &=\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (363 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{64 d}\\ &=\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (363 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}\\ &=\frac {363 (-1)^{3/4} a^{5/2} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{64 d}+\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {149 i a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{64 d}+\frac {107 a^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{96 d}+\frac {17 i a^2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{24 d}-\frac {a^2 \tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 4.08, size = 232, normalized size = 0.90 \[ \frac {a^2 \sqrt {a+i a \tan (c+d x)} \left (\frac {6 e^{-i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \left (512 \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-363 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}-i \sqrt {\tan (c+d x)} \sec ^3(c+d x) (70 i \sin (c+d x)+262 i \sin (3 (c+d x))+1205 \cos (c+d x)+583 \cos (3 (c+d x)))\right )}{768 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.83, size = 769, normalized size = 2.98 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.21, size = 492, normalized size = 1.91 \[ -\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (96 \left (\tan ^{3}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-272 i \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+384 i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +384 \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +1089 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a +894 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}-428 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+1536 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{384 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \tan \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________