Optimal. Leaf size=218 \[ -\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}+\frac{11 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 \sqrt{a} d}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}+\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]
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Rubi [A] time = 0.68934, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {3558, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}+\frac{11 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 \sqrt{a} d}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}+\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]
Antiderivative was successfully verified.
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Rule 3558
Rule 3597
Rule 3601
Rule 3544
Rule 205
Rule 3599
Rule 63
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{\tan ^{\frac{7}{2}}(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{5 a}{2}+3 i a \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}-\frac{\int \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-\frac{9 i a^2}{2}-\frac{7}{2} a^2 \tan (c+d x)\right ) \, dx}{2 a^3}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}-\frac{\int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{7 a^3}{4}-\frac{11}{4} i a^3 \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{2 a^4}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}-\frac{11 \int \frac{(a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^2}+\frac{\int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac{(i a) \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{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{4 d}\\ &=\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}-\frac{11 \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 )}{4 d}\\ &=\frac{11 \sqrt [4]{-1} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 \sqrt{a} d}+\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{\tan ^{\frac{5}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 \sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}{4 a d}-\frac{3 i \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{2 a d}\\ \end{align*}
Mathematica [A] time = 3.15688, size = 224, normalized size = 1.03 \[ \frac{\sqrt{\tan (c+d x)} \sec ^2(c+d x) (i \sin (2 (c+d x))+5 \cos (2 (c+d x))+9)-\frac{2 e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \left (4 \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-11 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{\left (-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^2}}{8 d \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.078, size = 667, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{\frac{7}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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