Optimal. Leaf size=80 \[ \frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.12, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3592, 3528, 3533, 205} \[ \frac {2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (B+i A) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 205
Rule 3528
Rule 3533
Rule 3592
Rubi steps
\begin {align*} \int \sqrt {\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\int \sqrt {\tan (c+d x)} (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx\\ &=\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\int \frac {-a (i A+B)+a (A-i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {\left (2 a^2 (i A+B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a (i A+B)-a (A-i B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {2 \sqrt [4]{-1} a (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 a (i A+B) \sqrt {\tan (c+d x)}}{d}+\frac {2 i a B \tan ^{\frac {3}{2}}(c+d x)}{3 d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 2.09, size = 112, normalized size = 1.40 \[ \frac {2 a \sqrt {\tan (c+d x)} \left (\sqrt {i \tan (c+d x)} (3 i A+i B \tan (c+d x)+3 B)+(-3 B-3 i A) \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{3 d \sqrt {i \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.60, size = 368, normalized size = 4.60 \[ -\frac {3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 3 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - {\left ({\left (24 i \, A + 32 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (24 i \, A + 16 \, B\right )} a\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )} \sqrt {\tan \left (d x + c\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.10, size = 475, normalized size = 5.94 \[ \frac {2 i a B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3 d}+\frac {2 i a A \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}+\frac {2 a B \left (\sqrt {\tan }\left (d x +c \right )\right )}{d}-\frac {i a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a A \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {i a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {a B \sqrt {2}\, \ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}-\frac {i a B \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}-\frac {i a B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d}+\frac {a A \sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )}{4 d}+\frac {a A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )}{2 d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 1.71, size = 170, normalized size = 2.12 \[ -\frac {-8 i \, B a \tan \left (d x + c\right )^{\frac {3}{2}} - 8 \, {\left (3 i \, A + 3 \, B\right )} a \sqrt {\tan \left (d x + c\right )} + 3 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 7.71, size = 99, normalized size = 1.24 \[ \frac {A\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {2\,B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {B\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}-\frac {2\,{\left (-1\right )}^{1/4}\,A\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx + \int \left (- i A \sqrt {\tan {\left (c + d x \right )}}\right )\, dx + \int \left (- i B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________